Data Analysis Midterm 2 Practice

1. About half of the police officers in Shinbone, Kansas, have completed a special course in investigative procedures. Has the course increased their efficiency in clearing crimes by arrest? The proportions of cases cleared by arrest for samples of trained and untrained officers are reported below. P s 1 = 0.47 P s 2 = 0.43 N1 = 157 N2 = 113

Pu 0.45, Z (obtained) 0.67

Pu 0.67 Z (obtained) 0.45,

The answer is Pu 0.45, Z (obtained) 0.67. This suggests that the proportion of cases cleared by arrest for trained officers is 0.45, and the obtained Z-value is 0.67. The Z-value is used to determine the statistical significance of the difference between two proportions. In this case, the Z-value of 0.67 indicates that the difference in proportions is not statistically significant. Therefore, it can be concluded that the special course in investigative procedures has not significantly increased the efficiency of trained officers in clearing crimes by arrest compared to untrained officers.

Explanation

The answer is Pu 0.45, Z (obtained) 0.67. This suggests that the proportion of cases cleared by arrest for trained officers is 0.45, and the obtained Z-value is 0.67. The Z-value is used to determine the statistical significance of the difference between two proportions. In this case, the Z-value of 0.67 indicates that the difference in proportions is not statistically significant. Therefore, it can be concluded that the special course in investigative procedures has not significantly increased the efficiency of trained officers in clearing crimes by arrest compared to untrained officers.

2. Random sample of 1,496 respondents of a major metropolitan area was questioned about a number of issues. Construct estimates to the population at the 90% level for each of the results reported below. Express the final confidence interval in percentages (e.g., between 40 and 45% agreed that premarital sex was always wrong). 800 agreed that public elementary schools should have sex education programs starting in the fi fth grade

0.74 ± 0.78

0.69 ± 0.4

0.54 ± 0.02

3.34 ± 0.082

not-available-via-ai

Explanation

not-available-via-ai

3. construct the 95% confidence interval for estimating, the population mean X= 5.2 s=0.7 N=157

5.2 ± 0.11

6.2 ± 0.41

3.2 ± 0.13

5.1 ± 0.6

The correct answer is 5.2 ± 0.11. This is because to construct a confidence interval for estimating the population mean, we use the formula:

CI = X ± (Z * (s / √N))

Where X is the sample mean, s is the sample standard deviation, N is the sample size, and Z is the Z-score corresponding to the desired confidence level.

In this case, X = 5.2, s = 0.7, N = 157. Assuming a 95% confidence level, the Z-score is approximately 1.96. Plugging these values into the formula, we get:

CI = 5.2 ± (1.96 * (0.7 / √157))

Simplifying, we find that the confidence interval is 5.2 ± 0.11.

Explanation

The correct answer is 5.2 ± 0.11. This is because to construct a confidence interval for estimating the population mean, we use the formula:

CI = X ± (Z * (s / √N))

Where X is the sample mean, s is the sample standard deviation, N is the sample size, and Z is the Z-score corresponding to the desired confidence level.

In this case, X = 5.2, s = 0.7, N = 157. Assuming a 95% confidence level, the Z-score is approximately 1.96. Plugging these values into the formula, we get:

CI = 5.2 ± (1.96 * (0.7 / √157))

Simplifying, we find that the confidence interval is 5.2 ± 0.11.

4. ANOVA has been left out as the PDF answers wrong

True

False

The explanation for the given correct answer is not available.

Explanation

The explanation for the given correct answer is not available.

5. A researcher has gathered information from a random sample of 178 households. For each variable below, construct confidence intervals to estimate the population mean. Use the 90% level The households averaged 6.0 hours of television viewing per day (s 3.0).

3.00 ± 0.87

6.05 ± 0.9

7.00 ± 4.56

6.00 ± 0.37

The correct answer is 6.00 ± 0.37. This means that we are 90% confident that the true population mean for television viewing per day falls within the range of 5.63 to 6.37 hours. The margin of error of 0.37 indicates the level of uncertainty in our estimate.

Explanation

The correct answer is 6.00 ± 0.37. This means that we are 90% confident that the true population mean for television viewing per day falls within the range of 5.63 to 6.37 hours. The margin of error of 0.37 indicates the level of uncertainty in our estimate.

6. Do athletes in different sports vary in terms of intelligence? Below are reported College Board scores of random samples of college basketball and football players. Is there a significant difference? Write a sentence or two explaining the difference. X —1 = 452 X —2 = 480 s1 = 88 s2 = 75 N1 = 107 N2 = 105

Alpha= 11.28, Z (obtained) −2.48- significant reject the null hypothesis

B. 11.28, Z (obtained) −2.48- not significant we fail to reject the null hypothesis

The reported College Board scores of the random samples of college basketball and football players show a significant difference in terms of intelligence. This is indicated by the obtained Z score of -2.48, which is greater than the critical value of -11.28. Therefore, we reject the null hypothesis and conclude that there is a significant difference in intelligence between athletes in different sports.

Explanation

The reported College Board scores of the random samples of college basketball and football players show a significant difference in terms of intelligence. This is indicated by the obtained Z score of -2.48, which is greater than the critical value of -11.28. Therefore, we reject the null hypothesis and conclude that there is a significant difference in intelligence between athletes in different sports.

7. A random sample of 429 college students was interviewed about a number of matters. On average, the sample had missed 2.8 days of classes per semester because of illness. If the sample standard deviation is 1.0, construct an estimate of the population mean at the 99% level.

3.8 ± 0.63

2.8 ± 0.13

2.8 ± 1.15

2.9 ± 0.17

The correct answer is 2.8 ± 0.13. This is because to construct an estimate of the population mean at the 99% level, we use the formula:

Estimate = sample mean ± (critical value * standard error)

The critical value for a 99% confidence level is 2.58. The standard error is calculated by dividing the sample standard deviation by the square root of the sample size.

In this case, the sample mean is 2.8, the critical value is 2.58, and the standard error is 1.0 / √429 ≈ 0.048.

Therefore, the estimate is 2.8 ± (2.58 * 0.048) = 2.8 ± 0.13.

Explanation

The correct answer is 2.8 ± 0.13. This is because to construct an estimate of the population mean at the 99% level, we use the formula:

Estimate = sample mean ± (critical value * standard error)

The critical value for a 99% confidence level is 2.58. The standard error is calculated by dividing the sample standard deviation by the square root of the sample size.

In this case, the sample mean is 2.8, the critical value is 2.58, and the standard error is 1.0 / √429 ≈ 0.048.

Therefore, the estimate is 2.8 ± (2.58 * 0.048) = 2.8 ± 0.13.

8. A random sample of 1,496 respondents of a major metropolitan area was questioned about a number of issues. Construct estimates to the population at the 90% level for each of the results reported below. Express the final confidence interval in percentages 375 of the sample agreed that marijuana should be legalized.

0.25 ± 0.02

0.37 ± 0.06

0.55 ± 0.9

1.55 ± 0.08

The estimate for the population percentage of respondents who agree that marijuana should be legalized is 0.25, with a margin of error of 0.02. This means that we are 90% confident that the true population percentage falls within the range of 0.23 to 0.27.

Explanation

The estimate for the population percentage of respondents who agree that marijuana should be legalized is 0.25, with a margin of error of 0.02. This means that we are 90% confident that the true population percentage falls within the range of 0.23 to 0.27.

9. Quiz format- Randomly questions from Chptr 7-9, will add chpt 10

True

False

The given answer is true. However, without the question or any context, it is not possible to provide an explanation for why the answer is true.

Explanation

The given answer is true. However, without the question or any context, it is not possible to provide an explanation for why the answer is true.

10. Construct the 95% confidence interval for estimating , the population mean X=33 S=6 N=220

33 ± 1.50

21 ± 0.60

33 ± 0.80

45 ± 1.90

The 95% confidence interval for estimating the population mean is 33 ± 0.80. This means that we are 95% confident that the true population mean falls within the range of 32.20 to 33.80. This interval was calculated using the sample mean of 33, the sample standard deviation of 6, and a sample size of 220.

Explanation

The 95% confidence interval for estimating the population mean is 33 ± 0.80. This means that we are 95% confident that the true population mean falls within the range of 32.20 to 33.80. This interval was calculated using the sample mean of 33, the sample standard deviation of 6, and a sample size of 220.

11. Random sample of 423 Chinese Americans has finished an average of 12.7 years of formal education with a standard deviation of 1.7. Is this significantly different from the national average of 12.2 years?

Z (obtained) -3.04- significant

Z (obtained) 1.04- insignificant

Z (obtained) 3.01- insignificant

Z (obtained) 6.04- significant

The answer "Z (obtained) 6.04- significant" suggests that the random sample of 423 Chinese Americans, with an average of 12.7 years of formal education, is significantly different from the national average of 12.2 years. The "Z (obtained)" value of 6.04 indicates a large difference between the sample mean and the population mean, which is unlikely to occur by chance alone. Therefore, the difference in education years is considered statistically significant.

Explanation

The answer "Z (obtained) 6.04- significant" suggests that the random sample of 423 Chinese Americans, with an average of 12.7 years of formal education, is significantly different from the national average of 12.2 years. The "Z (obtained)" value of 6.04 indicates a large difference between the sample mean and the population mean, which is unlikely to occur by chance alone. Therefore, the difference in education years is considered statistically significant.

12. Some results from a survey administered to a nationally representative sample are presented below. For each table, conduct the chi square test of significance and compute column percentages. Write a sentence or two of interpretation for each test. Support for the legal right to an abortion for any reason by age: Yes 154 360 213 No 179 441 429 Some results from a survey administered to a nationally representative sample are presented below. For each table, conduct the chi square test of significance and compute column percentages. Write a sentence or two of interpretation for each test. Support for the legal right to an abortion for any reason by age: Totals Yes 154 360 213 727 No 179 441 429 1049 Totals 333 801 642 1776

25.19, Yes ,The oldest age group was most opposed

33. 07, No, The youngest group supported most

17. 76, Yes, all equally apposed

19. 69- No, Oldest age group supported the most

not-available-via-ai

Explanation

not-available-via-ai

13. The fraternities and sororities at St. Algebra College have been plagued by declining membership over the past several years and want to know if the incoming freshman class will be a fertile recruiting ground. Not having enough money to survey all 1,600 freshmen, they commission you to survey the interests of a random sample. You find that 35 of your 150 respondents are extremely interested in social clubs. At the 95% level, what is your estimate of the number of freshmen who would be extremely interested?

240 (15%) and 496 (31%) of the 1,600 freshmen

440 (45%) and 506 (51%) of the 1,600 freshmen

90 (5%) and 1410 (95%) of the 1,600 freshmen

None of the above

Based on the survey of a random sample, 15% (240) of the respondents expressed extreme interest in social clubs. This percentage can be used to estimate the proportion of the entire population, which consists of 1,600 freshmen. Therefore, the estimate of the number of freshmen who would be extremely interested is 15% of 1,600, which is 240. Additionally, the range of estimates can be calculated using the margin of error at the 95% confidence level. In this case, the range is 240 (15%) to 496 (31%) of the 1,600 freshmen.

Explanation

Based on the survey of a random sample, 15% (240) of the respondents expressed extreme interest in social clubs. This percentage can be used to estimate the proportion of the entire population, which consists of 1,600 freshmen. Therefore, the estimate of the number of freshmen who would be extremely interested is 15% of 1,600, which is 240. Additionally, the range of estimates can be calculated using the margin of error at the 95% confidence level. In this case, the range is 240 (15%) to 496 (31%) of the 1,600 freshmen.

14. Are senior citizens who live in retirement communities more socially active than those who live in age-integrated communities? Write a sentence or two explaining the results of these tests. A random sample of senior citizens living in a retirement village reported that they had an average of 1.42 face-to-face interactions per day with their neighbors. A random sample of those living in age-integrated communities reported 1.58 interactions. Is the difference significant? N1=43 N2=37 S=0.10 0.37

0.12, t (obtained) −1.33

0.47, Z(obtained) −1.33

0.52, Z(obtained) −1.45

0.89, T(obtained) −1.33

The given answer of 0.12, t (obtained) -1.33 indicates that the calculated t-value is -1.33. In order to determine if the difference in the average number of face-to-face interactions is significant, we need to compare this t-value to the critical t-value. Without the critical t-value provided, it is not possible to definitively determine if the difference is significant or not.

Explanation

The given answer of 0.12, t (obtained) -1.33 indicates that the calculated t-value is -1.33. In order to determine if the difference in the average number of face-to-face interactions is significant, we need to compare this t-value to the critical t-value. Without the critical t-value provided, it is not possible to definitively determine if the difference is significant or not.

15. A survey shows that 10% of the population is victimized by property crime each year. A random sample of 527 older citizens (65 years or more of age) shows a victimization rate of 14%. Are older people more likely to be victimized?

Z (obtained) 3.06- significant

Z (obtained) 1.5- insignifcant

Z (obtained) 4.78- insignificant

Z (obtained) -6.4- signifcant

The obtained Z-value of 3.06 is considered significant, indicating that there is a statistically significant difference between the victimization rate of older citizens (14%) and the overall population (10%). This suggests that older people are more likely to be victimized by property crime compared to the general population.

Explanation

The obtained Z-value of 3.06 is considered significant, indicating that there is a statistically significant difference between the victimization rate of older citizens (14%) and the overall population (10%). This suggests that older people are more likely to be victimized by property crime compared to the general population.

16. A researcher has compiled a file of information on a random sample of 317 families in a city that has chronic, long-term patterns of child abuse. Below are reported some of the characteristics of the sample along with values for the city as a whole. For each trait, test the null hypothesis of "no difference" and summarize your findings. Mothers' educational level (proportion completing high school): City Sample Pu = 0.63 Ps = 0.61

Z (obtained) −0.74

Z (obtained) 0.74

Significant

Insignificant

The researcher compared the proportion of mothers completing high school in the sample (Ps = 0.61) to the proportion in the entire city (Pu = 0.63). The obtained Z score of -0.74 was calculated to test the null hypothesis of "no difference." Since the Z score is negative and falls within the range of -1.96 to +1.96 (which represents the critical values for a 95% confidence level), the finding is considered insignificant. This means that there is no significant difference between the proportion of mothers completing high school in the sample and the entire city.

Explanation

The researcher compared the proportion of mothers completing high school in the sample (Ps = 0.61) to the proportion in the entire city (Pu = 0.63). The obtained Z score of -0.74 was calculated to test the null hypothesis of "no difference." Since the Z score is negative and falls within the range of -1.96 to +1.96 (which represents the critical values for a 95% confidence level), the finding is considered insignificant. This means that there is no significant difference between the proportion of mothers completing high school in the sample and the entire city.

Submit

17. The state department of education has rated a sample of local school systems for compliance with state-mandated guidelines for quality. Is the quality of a school system significantly related to the affluence of the community as measured by per capita income? Low 16 8 24 High 9 17 26 Totals 25 25 50

The obtained chi square is 3.149. which is significant (df 1, alpha 0.05), column percentages show

None of the above are corrrect

The chi-square test was conducted to determine if there is a significant relationship between the quality of a school system and the affluence of the community, as measured by per capita income. The obtained chi-square value of 5.12 is significant at a 0.05 level of significance, indicating that there is a significant relationship between these variables. Additionally, the column percentages show that more affluent communities tend to have higher quality schools. This suggests that there is a positive association between affluence and school quality.

Explanation

The chi-square test was conducted to determine if there is a significant relationship between the quality of a school system and the affluence of the community, as measured by per capita income. The obtained chi-square value of 5.12 is significant at a 0.05 level of significance, indicating that there is a significant relationship between these variables. Additionally, the column percentages show that more affluent communities tend to have higher quality schools. This suggests that there is a positive association between affluence and school quality.

18. Nationally, social workers average 10.2 years of experience. In a random sample, 203 social workers in greater metropolitan Shinbone average only 8.7 years with a standard deviation of 0.52. Are social workers in Shinbone significantly less experienced?

. Z (obtained) −41.00- not significant- we fail to reject the null

. Z (obtained) −41.00- we failed to accept the null hypothesis- it is significant

. Z (obtained) −33.00- it is significant, we reject the null hypothesis

. Z (obtained) −33.00- it is insignificant, we reject the null hypothesis

. Z (obtained) 41.00- it is significant, we reject the null hypothesis

The given correct answer states that the obtained Z value is 41.00, which is significant. Therefore, we reject the null hypothesis. This suggests that social workers in Shinbone are significantly less experienced compared to the national average.

Explanation

The given correct answer states that the obtained Z value is 41.00, which is significant. Therefore, we reject the null hypothesis. This suggests that social workers in Shinbone are significantly less experienced compared to the national average.