Statistics Final Exam: MCQ Quiz!
-
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.
-
Do you think you could procure a good grade in the subject of statistics? Would you like to try it? In applying statistics to a question, it is a widespread practice to start a population or process to be studied. Statisticians compile data about the entire population, which is a procedure called a census. If you want to put your See moreknowledge to the test, get ready to take this final statistics exam.
(216).jpg "Statistics Final Exam: MCQ Quiz! - Quiz")
Quiz Preview
- 2.
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%
Correct Answer
A. 31%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.Rate this question:
-
- 3.
Appropriate graphical summary of the distribution of a categorical variable.
-
Bar graph
-
Box plot
-
Stemplot
-
Residual plot
-
Scatter plot
Correct Answer
A. Bar graphExplanation
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.Rate this question:
-
- 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
Correct Answer
A. All BYU Single studentsExplanation
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.Rate this question:
-
- 5.
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
Correct Answer
A. Make inferences about population parametersExplanation
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.Rate this question:
-
- 6.
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
Correct Answer
A. 100Explanation
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.Rate this question:
-
- 7.
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.
Correct Answer
A. A comparative experiment where each single student is randomly assigned to one of two treatmentsExplanation
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.Rate this question:
-
- 8.
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
Correct Answer
A. Will be greater thanExplanation
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.Rate this question:
-
- 9.
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
Correct Answer
A. Typical distance of the Final scores from their mean was about 10 pointsExplanation
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.Rate this question:
-
- 10.
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
Correct Answer
A. Jeremiah’s scoring is 2.5 standard deviations above the average scoring in the league.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.Rate this question:
-
- 11.
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.
Correct Answer
A. The distribution of the data is skewed to the leftExplanation
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.Rate this question:
-
- 12.
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
Correct Answer
A. 301, 304, 306, 308, 311Explanation
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.Rate this question:
-
- 13.
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.
Correct Answer
A. Where r is close to 1, there is a good evidence that x and y have strong positive linear relationship.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.Rate this question:
-
- 14.
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.
Correct Answer
A. Has a small probability (P-value < alpha) of occurring by chance.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 < alpha) that the observed result is due to random variation. This suggests that there may be a true effect or relationship between the variables being studied, rather than just a random fluctuation. Therefore, a statistically significant result provides evidence against the null hypothesis and supports the alternative hypothesis.Rate this question:
-
- 15.
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
Correct Answer
A. There is a curved association between x and yExplanation
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).Rate this question:
-
- 16.
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.
Correct Answer
A. Figure AExplanation
Figure A indicates that all the assumptions are met.Rate this question:
-
- 17.
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
Correct Answer
A. On the average, FS increases by about 1.4 units when the Test3 score increases by 1 unitExplanation
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.Rate this question:
-
- 18.
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.
Correct Answer
A. A lurking variable, such as higher socioeconomic condition, affects the association.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.Rate this question:
-
- 19.
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
Correct Answer
A. DExplanation
The conditional distribution for college Majors for students whose last Math class taken was College Algebra is represented by option D.Rate this question:
-
- 20.
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
Correct Answer
A. ParameterExplanation
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.Rate this question:
-
- 21.
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.
-
Use the standard normal table to compute probabilities about sample means and sample proportions from a large random samples without knowing the distribution of the population.
-
Determine whether the data are sampled from a population which is normally distributed.
Correct Answer
A. Use the standard normal table to compute probabilities about sample means and sample proportions from a large random samples without knowing the distribution of the population.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.Rate this question:
-
- 22.
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
Correct Answer
A. Approximately normal with mean=16 and a standard deviation of 0.5Explanation
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.Rate this question:
-
- 23.
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.
Correct Answer
A. The possible values of the statistic and their frequencies from all possible samples.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.Rate this question:
-
- 24.
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.
Correct Answer
A. .2729Explanation
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.Rate this question:
-
- 25.
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.
Correct Answer
A. To give a range of plausible values for the population mean.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.Rate this question:
-
- 26.
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.
-
For 95% of all possible samples, the procedure used to obtain the confidence interval provides an interval containing the population mean
Correct Answer
A. For 95% of all possible samples, the procedure used to obtain the confidence interval provides an interval containing the population meanExplanation
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.Rate this question:
-
- 27.
In hypothesis testing, what does the symbol αdenote?
-
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.
Correct Answer
A. The probability of Type I error.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.Rate this question:
-
- 28.
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.
Correct Answer
A. 0.0082Explanation
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.Rate this question:
-
- 29.
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
Correct Answer
A. 104.6Explanation
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.Rate this question:
-
- 30.
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
Correct Answer
A. 8Explanation
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.Rate this question:
-
- 31.
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
Correct Answer
A. Yes, because all expected count are greater than 5Explanation
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.Rate this question:
-
- 32.
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
Correct Answer
A. The proportions of people who are well satisfied financially are not all equal for all educational levelsExplanation
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.Rate this question:
-
- 33.
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: µ1 = µ2 = µ3 versus Ha: µ1 NE µ2 NE µ3 b. Ho: µ1 = µ2 = µ3 versus Ha: At least one of the means is different from the others.c. Ho: p1 = p2 = p3 versus Ha: Not all the proportions are equal.d. None of the above.
-
A
-
B
-
C
-
D
Correct Answer
A. CExplanation
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.Rate this question:
-
- 34.
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 < 20b. Ho: µ = 12 versus Ha: µ < 12c. Ho: µ = 12 versus Ha: µ > 12 d. Ho: µ = 12 versus Ha: µ NE 12e. None of the above.
-
A
-
B
-
C
-
D
-
E
Correct Answer
A. BExplanation
The appropriate Ho and Ha for this experiment are Ho: µ = 12 versus Ha: µ < 12. This is because the null hypothesis (Ho) states that there is no difference in the mean time it takes for dating students to get married with or without the extra credits. The alternative hypothesis (Ha) suggests that the mean time for students with the extra credits is less than 12 months, indicating that the extra credits may cause them to marry faster.Rate this question:
-
- 35.
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.
Correct Answer
A. Deciding that the incentive does speed up marriage when the incentive does not speed up marriage.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.Rate this question:
-
- 36.
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
Correct Answer
A. 2.9 to 4.1Explanation
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.Rate this question:
-
- 37.
Which of the following is an appropriate graphical summary for displaying the relationship between bivariate quantitative variables?
-
Boxplot
-
Histogram
-
Stemplot
-
Bar chart
-
Scatter plot
Correct Answer
A. Scatter plotExplanation
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.Rate this question:
-
- 38.
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.
Correct Answer
A. Using a confidence level of 99%.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.Rate this question:
-
- 39.
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.
Correct Answer
A. The probability of rejecting a false null hypothesis is 0.95.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.Rate this question:
-
- 40.
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.
Correct Answer
A. Time spent practicing free-throwsExplanation
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).Rate this question:
-
- 41.
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
Correct Answer
A. 12Explanation
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.Rate this question:
-
- 42.
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
Correct Answer
A. 9.63Explanation
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.Rate this question:
-
- 43.
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
Correct Answer
A. 0.0054Explanation
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.Rate this question:
-
- 44.
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.
Correct Answer
A. NO, the sample size is too small to apply the Central Limit theorem.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.Rate this question:
-
- 45.
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?
-
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.
-
Fail to reject Ho: there is sufficient evidence to conclude that the proportion of NBA players who complete college education is less than 0.60.
-
Reject Ho: there is not sufficient evidence to conclude that the proportion of NBA players who complete college education is less than 0.60.
-
Reject Ho: there is sufficient evidence to conclude that the proportion of NBA players who complete college education is less than 0.60.
-
These players have a lot of money so they do need a college education.
Correct Answer
A. 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.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.Rate this question:
-
- 46.
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
Correct Answer
A. DExplanation
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.Rate this question:
-
- 47.
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 andHa: Mean for New > Mean for Traditionalwith 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.
-
There is insufficient evidence to conclude that the New method has a significantly higher mean free-throw shooting percentage 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.
Correct Answer
A. The New method has a significantly higher mean free-throw shooting percentage than 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.Rate this question:
-
- 48.
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.
Correct Answer
A. Increase the sample size.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.Rate this question:
-
- 49.
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.
Correct Answer
A. Decrease the sample size.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.Rate this question:
-
Quiz Review Timeline (Updated): Mar 21, 2023 +
Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.
-
Current Version
-
Mar 21, 2023Quiz Edited by
ProProfs Editorial Team -
Jun 15, 2009Quiz Created by
Doriarg
Back to top