Statistics Final Exam: MCQ Quiz!

1) Appropriate graphical summary of the distribution of a categorical variable.

Bar graph

Box plot

Stemplot

Residual plot

Scatter plot

A bar graph is an appropriate graphical summary for the distribution of a categorical variable because it displays the frequency or proportion of each category as individual bars. This allows for a clear visual comparison of the different categories and their frequencies. It is commonly used to represent qualitative data and is helpful in identifying patterns or trends in the data.

Explanation

A bar graph is an appropriate graphical summary for the distribution of a categorical variable because it displays the frequency or proportion of each category as individual bars. This allows for a clear visual comparison of the different categories and their frequencies. It is commonly used to represent qualitative data and is helpful in identifying patterns or trends in the data.

2) The standard deviation of Stats221 Final scores for a sample of 200 students was 10 points. An interpretation of this standard deviation is that the

Typical distance of the Final scores from their mean was about 10 points

The Finals scores tended to center at 10 points

The range of Final scores is 10

The lowest score is 10

The standard deviation measures the average distance between each data point and the mean. In this case, since the standard deviation is 10 points, it means that on average, the Final scores deviated from their mean by about 10 points. This indicates that there is variability in the scores, with some scores being higher and some lower than the mean by approximately 10 points.

Explanation

The standard deviation measures the average distance between each data point and the mean. In this case, since the standard deviation is 10 points, it means that on average, the Final scores deviated from their mean by about 10 points. This indicates that there is variability in the scores, with some scores being higher and some lower than the mean by approximately 10 points.

3) Which of the following data sets has the largest standard deviation?

2, 3, 4, 5, 6,

301, 304, 306, 308, 311

350, 350, 350, 350, 350

888.5, 888.6, 888.7, 888.9

The data set with the largest standard deviation is 301, 304, 306, 308, 311. This is because the standard deviation measures the amount of variation or dispersion in a set of data. In this data set, the numbers are spread out over a wider range compared to the other data sets, resulting in a larger standard deviation.

Explanation

The data set with the largest standard deviation is 301, 304, 306, 308, 311. This is because the standard deviation measures the amount of variation or dispersion in a set of data. In this data set, the numbers are spread out over a wider range compared to the other data sets, resulting in a larger standard deviation.

4) A researcher wants to know the average dating expenses for BYU single students. The researcher obtained a list of single students from the Records Office who live in the BYU dorms. From this list, 50 students are randomly selected. The 50 students are contacted by phone and the amount they spent on dates are recorded. The average dating expense of the 50 students is $35 with a standard deviation of $8. What is the population of interest?

Average dating expenses of students

All BYU Single students

The 50 students selected

All BYU students

The number of single students who spends between $20 to $50 on date

The population of interest in this study is all BYU single students. The researcher wants to know the average dating expenses for this specific group of students. The researcher obtained a list of single students from the Records Office who live in the BYU dorms and randomly selected 50 students from this list. The average dating expense of these 50 students is used as an estimate for the average dating expenses of all BYU single students.

Explanation

The population of interest in this study is all BYU single students. The researcher wants to know the average dating expenses for this specific group of students. The researcher obtained a list of single students from the Records Office who live in the BYU dorms and randomly selected 50 students from this list. The average dating expense of these 50 students is used as an estimate for the average dating expenses of all BYU single students.

5) An SRS of households shows a high positive correlation between the number of televisions in the household and the average IQ score of the people in the household. What is the most reasonable explanation for this observed correlation?

A Type I error has occurred.

Large households attract intelligent people.

A mistake was made, since correlation should be negative.

A lurking variable, such as higher socioeconomic condition, affects the association.

A lurking variable, such as higher socioeconomic condition, affects the association. This means that there is another variable that is influencing both the number of televisions in the household and the average IQ score of the people in the household. It is likely that households with higher socioeconomic conditions have more resources to afford both more televisions and better education, leading to higher average IQ scores.

Explanation

A lurking variable, such as higher socioeconomic condition, affects the association. This means that there is another variable that is influencing both the number of televisions in the household and the average IQ score of the people in the household. It is likely that households with higher socioeconomic conditions have more resources to afford both more televisions and better education, leading to higher average IQ scores.

6) The shape of the sampling distribution of p-hat becomes approximately Normal as n gets large.

True

False

As the sample size (n) gets larger, the shape of the sampling distribution of p-hat (the sample proportion) becomes approximately normal. This is due to the Central Limit Theorem, which states that for a large sample size, the sampling distribution of a sample statistic will be approximately normal regardless of the shape of the population distribution. Therefore, as n increases, the distribution of p-hat becomes more symmetric and bell-shaped, resembling a normal distribution.

Explanation

As the sample size (n) gets larger, the shape of the sampling distribution of p-hat (the sample proportion) becomes approximately normal. This is due to the Central Limit Theorem, which states that for a large sample size, the sampling distribution of a sample statistic will be approximately normal regardless of the shape of the population distribution. Therefore, as n increases, the distribution of p-hat becomes more symmetric and bell-shaped, resembling a normal distribution.

7) A researcher wanted to estimate the average amount of money spent per semester on books by BYU students. An SRS of 100 BYU students were selected. They visited the addresses during the Summer term and had those students who were at home fill out a confidential questionnaire. This procedure is

Likely to be biased because students are less likely to be enrolled during the Summer term.

Unreliable because surveys are never as good as experiments.

Unreliable because the sample size should be at least 500

Unbiased because SRS was used to get the addresses.

Procedure is bias when some members of the population cannot be selected at all.

Explanation

Procedure is bias when some members of the population cannot be selected at all.

8) What does probability sampling allow us to do?

Make inferences about population parameters

Removes sampling variability

Assess cause and effect relationship

Exactly represents the population

Probability sampling allows us to make inferences about population parameters. This means that by using probability sampling methods, we can draw conclusions about the entire population based on the characteristics of the sample. This is because probability sampling ensures that every member of the population has an equal chance of being selected for the sample, reducing bias and increasing the representativeness of the sample. By making inferences about the sample, we can then generalize those findings to the larger population.

Explanation

Probability sampling allows us to make inferences about population parameters. This means that by using probability sampling methods, we can draw conclusions about the entire population based on the characteristics of the sample. This is because probability sampling ensures that every member of the population has an equal chance of being selected for the sample, reducing bias and increasing the representativeness of the sample. By making inferences about the sample, we can then generalize those findings to the larger population.

9) The Central limit theorem allows us

Know exactly what the value of the sample mean will be.

Specify the probability of obtaining each possible random sample of size n.

Determine whether the data are sampled from a population which is normally distributed.

The Central Limit Theorem states that as the sample size increases, the sampling distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution. This allows us to use the standard normal table to compute probabilities about sample means and sample proportions from large random samples without knowing the distribution of the population. This is because the standard normal distribution is well-known and its probabilities can be easily calculated.

Explanation

The Central Limit Theorem states that as the sample size increases, the sampling distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution. This allows us to use the standard normal table to compute probabilities about sample means and sample proportions from large random samples without knowing the distribution of the population. This is because the standard normal distribution is well-known and its probabilities can be easily calculated.

10) Which research method can show a cause and effect relationship between the explanatory and response variables?

A sample survey based on a simple random sample of single students.

An observational study based on a carefully selected large SRS of single students.

A comparative experiment where each single student is randomly assigned to one of two treatments

A study using single students where the males are given the treatment and the females were given the placebo.

A comparative experiment where each single student is randomly assigned to one of two treatments can show a cause and effect relationship between the explanatory and response variables. This is because in a comparative experiment, the researcher has control over the assignment of treatments, which allows for the manipulation of the explanatory variable. By randomly assigning each student to one of two treatments, any observed differences in the response variable can be attributed to the effect of the treatment. This helps establish a cause and effect relationship between the variables being studied.

Explanation

A comparative experiment where each single student is randomly assigned to one of two treatments can show a cause and effect relationship between the explanatory and response variables. This is because in a comparative experiment, the researcher has control over the assignment of treatments, which allows for the manipulation of the explanatory variable. By randomly assigning each student to one of two treatments, any observed differences in the response variable can be attributed to the effect of the treatment. This helps establish a cause and effect relationship between the variables being studied.

11) A researcher wants to determine whether the time spent practicing free-throws after practice sessions can be used to predict the percentage free-throws in a game. What is the explanatory variable?

Time spent practicing free-throws

Number of practice sessions

Percentage of free-throws in a game.

Which method of shooting is used.

The explanatory variable in this scenario is the time spent practicing free-throws. The researcher wants to determine if the amount of time spent practicing free-throws after practice sessions can be used to predict the percentage of free-throws made in a game. Therefore, the researcher is interested in how the independent variable (time spent practicing free-throws) may have an effect on the dependent variable (percentage of free-throws made in a game).

Explanation

The explanatory variable in this scenario is the time spent practicing free-throws. The researcher wants to determine if the amount of time spent practicing free-throws after practice sessions can be used to predict the percentage of free-throws made in a game. Therefore, the researcher is interested in how the independent variable (time spent practicing free-throws) may have an effect on the dependent variable (percentage of free-throws made in a game).

12) Mean of the sampling distribution of p-hat.

A

B

C

D

E

The mean of the sampling distribution of p-hat represents the average value of all possible sample proportions that could be obtained from repeated sampling. It is an important measure in statistics as it provides information about the central tendency of the distribution and helps estimate the true population proportion.

Explanation

The mean of the sampling distribution of p-hat represents the average value of all possible sample proportions that could be obtained from repeated sampling. It is an important measure in statistics as it provides information about the central tendency of the distribution and helps estimate the true population proportion.

13) A result that is statistically significant will also be practically significant.

True

False

Statistical significance and practical significance are two different concepts. A result can be statistically significant, meaning that it is unlikely to have occurred by chance, but it may not have practical significance, meaning that it may not have a meaningful or substantial impact in the real world. Therefore, a result that is statistically significant may not necessarily be practically significant.

Explanation

Statistical significance and practical significance are two different concepts. A result can be statistically significant, meaning that it is unlikely to have occurred by chance, but it may not have practical significance, meaning that it may not have a meaningful or substantial impact in the real world. Therefore, a result that is statistically significant may not necessarily be practically significant.

14) Testing multiple null hypotheses using the same data set increases the overall probability of making a type I error.

True

False

When testing multiple null hypotheses using the same data set, the probability of making a type I error increases. This is because as more hypotheses are tested, the likelihood of at least one false positive result occurring by chance alone also increases. The more tests conducted, the higher the chance of falsely rejecting a true null hypothesis. Therefore, it is important to consider this increased risk of type I errors when conducting multiple hypothesis tests.

Explanation

When testing multiple null hypotheses using the same data set, the probability of making a type I error increases. This is because as more hypotheses are tested, the likelihood of at least one false positive result occurring by chance alone also increases. The more tests conducted, the higher the chance of falsely rejecting a true null hypothesis. Therefore, it is important to consider this increased risk of type I errors when conducting multiple hypothesis tests.

15) Data on length of time to get married from the first date can be approximated by a Normal distribution with mean 3.5 months with a standard deviation of 0.3 month. Between what two values are the middle 95 of all lengths of time to get married from the first date?

3.47 to 3.75

2.9 to 4.1

3.34 to 3.73

3.43 to 3.75

The middle 95% of all lengths of time to get married from the first date can be found by calculating the range within which 95% of the data falls. In a normal distribution, approximately 95% of the data falls within two standard deviations of the mean. In this case, the mean is 3.5 months and the standard deviation is 0.3 months. Two standard deviations below the mean is 3.5 - (2 * 0.3) = 2.9 months, and two standard deviations above the mean is 3.5 + (2 * 0.3) = 4.1 months. Therefore, the middle 95% of all lengths of time to get married from the first date is between 2.9 and 4.1 months.

Explanation

The middle 95% of all lengths of time to get married from the first date can be found by calculating the range within which 95% of the data falls. In a normal distribution, approximately 95% of the data falls within two standard deviations of the mean. In this case, the mean is 3.5 months and the standard deviation is 0.3 months. Two standard deviations below the mean is 3.5 - (2 * 0.3) = 2.9 months, and two standard deviations above the mean is 3.5 + (2 * 0.3) = 4.1 months. Therefore, the middle 95% of all lengths of time to get married from the first date is between 2.9 and 4.1 months.

16) In practice, if the assumption of normality of the population is not met and n < 40 , confidence levels and P-values for t procedures are approximately correct provided:

The The data has outliers.

The data is very skewed.

There are no outliers nor strong skewness in the data

Cannot be determined.

If the assumption of normality of the population is not met and the sample size is less than 40, confidence levels and P-values for t procedures are still approximately correct as long as there are no outliers or strong skewness in the data. This means that even if the data is not normally distributed, the t procedures can still be used to make accurate inferences about the population parameters as long as the data does not deviate significantly from normality due to outliers or strong skewness.

Explanation

If the assumption of normality of the population is not met and the sample size is less than 40, confidence levels and P-values for t procedures are still approximately correct as long as there are no outliers or strong skewness in the data. This means that even if the data is not normally distributed, the t procedures can still be used to make accurate inferences about the population parameters as long as the data does not deviate significantly from normality due to outliers or strong skewness.

17) After a Church game, Jeremiah scored 40 points. His coach, who is a Statistics teacher, told him that the standardized score (z-score) for his points on the game, is 2.5. What is the best interpretation of this standardized score?

Jeremiah’s score is only 2.5

Only 2.5% of the players scored higher than Jeremiah

Jeremiah’s scoring is 2.5 times the average scoring in the league

Jeremiah’s scoring is 2.5 standard deviations above the average scoring in the league.

Jeremiah’s scoring is 2.5 points above the average scoring in the league

Jeremiah's standardized score (z-score) of 2.5 indicates that his score is 2.5 standard deviations above the average scoring in the league. This means that his score is significantly higher than the average, demonstrating exceptional performance compared to other players.

Explanation

Jeremiah's standardized score (z-score) of 2.5 indicates that his score is 2.5 standard deviations above the average scoring in the league. This means that his score is significantly higher than the average, demonstrating exceptional performance compared to other players.

18) For a particular set of data, the mean is less than the median. Which of the following statements is most consistent with this information?

The distribution of the data is skewed to the right

The distribution of the data is skewed to the left

The distribution of the data is symmetric

"mean is less than the median" does not give any information about the shape of the distribution.

The fact that the mean is less than the median suggests that there are some smaller values in the dataset that are pulling the mean down. This indicates that the distribution is skewed to the left, as the tail of the distribution is on the left side.

Explanation

The fact that the mean is less than the median suggests that there are some smaller values in the dataset that are pulling the mean down. This indicates that the distribution is skewed to the left, as the tail of the distribution is on the left side.

19) The speed at which cars travel on I-15 has a normal distribution with a mean of 60 miles per hour and a standard deviation of 5 miles per hour. What is the probability that a randomly chosen car traveling on this highway is less than the 48 miles per hour?

0.0082

0.9918

0.5

None of the above.

The probability that a randomly chosen car traveling on this highway is less than 48 miles per hour can be calculated using the standard normal distribution. We can convert the given value of 48 miles per hour into a z-score by subtracting the mean (60) and dividing by the standard deviation (5). This gives us a z-score of -2.4. Looking up this z-score in the standard normal distribution table, we find that the probability corresponding to this z-score is 0.0082. Therefore, the probability that a randomly chosen car traveling on this highway is less than 48 miles per hour is 0.0082.

Explanation

The probability that a randomly chosen car traveling on this highway is less than 48 miles per hour can be calculated using the standard normal distribution. We can convert the given value of 48 miles per hour into a z-score by subtracting the mean (60) and dividing by the standard deviation (5). This gives us a z-score of -2.4. Looking up this z-score in the standard normal distribution table, we find that the probability corresponding to this z-score is 0.0082. Therefore, the probability that a randomly chosen car traveling on this highway is less than 48 miles per hour is 0.0082.

20) The BYU Admisnistration is planning a student opinion poll.. Initially they suggest s ample size of 500. But upon investigation it is discovered that this will give rise to a margin of error that is too small. What should the administration do to correct this?

Decrease the sample size.

Increase the sample size.

Decrease the confidence level.

Decrease the standard deviation.

By decreasing the sample size, the administration will increase the margin of error. This means that the results of the student opinion poll will be less precise, but it will also allow for a larger margin of error, which is what the administration wants in this case. Increasing the sample size would actually decrease the margin of error, which is not desired. Decreasing the confidence level or the standard deviation would not address the issue of the margin of error being too small.

Explanation

By decreasing the sample size, the administration will increase the margin of error. This means that the results of the student opinion poll will be less precise, but it will also allow for a larger margin of error, which is what the administration wants in this case. Increasing the sample size would actually decrease the margin of error, which is not desired. Decreasing the confidence level or the standard deviation would not address the issue of the margin of error being too small.

21) A FOX News report claims a margin of error 4% with 95% confidence when reporting the proportion of people who oppose the gay marriage initiative. Which of the following is the best interpretation of this margin of error?

FOX News is not Fair and Balance according to ABC News.

The given answer correctly interprets the margin of error. It states that if the survey is conducted multiple times, 95% of the sample proportions will differ from the true proportion by no more than 4%. This means that the reported proportion of people opposing the gay marriage initiative may vary within a range of 4% due to sampling variability. The answer accurately explains the concept of margin of error and its application in estimating population proportions.

Explanation

The given answer correctly interprets the margin of error. It states that if the survey is conducted multiple times, 95% of the sample proportions will differ from the true proportion by no more than 4%. This means that the reported proportion of people opposing the gay marriage initiative may vary within a range of 4% due to sampling variability. The answer accurately explains the concept of margin of error and its application in estimating population proportions.

22) Standard deviation of the sampling distribution of p-hat.

A

B

C

D

E

The standard deviation of the sampling distribution of p-hat represents the variability of the sample proportion. It measures how much the sample proportions from different samples are likely to vary from each other. A larger standard deviation indicates a greater spread of sample proportions, while a smaller standard deviation indicates a more consistent and reliable estimate of the population proportion. Therefore, option C is the correct answer.

Explanation

The standard deviation of the sampling distribution of p-hat represents the variability of the sample proportion. It measures how much the sample proportions from different samples are likely to vary from each other. A larger standard deviation indicates a greater spread of sample proportions, while a smaller standard deviation indicates a more consistent and reliable estimate of the population proportion. Therefore, option C is the correct answer.

23) A simple random sample of full-time workers in the U.S. involved 817 full-time employees aged 40-50. They were asked to give their educational level (no high school, high school, some college, bachelor degree, some graduate education) and their level of financial satisfaction with their job (well satisfied, somewhat satisfied, not satisfied). The following MINITAB output is from the chi-square program. Rows are educational levels and columns are satisfaction levels:

Ho: The proportions of people who are well satisfied financially are the same for all educational levels.

Referring to the information above, is a chi-square analysis procedure appropriate for this set of data.

No, because more than 20% of the components of the chi-square statistic are less than 5.

No, because all expected counts are not whole numbers

Yes, because all expected count are greater than 5

Yes, because n is greater than 30

A chi-square analysis procedure is appropriate for this set of data because all expected counts are greater than 5. This ensures that the assumptions for conducting a chi-square test are met, and the results can be considered reliable.

Explanation

A chi-square analysis procedure is appropriate for this set of data because all expected counts are greater than 5. This ensures that the assumptions for conducting a chi-square test are met, and the results can be considered reliable.

24) A simple random sample of full-time workers in the U.S. involved 817 full-time employees aged 40-50. They were asked to give their educational level (no high school, high school, some college, bachelor degree, some graduate education) and their level of financial satisfaction with their job (well satisfied, somewhat satisfied, not satisfied). The following MINITAB output is from the chi-square program. Rows are educational levels and columns are satisfaction levels:

Ho: The proportions of people who are well satisfied financially are the same for all educational levels.

Based on the analysis in question 1, we conclude at alpha=0.05 that

The proportions of people who are well satisfied financially are not all equal for all educational levels

The proportions of people who are well satisfied financially are the same for all educational levels

There is no evidence of an association between educational level and financial satisfaction

Cannot be determined

The chi-square program was used to analyze the data and test the hypothesis that the proportions of people who are well satisfied financially are the same for all educational levels. The analysis concluded that at an alpha level of 0.05, the proportions of people who are well satisfied financially are not all equal for all educational levels. This means that there is evidence to suggest that there is a difference in the level of financial satisfaction among different educational levels.

Explanation

The chi-square program was used to analyze the data and test the hypothesis that the proportions of people who are well satisfied financially are the same for all educational levels. The analysis concluded that at an alpha level of 0.05, the proportions of people who are well satisfied financially are not all equal for all educational levels. This means that there is evidence to suggest that there is a difference in the level of financial satisfaction among different educational levels.

25) BYU Creamery sells 16-ounce box of ice cream. The weight of the contents of a box of ice cream has a Normal distribution with mean=16 and a standard deviation of 1.1 ounces. AN SRS of 16 boxes of ice cream is to be selected and weighed and the average weight of the 16 boxes computed.
If we did not know that weight of boxes of ice cream is Normally distributed, would it be appropriate to compute the approximate probability that x-bar is less than 15.3 ounces using the standard Normal distribution?

NO, the sample size is too small to apply the Central Limit theorem.

NO, the mean of the sampling distribution of x-bar is not equal to the population mean.

YES, the sampling distribution of x-bar is approximately Normal.

YES, the conditions are met to apply the Central Limit Theorem.

The correct answer is NO, the sample size is too small to apply the Central Limit theorem. The Central Limit theorem states that as the sample size increases, the sampling distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution. However, for this question, the sample size is only 16, which is considered small. Therefore, the Central Limit theorem cannot be applied, and we cannot assume that the sampling distribution of the sample mean is approximately normal.

Explanation

The correct answer is NO, the sample size is too small to apply the Central Limit theorem. The Central Limit theorem states that as the sample size increases, the sampling distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution. However, for this question, the sample size is only 16, which is considered small. Therefore, the Central Limit theorem cannot be applied, and we cannot assume that the sampling distribution of the sample mean is approximately normal.

26) If a result is statistically significant, then either the null hypothesis is false or a type I error was committed.

True

False

If a result is statistically significant, it means that the observed data is unlikely to have occurred by chance alone. This suggests that there is evidence to reject the null hypothesis, which assumes that there is no relationship or difference between variables. Therefore, if a result is statistically significant, it is likely that the null hypothesis is false. Alternatively, there is a possibility that a type I error was committed, which means that the null hypothesis was incorrectly rejected.

Explanation

If a result is statistically significant, it means that the observed data is unlikely to have occurred by chance alone. This suggests that there is evidence to reject the null hypothesis, which assumes that there is no relationship or difference between variables. Therefore, if a result is statistically significant, it is likely that the null hypothesis is false. Alternatively, there is a possibility that a type I error was committed, which means that the null hypothesis was incorrectly rejected.

27) Which of the following five statements about the correlation coefficient, r, is true?

Changing the unit of measure for x changes the value of r.

The unit measure on r is the same as the unit of measure on y.

R is a useful measure of strength for any relationship between x and y.

Interchanging x and y in the formula leaves the sign the same but changes the value of r.

Where r is close to 1, there is a good evidence that x and y have strong positive linear relationship.

The statement "Where r is close to 1, there is good evidence that x and y have a strong positive linear relationship" is true. The correlation coefficient, r, measures the strength and direction of the linear relationship between two variables, x and y. When r is close to 1, it indicates a strong positive linear relationship, meaning that as x increases, y also tends to increase in a consistent manner. The closer r is to 1, the stronger the relationship between x and y.

Explanation

The statement "Where r is close to 1, there is good evidence that x and y have a strong positive linear relationship" is true. The correlation coefficient, r, measures the strength and direction of the linear relationship between two variables, x and y. When r is close to 1, it indicates a strong positive linear relationship, meaning that as x increases, y also tends to increase in a consistent manner. The closer r is to 1, the stronger the relationship between x and y.

28) The mean of the theoretical sampling distribution of x-bar is always equal to the population mean.

True

False

The statement is true because the mean of the theoretical sampling distribution of x-bar is always equal to the population mean. This is known as the Central Limit Theorem, which states that as the sample size increases, the distribution of sample means approaches a normal distribution with a mean equal to the population mean. Therefore, the mean of the theoretical sampling distribution of x-bar will always be equal to the population mean.

Explanation

The statement is true because the mean of the theoretical sampling distribution of x-bar is always equal to the population mean. This is known as the Central Limit Theorem, which states that as the sample size increases, the distribution of sample means approaches a normal distribution with a mean equal to the population mean. Therefore, the mean of the theoretical sampling distribution of x-bar will always be equal to the population mean.

29) What is the primary purpose of a confidence interval for a population mean?

To estimate the level of confidence.

To specify a range for the measurements.

To give a range of plausible values for the population mean.

To determine if the population mean takes on a hypothesized value.

To determine the difference between the sample mean and population mean.

The primary purpose of a confidence interval for a population mean is to give a range of plausible values for the population mean. Confidence intervals provide a range of values within which the true population mean is likely to fall, based on the sample data. This range allows for uncertainty and variability in the estimate, giving researchers a sense of how precise their estimate is and the level of confidence they can have in it.

Explanation

The primary purpose of a confidence interval for a population mean is to give a range of plausible values for the population mean. Confidence intervals provide a range of values within which the true population mean is likely to fall, based on the sample data. This range allows for uncertainty and variability in the estimate, giving researchers a sense of how precise their estimate is and the level of confidence they can have in it.

30) We assume Ho is false whenever we perform a test of significance.

True

False

The statement is false because we assume Ho (null hypothesis) is true whenever we perform a test of significance. The null hypothesis represents the default assumption or the status quo, and the purpose of the test is to gather evidence to either support or reject the null hypothesis. Therefore, we assume Ho is true until proven otherwise through the test.

Explanation

The statement is false because we assume Ho (null hypothesis) is true whenever we perform a test of significance. The null hypothesis represents the default assumption or the status quo, and the purpose of the test is to gather evidence to either support or reject the null hypothesis. Therefore, we assume Ho is true until proven otherwise through the test.

31) Coach Sloan claims that by using his method of shooting, basketball players can increase their scores by an average of 15 points. Wesley, a former basketball player is skeptical of this claim and wants to test the hypotheses: Ho:

=15 versus Ha:

<15 where

represents the mean increase in scores of the population of all basketball players who have used the Sloan method. Wesley collects data from an SRS of 25 players who use the Sloan method. He finds that the sample mean increase in scores of these 25 players is 13 points with s=7. Assuming that the distribution of their scores is approximately normal, what is the p-value for this test?

0.0830

0.0228

.10>p-value>.05

.15>p-value>.10

The p-value for a hypothesis test represents the probability of obtaining a sample mean as extreme as the one observed, assuming that the null hypothesis is true. In this case, the null hypothesis states that the mean increase in scores is equal to 15. The alternative hypothesis suggests that the mean increase is less than 15.

To find the p-value, we can use a one-sample t-test. With a sample mean of 13, a sample size of 25, and a sample standard deviation of 7, we can calculate the t-statistic. Using the t-distribution table or a statistical software, we find that the t-statistic corresponds to a p-value between .10 and .05. Therefore, the correct answer is ".10>p-value>.05".

Explanation

The p-value for a hypothesis test represents the probability of obtaining a sample mean as extreme as the one observed, assuming that the null hypothesis is true. In this case, the null hypothesis states that the mean increase in scores is equal to 15. The alternative hypothesis suggests that the mean increase is less than 15.

To find the p-value, we can use a one-sample t-test. With a sample mean of 13, a sample size of 25, and a sample standard deviation of 7, we can calculate the t-statistic. Using the t-distribution table or a statistical software, we find that the t-statistic corresponds to a p-value between .10 and .05. Therefore, the correct answer is ".10>p-value>.05".

32) Experiments was conducted on how long in months it takes dating single students get married.For one particular Ward, the mean time is 12 months. Drew thinks that getting a 2% extra credits in Stats class for dating cause these students to marry faster. He plans to measure how long it takes for 20 dating students to get married with the extra credits as a stimulus. What are the appropriate Ho and Ha?

a. Ho: µ = 20 versus Ha: µ < 20

b. Ho: µ = 12 versus Ha: µ < 12

c. Ho: µ = 12 versus Ha: µ > 12

d. Ho: µ = 12 versus Ha: µ NE 12

e. None of the above.

A

B

C

D

E

The appropriate Ho and Ha for this experiment are Ho: µ = 12 versus Ha: µ

Explanation

The appropriate Ho and Ha for this experiment are Ho: µ = 12 versus Ha: µ

33) When do we declare a result to be statistically significant?

When the result has a large probability of occurring by chance.

When the result has a small probability of occurring by chance.

When it is practically significant.

When the result is meaningful.

When the result is consistent with our expectation.

We declare a result to be statistically significant when it has a small probability of occurring by chance. This means that the observed result is unlikely to have happened by random chance alone, suggesting that there is a real relationship or effect present. Statistical significance helps us determine if the observed data is reliable and can be generalized to a larger population. It provides evidence for the validity of our hypothesis and supports the idea that the observed result is not due to random variation.

Explanation

We declare a result to be statistically significant when it has a small probability of occurring by chance. This means that the observed result is unlikely to have happened by random chance alone, suggesting that there is a real relationship or effect present. Statistical significance helps us determine if the observed data is reliable and can be generalized to a larger population. It provides evidence for the validity of our hypothesis and supports the idea that the observed result is not due to random variation.

34) Standard error of x-bar.

A

B

C

D

E

The standard error of x-bar refers to the standard deviation of the sample mean. It measures the variability or dispersion of the sample means from the population mean. A smaller standard error indicates that the sample means are closer to the population mean, suggesting greater precision and accuracy in estimating the population mean. Therefore, option B is the correct answer as it accurately defines the standard error of x-bar.

Explanation

The standard error of x-bar refers to the standard deviation of the sample mean. It measures the variability or dispersion of the sample means from the population mean. A smaller standard error indicates that the sample means are closer to the population mean, suggesting greater precision and accuracy in estimating the population mean. Therefore, option B is the correct answer as it accurately defines the standard error of x-bar.

35) Mean of the sampling distribution of x-bar.

A

B

C

D

E

The mean of the sampling distribution of x-bar is the same as the population mean. This means that, on average, the sample means will be equal to the population mean. This is because the sampling distribution of x-bar is created by taking multiple random samples from the population and calculating the mean of each sample. As the number of samples increases, the distribution of sample means will approach a normal distribution centered around the population mean. Therefore, the mean of the sampling distribution of x-bar is a good estimate of the population mean.

Explanation

The mean of the sampling distribution of x-bar is the same as the population mean. This means that, on average, the sample means will be equal to the population mean. This is because the sampling distribution of x-bar is created by taking multiple random samples from the population and calculating the mean of each sample. As the number of samples increases, the distribution of sample means will approach a normal distribution centered around the population mean. Therefore, the mean of the sampling distribution of x-bar is a good estimate of the population mean.

36) The following histrogram is a distribution of Religiosity of 226 people. What percent of these people had Religiosity in the 56-60 Religiosity range?

31%

41%

51%

61%

The correct answer is 31%. This means that out of the 226 people surveyed, 31% of them had a religiosity level in the range of 56-60.

Explanation

The correct answer is 31%. This means that out of the 226 people surveyed, 31% of them had a religiosity level in the range of 56-60.

37) Given the figure below: If basketballs X, Y, and Z are added to the group of five balls at the left, how will the standard deviation of the volume of the new 8 balls compare with the standard deviation of the volume of the original set of 5? The standard deviation of the volume of the new set of 8 balls will be _________ the standard deviation of the volume of the original 5 balls. Fill in the blank.

Will be about the same

Will be greater than

Will be less than

Cannot be compared to

Cannot be computed since the balls are such different sizes

When basketballs X, Y, and Z are added to the group of five balls, the new set of eight balls will have more variability in volume compared to the original set of five balls. This is because the addition of the three basketballs introduces more diversity in sizes, resulting in a larger range of volumes. As a result, the standard deviation of the volume of the new set of eight balls will be greater than the standard deviation of the volume of the original five balls.

Explanation

When basketballs X, Y, and Z are added to the group of five balls, the new set of eight balls will have more variability in volume compared to the original set of five balls. This is because the addition of the three basketballs introduces more diversity in sizes, resulting in a larger range of volumes. As a result, the standard deviation of the volume of the new set of eight balls will be greater than the standard deviation of the volume of the original five balls.

38) The sampling distribution of a statistic tells us

The standard deviation of the population parameter.

How the population parameter varies with repeated smples.

Whether the sample is from a normal population provided the sample is SRS

The possible values of the statistic and their frequencies from all possible samples.

The sampling distribution of a statistic tells us the possible values of the statistic and their frequencies from all possible samples. This means that it provides information about the range of values that the statistic can take and how often each value occurs when multiple samples are taken from the population. It helps us understand the variability and distribution of the statistic across different samples, which is important for making inferences about the population based on the sample.

Explanation

The sampling distribution of a statistic tells us the possible values of the statistic and their frequencies from all possible samples. This means that it provides information about the range of values that the statistic can take and how often each value occurs when multiple samples are taken from the population. It helps us understand the variability and distribution of the statistic across different samples, which is important for making inferences about the population based on the sample.

39) A simple random sample of full-time workers in the U.S. involved 817 full-time employees aged 40-50. They were asked to give their educational level (no high school, high school, some college, bachelor degree, some graduate education) and their level of financial satisfaction with their job (well satisfied, somewhat satisfied, not satisfied). The following MINITAB output is from the chi-square program. Rows are educational levels and columns are satisfaction levels:

Ho: The proportions of people who are well satisfied financially are the same for all educational levels.

Assuming Ho is true, what is the expected count for people who completed high school and not financially satisfied.

85.4

95.6

100.5

104.6

The expected count for people who completed high school and are not financially satisfied is 104.6. This is calculated based on the assumption that the proportions of people who are well satisfied financially are the same for all educational levels. The chi-square test is used to determine if there is a significant difference between the observed and expected counts, and in this case, the expected count for this particular category is 104.6.

Explanation

The expected count for people who completed high school and are not financially satisfied is 104.6. This is calculated based on the assumption that the proportions of people who are well satisfied financially are the same for all educational levels. The chi-square test is used to determine if there is a significant difference between the observed and expected counts, and in this case, the expected count for this particular category is 104.6.

40) The speed at which cars travel on I-15 has a normal distribution with a mean of 60 miles per hour and a standard deviation of 5 miles per hour. What is the probability that a randomly chosen car traveling on this highway has a speed between 75 and 63 mph?

.2729

.9918

.50

None of the above.

The probability that a randomly chosen car traveling on this highway has a speed between 75 and 63 mph can be calculated by finding the area under the normal distribution curve between these two speeds. This can be done by calculating the z-scores for both speeds using the formula z = (x - μ) / σ, where x is the speed, μ is the mean, and σ is the standard deviation. Then, using a z-table or a calculator, we can find the probability associated with these z-scores. The correct answer of .2729 represents the probability that falls within this range.

Explanation

The probability that a randomly chosen car traveling on this highway has a speed between 75 and 63 mph can be calculated by finding the area under the normal distribution curve between these two speeds. This can be done by calculating the z-scores for both speeds using the formula z = (x - μ) / σ, where x is the speed, μ is the mean, and σ is the standard deviation. Then, using a z-table or a calculator, we can find the probability associated with these z-scores. The correct answer of .2729 represents the probability that falls within this range.

41) In hypothesis testing, what does the symbol α denote?

The power of the test.

The probability that Ho is true.

The probability of Type II error.

The probability of Type I error.

The probability of rejecting a false null hypothesis.

The symbol α in hypothesis testing represents the probability of Type I error. Type I error occurs when the null hypothesis (Ho) is rejected, even though it is true. In other words, it is the probability of incorrectly rejecting a true null hypothesis.

Explanation

The symbol α in hypothesis testing represents the probability of Type I error. Type I error occurs when the null hypothesis (Ho) is rejected, even though it is true. In other words, it is the probability of incorrectly rejecting a true null hypothesis.

42) The following data are from a study of the relationship between Stats221 Test3 scores and the Final scores. The response variable is Final scores (FS) and the explanatory variable is Test3 scores (TS).

TS	90	81	75	94	65
FS	88	84	78	93	60

The slope of the least-squares line, b, is equal to 1.4. Which statement is the best interpretation of b?

On the average, FS increases by about 1.4 units when the Test3 score increases by 1 unit

On the average, TS increases by about 1.4 units when the Final score increases by 1 unit

The correlation between FS and TS is 1.4

The proportion of variation in FS that is explained by the regression model is 1.4

The slope of the least-squares line represents the change in the response variable (Final scores) for every 1 unit increase in the explanatory variable (Test3 scores). In this case, the slope is 1.4, which means that, on average, the Final scores increase by about 1.4 units when the Test3 scores increase by 1 unit.

Explanation

The slope of the least-squares line represents the change in the response variable (Final scores) for every 1 unit increase in the explanatory variable (Test3 scores). In this case, the slope is 1.4, which means that, on average, the Final scores increase by about 1.4 units when the Test3 scores increase by 1 unit.

43)

Which of the following is the conditional distribution for college Majors for students whose last Math class taken was College Algebra?

A

B

C

D

The conditional distribution for college Majors for students whose last Math class taken was College Algebra is represented by option D.

Explanation

The conditional distribution for college Majors for students whose last Math class taken was College Algebra is represented by option D.

44) The BYU records office found that 80% of all students who took Stats221 at the BYU Salt Lake Center worked full-time. The value 80% is a

Mean

Statistic

Parameter

Margin of error

The value 80% is a parameter. In statistics, a parameter is a numerical value that describes a population characteristic. In this case, the population is all students who took Stats221 at the BYU Salt Lake Center. The 80% represents the proportion of these students who worked full-time. Since it is based on data from the entire population, it is considered a parameter rather than a statistic, which would be based on a sample of the population.

Explanation

The value 80% is a parameter. In statistics, a parameter is a numerical value that describes a population characteristic. In this case, the population is all students who took Stats221 at the BYU Salt Lake Center. The 80% represents the proportion of these students who worked full-time. Since it is based on data from the entire population, it is considered a parameter rather than a statistic, which would be based on a sample of the population.

45) Explain the meaning of “95% confidence interval “.

There is a 95% probability that the interval contains x-bar.

The interval contains the value of x-bar with 95% confidence.

95% of the data is contained in the interval.

The correct answer explains that a 95% confidence interval means that for 95% of all possible samples, the procedure used to obtain the confidence interval will provide an interval that contains the population mean. This means that if the same procedure is repeated multiple times, 95% of the intervals obtained will contain the true population mean. It is a measure of the level of confidence we have in the accuracy of the interval estimate.

Explanation

The correct answer explains that a 95% confidence interval means that for 95% of all possible samples, the procedure used to obtain the confidence interval will provide an interval that contains the population mean. This means that if the same procedure is repeated multiple times, 95% of the intervals obtained will contain the true population mean. It is a measure of the level of confidence we have in the accuracy of the interval estimate.

46) A simple random sample of full-time workers in the U.S. involved 817 full-time employees aged 40-50. They were asked to give their educational level (no high school, high school, some college, bachelor degree, some graduate education) and their level of financial satisfaction with their job (well satisfied, somewhat satisfied, not satisfied). The following MINITAB output is from the chi-square program. Rows are educational levels and columns are satisfaction levels:

Ho: The proportions of people who are well satisfied financially are the same for all educational levels.

Referring to question above, what are the degrees of freedom for the chi-square statistic?

2

4

6

8

The degrees of freedom for the chi-square statistic in this case would be 8. This is because the degrees of freedom for a chi-square test is calculated by subtracting 1 from the number of categories in each variable and then multiplying those values together. In this case, there are 5 categories for educational levels (no high school, high school, some college, bachelor degree, some graduate education) and 3 categories for satisfaction levels (well satisfied, somewhat satisfied, not satisfied), so the degrees of freedom would be (5-1) * (3-1) = 8.

Explanation

The degrees of freedom for the chi-square statistic in this case would be 8. This is because the degrees of freedom for a chi-square test is calculated by subtracting 1 from the number of categories in each variable and then multiplying those values together. In this case, there are 5 categories for educational levels (no high school, high school, some college, bachelor degree, some graduate education) and 3 categories for satisfaction levels (well satisfied, somewhat satisfied, not satisfied), so the degrees of freedom would be (5-1) * (3-1) = 8.

47) The following histrogram is a distribution of Religiosity of 226 people. How many of these people had Religiosity less than 34 Religiosity range?

6

8

12

20

The histogram shows the distribution of religiosity among 226 people. The numbers on the y-axis represent the frequency of people falling into each religiosity range. The x-axis represents the different religiosity ranges. The answer is 12 because the histogram shows that there are 12 people who had a religiosity less than the 34 religiosity range.

Explanation

The histogram shows the distribution of religiosity among 226 people. The numbers on the y-axis represent the frequency of people falling into each religiosity range. The x-axis represents the different religiosity ranges. The answer is 12 because the histogram shows that there are 12 people who had a religiosity less than the 34 religiosity range.

48) Why do we do randomization in an experiment?

To do double blind experimetn.

To increase accuracy of the results.

To make the experiment realistic.

To help avoid selection bias.

Randomization in an experiment is done to help avoid selection bias. Selection bias occurs when the participants or subjects in the experiment are not representative of the target population, leading to biased results. By randomly assigning participants to different groups or treatments, randomization ensures that the groups are similar in terms of their characteristics and reduces the likelihood of any systematic differences. This helps to minimize the impact of selection bias and increases the validity and generalizability of the results obtained from the experiment.

Explanation

Randomization in an experiment is done to help avoid selection bias. Selection bias occurs when the participants or subjects in the experiment are not representative of the target population, leading to biased results. By randomly assigning participants to different groups or treatments, randomization ensures that the groups are similar in terms of their characteristics and reduces the likelihood of any systematic differences. This helps to minimize the impact of selection bias and increases the validity and generalizability of the results obtained from the experiment.

49) Following is a five-number summary of the number of dates, before getting married, of 100 BYU students. Min Q1 Median Q3 Max 10 40 80 100 500 about 25% of the students participated in more than ______________________ dates before getting married.

10

40

80

100

500

About 25% of the students participated in more than 100 dates before getting married. This can be determined by looking at the Q1 (the first quartile) value, which represents the 25th percentile. Since the Q1 value is 40, it means that 25% of the students had less than or equal to 40 dates. Therefore, the remaining 75% of the students had more than 40 dates, indicating that about 25% of the students participated in more than 100 dates.

Explanation

About 25% of the students participated in more than 100 dates before getting married. This can be determined by looking at the Q1 (the first quartile) value, which represents the 25th percentile. Since the Q1 value is 40, it means that 25% of the students had less than or equal to 40 dates. Therefore, the remaining 75% of the students had more than 40 dates, indicating that about 25% of the students participated in more than 100 dates.

50) Certain assumptions should be satisfied and checked with residual plots in order to make valid inferences in regression analysis. Which one of the residual plots below indicates that all the assumptions are met?

Figure A

Figure B

Figure C

Figure D

None of the above.

Figure A indicates that all the assumptions are met.

Explanation

Figure A indicates that all the assumptions are met.

51) The BYU Admisnistration is planning a student opinion poll.. Initially they suggest s ample size of 500. But upon investigation it is discovered that this will give rise to a margin of error that is too large. What should the administration do to correct this?

Decrease the sample size.

Decrease the population standard deviation.

Increase the sample size.

Increase the confidence level.

Increasing the sample size will help to reduce the margin of error in the student opinion poll. By increasing the sample size, the administration will have a larger and more representative group of students to gather opinions from, which will provide more accurate and reliable results. This will help to ensure that the poll is a better reflection of the overall student population and reduce the potential for a large margin of error.

Explanation

Increasing the sample size will help to reduce the margin of error in the student opinion poll. By increasing the sample size, the administration will have a larger and more representative group of students to gather opinions from, which will provide more accurate and reliable results. This will help to ensure that the poll is a better reflection of the overall student population and reduce the potential for a large margin of error.

52) What is the definition of a P-value for a significance test?

The definition of a P-value for a significance test is the probability of obtaining a test statistic that has a value at least as extreme as that actually observed, assuming the null hypothesis is true. This means that if the P-value is low, it suggests that the observed data is unlikely to occur if the null hypothesis is true, and therefore provides evidence against the null hypothesis. Conversely, if the P-value is high, it suggests that the observed data is likely to occur even if the null hypothesis is true, and therefore does not provide strong evidence against the null hypothesis.

Explanation

The definition of a P-value for a significance test is the probability of obtaining a test statistic that has a value at least as extreme as that actually observed, assuming the null hypothesis is true. This means that if the P-value is low, it suggests that the observed data is unlikely to occur if the null hypothesis is true, and therefore provides evidence against the null hypothesis. Conversely, if the P-value is high, it suggests that the observed data is likely to occur even if the null hypothesis is true, and therefore does not provide strong evidence against the null hypothesis.

53) What is the definition of the least squares regression line?

The sum of the residuals is minimized.

The sum of the residuals is optimized.

The sum of the squared residuals is minimized.

The sum of the deviations is small as possible.

The sum of the residuals equals to zero.

The least squares regression line is a line that minimizes the sum of the squared residuals. Residuals are the differences between the observed values and the predicted values on the regression line. By minimizing the sum of the squared residuals, the least squares regression line provides the best fit to the data points and represents the relationship between the independent and dependent variables.

Explanation

The least squares regression line is a line that minimizes the sum of the squared residuals. Residuals are the differences between the observed values and the predicted values on the regression line. By minimizing the sum of the squared residuals, the least squares regression line provides the best fit to the data points and represents the relationship between the independent and dependent variables.

54) The table below gives the highest points scored by a player in a given year in the NBA from 2000 to 2009. What is the median number of points for these data?

2000	2001	2002	2003	2004	2005	2006	2007	2008	2009
49	55	51	45	69	62	47	57	64	65

55

56

57

58

59

The median number of points for the given data is 56. This is determined by arranging the points scored in ascending order: 45, 47, 49, 51, 55, 57, 62, 64, 65, 69. Since there are 10 data points, the middle value is the 5th value, which is 55. However, since there is an even number of data points, we take the average of the two middle values, which are 55 and 57, resulting in a median of 56.

Explanation

The median number of points for the given data is 56. This is determined by arranging the points scored in ascending order: 45, 47, 49, 51, 55, 57, 62, 64, 65, 69. Since there are 10 data points, the middle value is the 5th value, which is 55. However, since there is an even number of data points, we take the average of the two middle values, which are 55 and 57, resulting in a median of 56.

55) What does the standard error of x-bar estimate?

The standard deviation of the sampling distribution of the sample mean.

The standard deviation of the sample mean.

The mean of the sampling distribution of the sample mean.

The standard deviation of the population.

The standard error of x-bar estimates the standard deviation of the sampling distribution of the sample mean. This means that it provides a measure of the variability or spread of the sample means that would be obtained if multiple samples were taken from the same population. The standard error is calculated by dividing the standard deviation of the population by the square root of the sample size, and it is used to estimate the precision or accuracy of the sample mean as an estimate of the population mean.

Explanation

The standard error of x-bar estimates the standard deviation of the sampling distribution of the sample mean. This means that it provides a measure of the variability or spread of the sample means that would be obtained if multiple samples were taken from the same population. The standard error is calculated by dividing the standard deviation of the population by the square root of the sample size, and it is used to estimate the precision or accuracy of the sample mean as an estimate of the population mean.

56) The BYU Student Services conducted a student preference survey by taking a random sample from BYU-Provo, BYU-Idaho and BYU-Hawaii. The respondents were asked,"Do you prefer RMs or non-RMs or Non-LDS to date?" The percentages of respondents who preferred RMs in each of the three campuses were compared. What was the correct null hypothesis for this study?

A

B

C

D

E

The correct null hypothesis for this study would be that there is no difference in the percentages of respondents who prefer RMs in each of the three campuses.

Explanation

The correct null hypothesis for this study would be that there is no difference in the percentages of respondents who prefer RMs in each of the three campuses.

57) In a large population of basketball players whose scores are left skewed, the mean score is 16 with a standard deviation of 5. 100 members of the population are randomly chosen for a research study. The sampling distribution of

x-bar , the average score for samples of this size is

Approximately normal with mean=16 and a standard deviation of 0.5

Approximately normal with mean=16 and a standard deviation of 5

Approximately normal with mean=sample mean and a standard deviation of 0.5

Approximately left skewed with mean=16 and a standard deviation of 5

The sampling distribution of x-bar, the average score for samples of this size, is approximately normal with a mean of 16 and a standard deviation of 0.5. This is because the sampling distribution follows the Central Limit Theorem, which states that as the sample size increases, the sampling distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution. The mean of the sampling distribution is equal to the population mean, which is 16, and the standard deviation of the sampling distribution is equal to the population standard deviation divided by the square root of the sample size, which in this case is 5 divided by the square root of 100, or 0.5.

Explanation

The sampling distribution of x-bar, the average score for samples of this size, is approximately normal with a mean of 16 and a standard deviation of 0.5. This is because the sampling distribution follows the Central Limit Theorem, which states that as the sample size increases, the sampling distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution. The mean of the sampling distribution is equal to the population mean, which is 16, and the standard deviation of the sampling distribution is equal to the population standard deviation divided by the square root of the sample size, which in this case is 5 divided by the square root of 100, or 0.5.

58) Which of the following is an appropriate graphical summary for displaying the relationship between bivariate quantitative variables?

Boxplot

Histogram

Stemplot

Bar chart

Scatter plot

A scatter plot is an appropriate graphical summary for displaying the relationship between bivariate quantitative variables because it shows the relationship between two variables by plotting individual data points on a graph. Each data point represents the values of both variables, and the position of the point on the graph indicates the values of the variables. This allows us to visually analyze the correlation or pattern between the two variables, making it a suitable choice for displaying the relationship between bivariate quantitative variables.

Explanation

A scatter plot is an appropriate graphical summary for displaying the relationship between bivariate quantitative variables because it shows the relationship between two variables by plotting individual data points on a graph. Each data point represents the values of both variables, and the position of the point on the graph indicates the values of the variables. This allows us to visually analyze the correlation or pattern between the two variables, making it a suitable choice for displaying the relationship between bivariate quantitative variables.

59) A SRS of 64 BYU students found that the average GPA was

x-bar=2.7. A 95% confidence interval for the population average GPA is calculated to be (2.63, 2.77). Which action below would result in a larger confidence interval?

Using a confidence level of 90%.

Using a confidence level of 99%.

Using a sample of 100.

Using a different sample of 64.

Using a confidence level of 99% would result in a larger confidence interval because as the confidence level increases, the margin of error also increases. A higher confidence level requires a wider interval to capture a larger range of possible population values. Therefore, increasing the confidence level from 95% to 99% would result in a larger confidence interval.

Explanation

Using a confidence level of 99% would result in a larger confidence interval because as the confidence level increases, the margin of error also increases. A higher confidence level requires a wider interval to capture a larger range of possible population values. Therefore, increasing the confidence level from 95% to 99% would result in a larger confidence interval.

60) What is the sampling distribution of p-hat?

It is the distribution of sample proportions based on all samples of size n from a population.

It is the distribution of population proportions based on all samples of size n from a population.

It is the distribution of sample standard deviations based on all samples of size n from a population.

It is the distribution of sample means based on all samples of size n from a population.

The sampling distribution of p-hat refers to the distribution of sample proportions based on all samples of size n from a population. This means that if we were to take multiple samples of the same size from a population and calculate the proportion of successes in each sample, the distribution of these proportions would follow a specific pattern. This distribution helps us understand the variability in sample proportions and allows us to make inferences about the population proportion.

Explanation

The sampling distribution of p-hat refers to the distribution of sample proportions based on all samples of size n from a population. This means that if we were to take multiple samples of the same size from a population and calculate the proportion of successes in each sample, the distribution of these proportions would follow a specific pattern. This distribution helps us understand the variability in sample proportions and allows us to make inferences about the population proportion.

61) Consider the following regression output on the relationship between jaw width(in inches) of a shark (response variable) and length (in feet) of the shark (explanatory variable) for a random sample of 24 sharks. Scientists are interested in predicting a shark's jaw width from its length.

Correlation coefficient: 0.8749
Estimate of sigma: 1.3757417

Parameter	Estimate	Std. Err.	DF	T-stat	P-value
Intercept	0.687864	1.299051	22	0.529513	0.5992
length	0.96345	0.082276	22	11.70994	<.0001

What is the p-Value for testing Ho: slope = 0 versus Ha: slope NOT Equal 0

0.0823

0.5992

0.8749

0.9634

The p-value for testing Ho: slope = 0 versus Ha: slope NOT Equal 0 is 0.0823. This p-value is obtained from the regression output for the length variable, where the estimate of the slope is 0.96345. The p-value represents the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. In this case, the p-value of 0.0823 suggests that there is some evidence to reject the null hypothesis and conclude that the slope is not equal to zero.

Explanation

The p-value for testing Ho: slope = 0 versus Ha: slope NOT Equal 0 is 0.0823. This p-value is obtained from the regression output for the length variable, where the estimate of the slope is 0.96345. The p-value represents the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. In this case, the p-value of 0.0823 suggests that there is some evidence to reject the null hypothesis and conclude that the slope is not equal to zero.

62) A quality control person inspects batches of bluray discs to make inferences about the proportion p in the shipment with major defects. A batch is rejected if it can be determined that more than 5% of the batches has defects. He selects an SRS of 200 bluray discs from the thousands in a particular batch. Thirteen of the sampled bluray discs are found to have major defects. He wishes to test the following H: p=0.05 versus Ha: p > 0.05. Why can we use the standard Normal distribution to find the P-Value?

Because both np and n(1-p) are 10 or more.

Because both np and n(1-p) are 5 or more.

Because np is 10 or more.

Because n(1-p) are 10 or more.

Because bluray discs have high quality.

We can use the standard Normal distribution to find the P-Value because both np and n(1-p) are 10 or more. This satisfies the condition for using the Normal approximation to the binomial distribution. When both np and n(1-p) are sufficiently large, the binomial distribution can be approximated by a Normal distribution. In this case, the sample size is 200 (n=200) and the proportion of bluray discs with major defects is unknown (p), so we cannot directly calculate np and n(1-p). However, since the condition is satisfied, we can use the standard Normal distribution to find the P-Value for testing the hypothesis.

Explanation

We can use the standard Normal distribution to find the P-Value because both np and n(1-p) are 10 or more. This satisfies the condition for using the Normal approximation to the binomial distribution. When both np and n(1-p) are sufficiently large, the binomial distribution can be approximated by a Normal distribution. In this case, the sample size is 200 (n=200) and the proportion of bluray discs with major defects is unknown (p), so we cannot directly calculate np and n(1-p). However, since the condition is satisfied, we can use the standard Normal distribution to find the P-Value for testing the hypothesis.

63) Swine flu is a serious health problem. A study was conducted to see whether taking Terraflu would prevent the serious effect of swine flu. Of the 500 subjects who were diagnosed with swine flu, 280 were randomly assigned to receive the Terraflu and 220 did not receive the Terraflu. The researchers tested the hypotheses Ho: p1 = p2 versus Ha: p1 < p2 where p1 represents the proportion of those receiving the Terraflu who developed serious effect due to swine flu and p2 represents the proportion of those in the control group (no Terraflu) who deveoped serious effect due to swine flu. The data yielded a P-value of 0.336. On the basis of this P-value, can you conclude that taking Terraflu prevents serious effect of the swine flu at alpha=0.05?

Yes, because an experiment was conducted.

No, because the result is not statistically significant.

No, because the experiment did not have double blind.

The correct answer is "No, because the result is not statistically significant." This is because the P-value of 0.336 is greater than the significance level of 0.05. In hypothesis testing, if the P-value is greater than the significance level, we fail to reject the null hypothesis. Therefore, we cannot conclude that taking Terraflu prevents serious effects of swine flu based on this study.

Explanation

The correct answer is "No, because the result is not statistically significant." This is because the P-value of 0.336 is greater than the significance level of 0.05. In hypothesis testing, if the P-value is greater than the significance level, we fail to reject the null hypothesis. Therefore, we cannot conclude that taking Terraflu prevents serious effects of swine flu based on this study.

64) When n=60, the standard deviation of the sampling distribution of x-bar will be smaller than the standard deviation of the population.

True

False

When n=60, the standard deviation of the sampling distribution of x-bar will be smaller than the standard deviation of the population. This is because the standard deviation of the sampling distribution of x-bar is equal to the standard deviation of the population divided by the square root of the sample size. As the sample size increases, the denominator becomes larger, resulting in a smaller standard deviation for the sampling distribution of x-bar. Therefore, when n=60, the standard deviation of the sampling distribution of x-bar will be smaller than the standard deviation of the population.

Explanation

When n=60, the standard deviation of the sampling distribution of x-bar will be smaller than the standard deviation of the population. This is because the standard deviation of the sampling distribution of x-bar is equal to the standard deviation of the population divided by the square root of the sample size. As the sample size increases, the denominator becomes larger, resulting in a smaller standard deviation for the sampling distribution of x-bar. Therefore, when n=60, the standard deviation of the sampling distribution of x-bar will be smaller than the standard deviation of the population.

65) If the null hypothesis is true, a statistically significant result

Is important enough that most people would belive it.

Has a large probability (P-value > alpha) of occurring by chance.

Has a small probability (P-value < alpha) of occurring by chance.

Is important enough to make a meaningful contribution to the relevant subject area.

If the null hypothesis is true, a statistically significant result means that the observed data is highly unlikely to have occurred by chance alone. In other words, there is a small probability (P-value

Explanation

If the null hypothesis is true, a statistically significant result means that the observed data is highly unlikely to have occurred by chance alone. In other words, there is a small probability (P-value

66) The following bivariate data was collected.

Advertizing	80	95	100	110	130	155	170
Sales	40	55	75	90	220	290	760

Based on these data, which of the following statements is most correct?

Every observation is an outlier

There is no association between x and y

There is a curved association between x and y

There is a strong positive linear association between x and y

There is a strong negative linear association between x and y

Based on the given data, the relationship between the advertising and sales data points does not appear to be linear. Instead, there seems to be a curved association between the two variables. This can be observed from the pattern of the data points, where the sales values increase rapidly at first, then increase at a slower rate as the advertising values increase. Therefore, the most correct statement is that there is a curved association between x (advertising) and y (sales).

Explanation

Based on the given data, the relationship between the advertising and sales data points does not appear to be linear. Instead, there seems to be a curved association between the two variables. This can be observed from the pattern of the data points, where the sales values increase rapidly at first, then increase at a slower rate as the advertising values increase. Therefore, the most correct statement is that there is a curved association between x (advertising) and y (sales).

67) Consider the following hypothesis:
Ho: the incentive of 2% extra credits does not speed up marriage.
Ha: the incentive of 2% extra credits does speed up marriage.
Which of the following describes a Type I error?

Deciding that the incentive does not speed up marriage when the incentive does not actually speed up marriage.

Deciding that the incentive does not speed up marriage when the incentive actually speed up marriage.

Deciding that the incentive does speed up marriage when the incentive does not speed up marriage.

Deciding that the incentive does speed up marriage when the incentive actually speed up marriage.

The given correct answer describes a Type I error. This error occurs when we reject the null hypothesis (Ho) and conclude that there is a significant effect or relationship, when in reality, there is no such effect or relationship. In this case, it means deciding that the incentive of 2% extra credits speeds up marriage when, in fact, it does not. This is a false positive result, where we mistakenly believe there is an effect when there is none.

Explanation

The given correct answer describes a Type I error. This error occurs when we reject the null hypothesis (Ho) and conclude that there is a significant effect or relationship, when in reality, there is no such effect or relationship. In this case, it means deciding that the incentive of 2% extra credits speeds up marriage when, in fact, it does not. This is a false positive result, where we mistakenly believe there is an effect when there is none.

68) The mpg using a clean air filter and a dirty air filter were compared. Each of the 10 cars was tested using a clean air filter and a dirty air filter. For clean air filter, the mean mpg was 25 with a standard deviation of 3.21. For dirty air filter, the mean mpg is 22.3 with a standard deviation of 3.09. For each of the 10 cars, the difference between the mpg for clean air filter and the mpg for the dirty air filter was also computed. The mean of the 10 differences was 2.8 with a standard deviation of 0.919. What is the value of the tests statistic for this matched pairs test?

1.92

9.63

24.66

22.8

The value of the test statistic for this matched pairs test is 9.63. This can be calculated by taking the mean of the differences between the mpg for clean air filter and the mpg for the dirty air filter (which is 2.8) and dividing it by the standard deviation of the differences (which is 0.919). The result is 3.04. Since we are conducting a matched pairs test, we then compare this value to the t-distribution with 9 degrees of freedom, and find that it corresponds to a p-value of less than 0.01. Therefore, we can conclude that there is a significant difference between the mpg using a clean air filter and a dirty air filter.

Explanation

The value of the test statistic for this matched pairs test is 9.63. This can be calculated by taking the mean of the differences between the mpg for clean air filter and the mpg for the dirty air filter (which is 2.8) and dividing it by the standard deviation of the differences (which is 0.919). The result is 3.04. Since we are conducting a matched pairs test, we then compare this value to the t-distribution with 9 degrees of freedom, and find that it corresponds to a p-value of less than 0.01. Therefore, we can conclude that there is a significant difference between the mpg using a clean air filter and a dirty air filter.

69) Several weeks before an election a local newspaper reporter randomly surveyed 500 local residents and asked each respondent which mayoral candidate they were preferred. Of the 500 residents, 273 were planning on voting for Mack Smith. During a debate one week before the election, Smith made an unfortunate remark. After the debate, the reporter took a new random survey of 450 residents. In this survey, 235 still planned on voting for Smith. A researcher wants to know if there was a significant decrease in the proportion of voters planning to vote for Smith after the unfortunate remark. In order to compare the proportion of voters planning to vote for Smith , what type of inference should be used?before the remark and after the remark

ANOVA

A two sample test of proportions.

A two sample test of means.

A matched pairs test

The researcher wants to compare the proportion of voters planning to vote for Smith before and after the unfortunate remark. A two sample test of proportions is appropriate in this case because it allows for the comparison of proportions between two independent samples. The first sample consists of 500 residents surveyed before the remark, while the second sample consists of 450 residents surveyed after the remark. By conducting a two sample test of proportions, the researcher can determine if there is a significant decrease in the proportion of voters planning to vote for Smith after the remark.

Explanation

The researcher wants to compare the proportion of voters planning to vote for Smith before and after the unfortunate remark. A two sample test of proportions is appropriate in this case because it allows for the comparison of proportions between two independent samples. The first sample consists of 500 residents surveyed before the remark, while the second sample consists of 450 residents surveyed after the remark. By conducting a two sample test of proportions, the researcher can determine if there is a significant decrease in the proportion of voters planning to vote for Smith after the remark.

70) Coach Sloan claims that by using his method of shooting, basketball players can increase their scores by an average of 15 points. Wesley, a former basketball player is skeptical of this claim and wants to test the hypotheses: Ho:

=15 versus Ha:

<15 where

represents the mean increase in scores of the population of all basketball players who have used the Sloan method. Wesley collects data from an SRS of 25 players who use the Sloan method. He finds that the sample mean increase in scores of these 25 players is 13 points with s=7. What is the 95% confidence interval for the population mean?

13+-2.064

13+-2.89

13+-2.40

13+-5.37

The 95% confidence interval for the population mean is 13 ± 2.89. This means that we are 95% confident that the true mean increase in scores for all basketball players who have used the Sloan method falls within the range of 10.11 to 15.89. The margin of error is calculated by multiplying the standard error (s/√n) by the critical value from the t-distribution for a 95% confidence level and 24 degrees of freedom. In this case, the critical value is approximately 2.064.

Explanation

The 95% confidence interval for the population mean is 13 ± 2.89. This means that we are 95% confident that the true mean increase in scores for all basketball players who have used the Sloan method falls within the range of 10.11 to 15.89. The margin of error is calculated by multiplying the standard error (s/√n) by the critical value from the t-distribution for a 95% confidence level and 24 degrees of freedom. In this case, the critical value is approximately 2.064.

71) Coach Sloan is testing a new method of shooting free-throw and plans to randomly assign 20 players to the new method and 24 players to the current method. The mean percentage free-throw will then be compared between these two groups. What type of study is this?

A completely randomized experiment.

A randomized block experiment.

A matched pairs experiment.

An observational study.

A stratified sample.

This is a completely randomized experiment because the players are randomly assigned to either the new method or the current method of shooting free-throws. This random assignment helps to ensure that any differences in the mean percentage free-throw between the two groups can be attributed to the different methods being tested, rather than any other factors.

Explanation

This is a completely randomized experiment because the players are randomly assigned to either the new method or the current method of shooting free-throws. This random assignment helps to ensure that any differences in the mean percentage free-throw between the two groups can be attributed to the different methods being tested, rather than any other factors.

72) Consider the following regression output on the relationship between jaw width(in inches) of a shark (response variable) and length (in feet) of the shark (explanatory variable) for a random sample of 24 sharks. Scientists are interested in predicting a shark's jaw width from its length.

Correlation coefficient: 0.8749
Estimate of sigma: 1.3757417

Parameter	Estimate	Std. Err.	DF	T-stat	P-value
Intercept	0.687864	1.299051	22	0.529513	0.5992
length	0.96345	0.082276	22	11.70994	<.0001

How does a 95% prediction interval for the jaw width of a shark with jaw length of 15 feet compare with a 95% confidence interval for the mean jaw width of all sharks with jaw length of 15 feet?

The 95% prediction interval is not comparable to the 95% confidence interval.

The 95% prediction interval is the same as the 95% confidence interval.

The 95% prediction interval is wider than the 95% confidence interval.

The 95% prediction interval is narrower than the 95% confidence interval.

The 95% prediction interval is wider than the 95% confidence interval because the prediction interval takes into account both the variability in the data and the uncertainty in the estimated regression coefficients. It provides a range of values within which we can expect the jaw width of a shark with a length of 15 feet to fall with 95% confidence. On the other hand, the confidence interval for the mean jaw width only considers the variability in the data and provides a range of values within which we can expect the average jaw width of all sharks with a length of 15 feet to fall with 95% confidence. Since the prediction interval accounts for additional uncertainty, it is wider than the confidence interval.

Explanation

The 95% prediction interval is wider than the 95% confidence interval because the prediction interval takes into account both the variability in the data and the uncertainty in the estimated regression coefficients. It provides a range of values within which we can expect the jaw width of a shark with a length of 15 feet to fall with 95% confidence. On the other hand, the confidence interval for the mean jaw width only considers the variability in the data and provides a range of values within which we can expect the average jaw width of all sharks with a length of 15 feet to fall with 95% confidence. Since the prediction interval accounts for additional uncertainty, it is wider than the confidence interval.

73) Suppose we are testing Ho: mean=70 versus Ha: mean > 70, and we obtain x-bar=64 and s=10 for a random sample of size 36. What is the value of the t test statistic?

3.6

6.3

-2.6

-3.6

-6.3

The value of the t test statistic is calculated by subtracting the hypothesized mean (70) from the sample mean (64) and dividing it by the standard deviation of the sample (10) divided by the square root of the sample size (36). This results in a t test statistic of -3.6.

Explanation

The value of the t test statistic is calculated by subtracting the hypothesized mean (70) from the sample mean (64) and dividing it by the standard deviation of the sample (10) divided by the square root of the sample size (36). This results in a t test statistic of -3.6.

74) The sampling distribution of p-hat is more spread out when n=300 than when n=150.

True

False

When the sample size (n) increases, the sampling distribution of p-hat becomes less spread out. This is because as the sample size increases, the sample proportion (p-hat) becomes more representative of the population proportion, resulting in less variability. Therefore, the statement that the sampling distribution of p-hat is more spread out when n=300 than when n=150 is false.

Explanation

When the sample size (n) increases, the sampling distribution of p-hat becomes less spread out. This is because as the sample size increases, the sample proportion (p-hat) becomes more representative of the population proportion, resulting in less variability. Therefore, the statement that the sampling distribution of p-hat is more spread out when n=300 than when n=150 is false.

75) The more serious the consequences of Type II error, the smaller alpha should be set.

True

False

Setting a smaller alpha level does not affect the consequences of Type II error. The consequences of Type II error are related to the power of the statistical test and the effect size. A smaller alpha level only affects the likelihood of making a Type I error, which is rejecting a true null hypothesis. Therefore, the statement is false.

Explanation

Setting a smaller alpha level does not affect the consequences of Type II error. The consequences of Type II error are related to the power of the statistical test and the effect size. A smaller alpha level only affects the likelihood of making a Type I error, which is rejecting a true null hypothesis. Therefore, the statement is false.

76) What is the advantage of randomized block design over a completely randomized design?

Variation associated with the blocking variable is removed.

Blocking is more accurate than the completely randomized design.

Blocking increases the precision of the experiment.

Blocking removes selection bias.

Blocking is illegal in basketball but not in football.

Randomized block design is advantageous over a completely randomized design because it removes the variation associated with the blocking variable. By blocking, or grouping similar experimental units together, the effect of the blocking variable is accounted for and controlled. This helps to reduce the overall variability in the experiment and increases the precision of the results. It ensures that any observed differences between treatments are not due to the blocking variable but rather to the treatment itself.

Explanation

Randomized block design is advantageous over a completely randomized design because it removes the variation associated with the blocking variable. By blocking, or grouping similar experimental units together, the effect of the blocking variable is accounted for and controlled. This helps to reduce the overall variability in the experiment and increases the precision of the results. It ensures that any observed differences between treatments are not due to the blocking variable but rather to the treatment itself.

77) After a statistical investigation, a researcher reports that “ in our sample, average expenses for a date was significantly higher (P-value=0.02) for married BYU students than for single BYU students.” What does the p-value in this statement tell us?

Only 2% of the BYU married students spend more on a date than the BYU single students.

The probability that Ho is true is 0.02.

The p-value in this statement tells us that if there were no difference in average expenses for a date between married and single students, the probability of observing a sample difference in average expenses for a date between the two groups is at least as large as that which was actually obtained, is only 0.02. In other words, the p-value suggests that the difference in average expenses for a date between married and single BYU students is statistically significant, indicating that there is likely a true difference in spending habits between the two groups.

Explanation

The p-value in this statement tells us that if there were no difference in average expenses for a date between married and single students, the probability of observing a sample difference in average expenses for a date between the two groups is at least as large as that which was actually obtained, is only 0.02. In other words, the p-value suggests that the difference in average expenses for a date between married and single BYU students is statistically significant, indicating that there is likely a true difference in spending habits between the two groups.

78) There are 3 main campuses fro BYU: Provo, Idaho, and Hawaii. The BYU Student Services conducted a student preference survey by taking a random sample from BYU-Provo, BYU-Idaho and BYU-Hawaii. The respondents were asked,"Do you prefer RMs or non-RMs or Non-LDS to date?" The percentages of respondents who preferred RMs in each of the three campuses were compared. What is the sampling design used in this study?

A completely randomized block experiment.

An SRS.

A convenience sample.

A stratified sample.

A multistage sample.

The sampling design used in this study is a stratified sample. This is because the respondents were divided into three main campuses (BYU-Provo, BYU-Idaho, and BYU-Hawaii), and the percentages of respondents who preferred RMs in each campus were compared. By dividing the population into distinct groups (strata) based on the campuses, the researchers ensured that each campus was represented proportionally in the sample, allowing for more accurate comparisons between the campuses.

Explanation

The sampling design used in this study is a stratified sample. This is because the respondents were divided into three main campuses (BYU-Provo, BYU-Idaho, and BYU-Hawaii), and the percentages of respondents who preferred RMs in each campus were compared. By dividing the population into distinct groups (strata) based on the campuses, the researchers ensured that each campus was represented proportionally in the sample, allowing for more accurate comparisons between the campuses.

79) A researcher from the A researcher from the NBA randomly sampled professional basketball players and obtained information about each player's attitude toward gambling restrictions (agree, strongly agree, disagree, etch.) and the player's division (whether pacific, west, east, or south) Testing the relationship between player attitudes and player's division, a P-value of 0.027 was computed. Using alpha=0.05, the researcher decided to reject the null hypothesis. Which one of the following conclusions is most appropriate for these results.

Player's attitudes toward gambling restrictions are associated with which division they play.

Player's attitudes toward gambling restrictions are NOT associated with which division they play.

Player's attitudes toward gambling restrictions are NOT correlated with which division they play.

Player's attitudes toward gambling restrictions are linearly correlated with which division they play.

Professional basketball players are role models. Therefore they should not gamble.

The P-value of 0.027 is less than the significance level of 0.05, indicating that there is a significant relationship between player attitudes toward gambling restrictions and the division they play in. Therefore, the researcher decided to reject the null hypothesis, suggesting that player's attitudes are associated with their division.

Explanation

The P-value of 0.027 is less than the significance level of 0.05, indicating that there is a significant relationship between player attitudes toward gambling restrictions and the division they play in. Therefore, the researcher decided to reject the null hypothesis, suggesting that player's attitudes are associated with their division.

80) A study of free-throw shooting technique in JR. Jazz leagues compared random samples of players who choose different techniques of free-throw shooting. One group only shot using the traditional method. The players in another group were asked to shoot using the new method. Here are the results on free-throw shooting percentages:

The hypotheses for this test were : Ho: Mean for New= Mean for Traditional and
Ha: Mean for New > Mean for Traditional
with P-value = 0.0018. If alpha=0.05, what can the researchers conclude?

The New method has a significantly higher mean game scores than the traditional method.

The Traditional method has a significantly higher mean free-throw shooting percentage than the New method.

The New method has a significantly higher mean free-throw shooting percentage than the Traditional method.

The researchers can conclude that the New method has a significantly higher mean free-throw shooting percentage than the Traditional method. This conclusion is based on the fact that the p-value (0.0018) is less than the significance level (0.05). Therefore, there is sufficient evidence to reject the null hypothesis (Ho) and accept the alternative hypothesis (Ha), which states that the mean for the New method is greater than the mean for the Traditional method.

Explanation

The researchers can conclude that the New method has a significantly higher mean free-throw shooting percentage than the Traditional method. This conclusion is based on the fact that the p-value (0.0018) is less than the significance level (0.05). Therefore, there is sufficient evidence to reject the null hypothesis (Ho) and accept the alternative hypothesis (Ha), which states that the mean for the New method is greater than the mean for the Traditional method.

81) Consider the following regression output on the relationship between jaw width(in inches) of a shark (response variable) and length (in feet) of the shark (explanatory variable) for a random sample of 24 sharks. Scientists are interested in predicting a shark's jaw width from its length.

Correlation coefficient: 0.8749
Estimate of sigma: 1.3757417

Parameter	Estimate	Std. Err.	DF	T-stat	P-value
Intercept	0.687864	1.299051	22	0.529513	0.5992
length	0.96345	0.082276	22	11.70994	<.0001

What is the percentage of variation in jaw width explained by the least squares regression on length?

8.2%

59.9%

76.5%

87.5%

96.3%

The correct answer is 76.5%. This percentage represents the amount of variation in jaw width that can be explained by the least squares regression on length. In other words, approximately 76.5% of the variability in jaw width can be attributed to the relationship with length. This indicates a strong and significant association between the two variables.

Explanation

The correct answer is 76.5%. This percentage represents the amount of variation in jaw width that can be explained by the least squares regression on length. In other words, approximately 76.5% of the variability in jaw width can be attributed to the relationship with length. This indicates a strong and significant association between the two variables.

82) The BYU Administration plans to conduct a survey among BYU single students to determine the proportion of single students who go out on a date every week. How many single students must be polled to estimate this proportion to within 0.04 with 95% confidence?

600

601

422

423

1037

To estimate a proportion with a certain level of confidence, the sample size can be determined using the formula n = (Z^2 * p * (1-p)) / E^2, where n is the sample size, Z is the Z-score corresponding to the desired confidence level (in this case, 95% confidence corresponds to a Z-score of approximately 1.96), p is the estimated proportion (unknown in this case), and E is the desired margin of error (0.04 in this case). Since the estimated proportion is unknown, we can assume a conservative estimate of 0.5 (maximum variability) to get the largest possible sample size. Plugging in the values, we get n = (1.96^2 * 0.5 * (1-0.5)) / 0.04^2, which simplifies to approximately 600.25. Since we cannot have a fraction of a person, we round up to the nearest whole number, which is 601. Therefore, 601 single students must be polled to estimate the proportion to within 0.04 with 95% confidence.

Explanation

To estimate a proportion with a certain level of confidence, the sample size can be determined using the formula n = (Z^2 * p * (1-p)) / E^2, where n is the sample size, Z is the Z-score corresponding to the desired confidence level (in this case, 95% confidence corresponds to a Z-score of approximately 1.96), p is the estimated proportion (unknown in this case), and E is the desired margin of error (0.04 in this case). Since the estimated proportion is unknown, we can assume a conservative estimate of 0.5 (maximum variability) to get the largest possible sample size. Plugging in the values, we get n = (1.96^2 * 0.5 * (1-0.5)) / 0.04^2, which simplifies to approximately 600.25. Since we cannot have a fraction of a person, we round up to the nearest whole number, which is 601. Therefore, 601 single students must be polled to estimate the proportion to within 0.04 with 95% confidence.

83) An An NBA researcher believes that less than less than 60% of the professional players complete college education. A random sample of 100 players yields 58 who did not complete college education. The test statistic for testing Ho: p = 0.60 versus Ha: p < 0.60 is z= -1.94. What is the correct conclusion at the 0.01 significance level?

These players have a lot of money so they do need a college education.

The correct conclusion at the 0.01 significance level is to fail to reject Ho: there is not sufficient evidence to conclude that the proportion of NBA players who complete college education is less than 0.60. This means that the researcher's belief that less than 60% of professional players complete college education cannot be supported by the sample data.

Explanation

The correct conclusion at the 0.01 significance level is to fail to reject Ho: there is not sufficient evidence to conclude that the proportion of NBA players who complete college education is less than 0.60. This means that the researcher's belief that less than 60% of professional players complete college education cannot be supported by the sample data.

84) Consider the following regression output on the relationship between jaw width(in inches) of a shark (response variable) and length (in feet) of the shark (explanatory variable) for a random sample of 24 sharks. Scientists are interested in predicting a shark's jaw width from its length.

Correlation coefficient: 0.8749
Estimate of sigma: 1.3757417

Parameter	Estimate	Std. Err.	DF	T-stat	P-value
Intercept	0.687864	1.299051	22	0.529513	0.5992
length	0.96345	0.082276	22	11.70994	<.0001

A

B

C

D

E

The estimate for the intercept in the regression equation is 0.687864. This means that when the length of the shark is 0, the estimated jaw width is 0.687864 inches. The estimate for the coefficient of the length variable is 0.96345. This means that for every 1-foot increase in the length of the shark, the estimated increase in the jaw width is 0.96345 inches. Therefore, the correct answer (B) is supported by the regression output.

Explanation

The estimate for the intercept in the regression equation is 0.687864. This means that when the length of the shark is 0, the estimated jaw width is 0.687864 inches. The estimate for the coefficient of the length variable is 0.96345. This means that for every 1-foot increase in the length of the shark, the estimated increase in the jaw width is 0.96345 inches. Therefore, the correct answer (B) is supported by the regression output.

85) A researcher for Presidential approval rating periodically conducts polls to estimate the proportion of people who agree that Pres. Obama is going a good job on the economy. The . researcher plans to triple the sample size in the polls. Why does the researcher plan to triple the sample size?

To decrease the standard deviation of the sampling distribution of the sample proportion.

To decrease the variation in the population.

To reduce the interviewer effect.

To reduce non-response bias.

To reduce confounding effects.

By tripling the sample size, the researcher increases the number of observations in the sample. This larger sample size helps to reduce the standard deviation of the sampling distribution of the sample proportion. With a smaller standard deviation, the estimates of the proportion of people who agree that Pres. Obama is doing a good job on the economy will be more precise and reliable.

Explanation

By tripling the sample size, the researcher increases the number of observations in the sample. This larger sample size helps to reduce the standard deviation of the sampling distribution of the sample proportion. With a smaller standard deviation, the estimates of the proportion of people who agree that Pres. Obama is doing a good job on the economy will be more precise and reliable.

86) The level of good cholesterol in blood samples of 40 randomly selected middle-aged men is measured at the start of a two-month study. The men participated in the same exercise routine for the two-month period and then the level of good cholesterol is measured again. Researchers want to compare the mean levels of the good cholesterol and in order to do this, they should perform

A one sample z-test for mean

A one sample t-test for mean

A two sample z-test for means

A two sample t-test for means

A matched pairs t-test for means

The correct answer is a matched pairs t-test for means. This is because the study involves measuring the level of good cholesterol in the same group of men before and after a two-month period. The purpose is to compare the mean levels of the good cholesterol within the same group, rather than comparing the means of two different groups. A matched pairs t-test is appropriate for analyzing the differences in means within the same group over time.

Explanation

The correct answer is a matched pairs t-test for means. This is because the study involves measuring the level of good cholesterol in the same group of men before and after a two-month period. The purpose is to compare the mean levels of the good cholesterol within the same group, rather than comparing the means of two different groups. A matched pairs t-test is appropriate for analyzing the differences in means within the same group over time.

87) An experiment is performed to examine the effect of three different dating activities on the rate of marriage of BYU single students. Twenty one subjects are randomly assigned to one of the three dating habits. What are the appropriate null and alternative hypotheses.

a. Ho: µ_{1 =}µ_{2 =}µ₃versus Ha: µ₁NEµ₂NEµ₃ b. Ho: µ_{1 =}µ_{2 =}µ₃versus Ha: At least one of the means is different from the others. c. Ho: p_{1 =}p_{2 =}p₃versus Ha: Not all the proportions are equal. d. None of the above.

A

B

C

D

The appropriate null and alternative hypotheses for the experiment are: Ho: p1 = p2 = p3 versus Ha: Not all the proportions are equal. This is because the experiment is examining the effect of three different dating activities on the rate of marriage, which involves proportions rather than means. The null hypothesis states that the proportions of marriage for all three dating activities are equal, while the alternative hypothesis states that at least one of the proportions is different from the others.

Explanation

The appropriate null and alternative hypotheses for the experiment are: Ho: p1 = p2 = p3 versus Ha: Not all the proportions are equal. This is because the experiment is examining the effect of three different dating activities on the rate of marriage, which involves proportions rather than means. The null hypothesis states that the proportions of marriage for all three dating activities are equal, while the alternative hypothesis states that at least one of the proportions is different from the others.

88) If we want to determine whether political party affiliation and opinion regarding health care reforms are related, what statistical procedure should we sue?

ANOVA

Chi-square test for independence

Correlation

Two-sample proportion z test

Two-sample t test for means

The Chi-square test for independence is the appropriate statistical procedure to determine whether political party affiliation and opinion regarding health care reforms are related. This test is used to analyze the relationship between two categorical variables and assess if they are independent of each other. In this case, we can use the Chi-square test to determine if there is a significant association between political party affiliation and opinions on health care reforms.

Explanation

The Chi-square test for independence is the appropriate statistical procedure to determine whether political party affiliation and opinion regarding health care reforms are related. This test is used to analyze the relationship between two categorical variables and assess if they are independent of each other. In this case, we can use the Chi-square test to determine if there is a significant association between political party affiliation and opinions on health care reforms.

89) BYU Creamery sells 16-ounce box of ice cream. The weight of the contents of a box of ice cream has a Normal distribution with mean=16 and a standard deviation of 1.1 ounces. AN SRS of 16 boxes of ice cream is to be selected and weighed and the average weight of the 16 boxes computed. What is the probability that the average weight will be less than 15.3 ounces?

0.0054

0.9946

0.2623

.0540

The probability that the average weight of the 16 boxes will be less than 15.3 ounces is 0.0054. This can be calculated using the properties of the Normal distribution and the given mean and standard deviation. By finding the z-score for 15.3 ounces and looking up the corresponding probability in a standard Normal distribution table, we can determine that the probability is 0.0054.

Explanation

The probability that the average weight of the 16 boxes will be less than 15.3 ounces is 0.0054. This can be calculated using the properties of the Normal distribution and the given mean and standard deviation. By finding the z-score for 15.3 ounces and looking up the corresponding probability in a standard Normal distribution table, we can determine that the probability is 0.0054.

90) Consider the following regression output on the relationship between jaw width(in inches) of a shark (response variable) and length (in feet) of the shark (explanatory variable) for a random sample of 24 sharks. Scientists are interested in predicting a shark's jaw width from its length.

Correlation coefficient: 0.8749
Estimate of sigma: 1.3757417

Parameter	Estimate	Std. Err.	DF	T-stat	P-value
Intercept	0.687864	1.299051	22	0.529513	0.5992
length	0.96345	0.082276	22	11.70994	<.0001

Which interval should we use to estimate the jaw width of a newly born shark?

A confidence interval for the mean jaw width of newly born sharks.

A confidence interval for the slope of the regression line.

A confidence interval for the y-intercept of the regression line.

A prediction interval for the jaw width of a newly born shark.

Based on the given regression output, the estimate of the sigma (standard deviation) is provided. This indicates that there is variability in the jaw width of sharks with the same length. A prediction interval takes into account this variability and provides a range within which the jaw width of a newly born shark is likely to fall. Therefore, we should use a prediction interval for estimating the jaw width of a newly born shark.

Explanation

Based on the given regression output, the estimate of the sigma (standard deviation) is provided. This indicates that there is variability in the jaw width of sharks with the same length. A prediction interval takes into account this variability and provides a range within which the jaw width of a newly born shark is likely to fall. Therefore, we should use a prediction interval for estimating the jaw width of a newly born shark.

91) Suppose that you were told that the statistical power of a test is 0.95. What does this mean?

The probability of failing to reject a true null hypothesis is 0.95.

The probability of rejecting a true null hypothesis is 0.95.

The probability of rejecting a false null hypothesis is 0.95.

The probability of Type I error.

A statistical power of 0.95 means that there is a 95% probability of correctly rejecting a false null hypothesis. This indicates that the test has a high chance of correctly identifying when there is a real effect or difference present.

Explanation

A statistical power of 0.95 means that there is a 95% probability of correctly rejecting a false null hypothesis. This indicates that the test has a high chance of correctly identifying when there is a real effect or difference present.

92) Certain assumptions should be satisfied and checked with residual plots in order to make valid inferences in regression analysis. Which one of the residual plots below indicates that the condition of equal variances in NOT met?

A

B

C

D

The correct answer is D. In regression analysis, one of the assumptions is that the residuals have equal variances, also known as homoscedasticity. Residual plots are used to check this assumption. Plot D indicates that the condition of equal variances is not met because it shows a pattern where the residuals are not randomly scattered around the zero line. This could suggest that the variability of the residuals is not constant across the range of predicted values, violating the assumption of equal variances.

Explanation

The correct answer is D. In regression analysis, one of the assumptions is that the residuals have equal variances, also known as homoscedasticity. Residual plots are used to check this assumption. Plot D indicates that the condition of equal variances is not met because it shows a pattern where the residuals are not randomly scattered around the zero line. This could suggest that the variability of the residuals is not constant across the range of predicted values, violating the assumption of equal variances.

93) Consider the following regression output on the relationship between jaw width(in inches) of a shark (response variable) and length (in feet) of the shark (explanatory variable) for a random sample of 24 sharks. Scientists are interested in predicting a shark's jaw width from its length.

Correlation coefficient: 0.8749
Estimate of sigma: 1.3757417

Parameter	Estimate	Std. Err.	DF	T-stat	P-value
Intercept	0.687864	1.299051	22	0.529513	0.5992
length	0.96345	0.082276	22	11.70994	<.0001

What is the predicted jaw width of a shark that is 15 feet in length?

15.1 inches

25.1 inches

35.1 inches

45.1 inches

55.1 inches

The estimated coefficient for the length variable is 0.96345, which means that for every one unit increase in length (in this case, one foot), the jaw width is predicted to increase by approximately 0.96345 units (in this case, inches). Therefore, the predicted jaw width of a shark that is 15 feet in length would be 15 * 0.96345 = 14.45175 inches, which can be rounded to 15.1 inches.

Explanation

The estimated coefficient for the length variable is 0.96345, which means that for every one unit increase in length (in this case, one foot), the jaw width is predicted to increase by approximately 0.96345 units (in this case, inches). Therefore, the predicted jaw width of a shark that is 15 feet in length would be 15 * 0.96345 = 14.45175 inches, which can be rounded to 15.1 inches.

94) Consider the following regression output on the relationship between jaw width(in inches) of a shark (response variable) and length (in feet) of the shark (explanatory variable) for a random sample of 24 sharks. Scientists are interested in predicting a shark's jaw width from its length.

Correlation coefficient: 0.8749
Estimate of sigma: 1.3757417

Parameter	Estimate	Std. Err.	DF	T-stat	P-value
Intercept	0.687864	1.299051	22	0.529513	0.5992
length	0.96345	0.082276	22	11.70994	<.0001

What is the 95% confidence interval for the slope?

0.96+-0.171

0.96+-0.271

0.69+-0.171

0.69+-0.271

0.69+-0.371

The 95% confidence interval for the slope is 0.96 ± 0.171. This means that we are 95% confident that the true slope of the relationship between jaw width and length of the shark falls within the range of 0.789 and 1.131.

Explanation

The 95% confidence interval for the slope is 0.96 ± 0.171. This means that we are 95% confident that the true slope of the relationship between jaw width and length of the shark falls within the range of 0.789 and 1.131.