**1. **Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis.

Alpha = 0.09 for a right-tailed test. (Points : 5)

±1.96

1.34

±1.34

1.96

**2. **Find the value of the test statistic z using z = W4Q3

The claim is that the proportion of drowning deaths of children attributable to beaches is more than 0.25, and the sample statistics include n = 681 drowning deaths of children with 30% of them attributable to beaches. (Points : 5)

3.01

2.85

-2.85

-3.01

3. Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).The test statistic in a left-tailed test is z = -1.83. (Points : 5) 0.0672; reject the null hypothesis 0.0336; reject the null hypothesis 0.9664; fail to reject the null hypothesis 0.0672; fail to reject the null hypothesis |

4. Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).With H1: p < 3/5, the test statistic is z = -1.68. (Points : 5) 0.093; fail to reject the null hypothesis 0.0465; fail to reject the null hypothesis 0.0465; reject the null hypothesis 0.9535; fail to reject the null hypothesis |

5. Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim.The owner of a football team claims that the average attendance at games is over 694, and he is therefore justified in moving the team to a city with a larger stadium. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is failure to reject the null hypothesis, state the conclusion in nontechnical terms. (Points : 5) There is not sufficient evidence to support the claim that the mean attendance is less than 694. There is sufficient evidence to support the claim that the mean attendance is greater than 694. There is sufficient evidence to support the claim that the mean attendance is less than 694. There is not sufficient evidence to support the claim that the mean attendance is greater than 694. |

6. Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test.A consumer advocacy group claims that the mean mileage for the Carter Motor Company’s new sedan is less than 32 miles per gallon. Identify the type I error for the test. (Points : 5) Fail to reject the claim that the mean is equal to 32 miles per gallon when it is actually greater than 32 miles per gallon. Reject the claim that the mean is equal to 32 miles per gallon when it is actually less than 32 miles per gallon. Reject the claim that the mean is equal to 32 miles per gallon when it is actually 32 miles per gallon. Fail to reject the claim that the mean is equal to 32 miles per gallon when it is actually less than 32 miles per gallon. |

7. Find the P-value for the indicated hypothesis test.In a sample of 88 children selected randomly from one town, it is found that 8 of them suffer from asthma. Find the P-value for a test of the claim that the proportion of all children in the town who suffer from asthma is equal to 11%. (Points : 5) 0.2843 -0.2843 0.2157 0.5686 |

8. Find the P-value for the indicated hypothesis test.An article in a journal reports that 34% of American fathers take no responsibility for child care. A researcher claims that the figure is higher for fathers in the town of Littleton. A random sample of 225 fathers from Littleton, yielded 97 who did not help with child care. Find the P-value for a test of the researcher’s claim. (Points : 5) 0.0019 0.0015 0.0038 0.0529 |

**9. **Find the critical value or values of CRitVALX2 based on the given information.

H1: sigma > 3.5

n = 14

Alpha= 0.05 (Points : 5)

22.362

5.892

24.736

23.685

10. Find the critical value or values of CritVALX2 based on the given information.H1: Sigma> 26.1 n = 9 Alpha= 0.01 (Points : 5) 1.646 21.666 20.090 2.088 |

11. Find the number of successes x suggested by the given statement.A computer manufacturer randomly selects 2850 of its computers for quality assurance and finds that 1.79% of these computers are found to be defective. (Points : 5) 51 56 54 49 |

12. Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2, Use the given sample sizes and numbers of successes to find the pooled estimate p-bar Round your answer to the nearest thousandth.n1 = 570; n2 = 1992 x1 = 143; x2 = 550 (Points : 5) 0.541 0.270 0.520 0.216 |

13. Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numbers of successes to find the z test statistic for the hypothesis test.A report on the nightly news broadcast stated that 10 out of 108 households with pet dogs were burglarized and 20 out of 208 without pet dogs were burglarized. (Points : 5) z = -0.041 z = -0.102 z = 0.000 z = -0.173 |

14. Solve the problem.
The table shows the number of smokers in a random sample of 500 adults aged 20-24 and the number of smokers in a random sample of 450 adults aged 25-29. Assume that you plan to use a significance level of alpha = 0.10 to test the claim that P1not =P2Find the critical value(s) for this hypothesis test. Do the data provide sufficient evidence that the proportion of smokers in the 20-24 age group is different from the proportion of smokers in the 25-29 age group? |

15. Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numbers of successes to find the P-value for the hypothesis test.n1 = 50; n2 = 75 x1 = 20; x2 = 15 (Points : 5) 0.0032 0.0146 0.1201 0.0001 |

16. Construct the indicated confidence interval for the difference between population proportions p1 – p2. Assume that the samples are independent and that they have been randomly selected.x1 = 61, n1 = 105 and x2 = 82, n2 = 120; Construct a 98% confidence interval for the difference between population proportions p1 – p2. (Points : 5) 0.456 < p1 – p2 < 0.707 0.432 < p1 – p2 < 0.730 -0.228 < p1 – p2 < 0.707 -0.252 < p1 – p2 < 0.047 |

17. Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal.A researcher was interested in comparing the resting pulse rates of people who exercise regularly and the pulse rates of people who do not exercise regularly. She obtained independent simple random samples of 16 people who do not exercise regularly and 12 people who do exercise regularly. The resting pulse rates (in beats per minute) were recorded and the summary statistics are as follows. Construct a 95% confidence interval for mu1 – mu2, the difference between the mean pulse rate of people who do not exercise regularly and the mean pulse rate of people who exercise regularly. (Points : 5) |

18. State what the given confidence interval suggests about the two population means.A researcher was interested in comparing the heights of women in two different countries. Independent simple random samples of 9 women from country A and 9 women from country B yielded the following heights (in inches). The following 90% confidence interval was obtained for mu1 – mu2, the difference between the mean height of women in country A and the mean height of women in country B. -4.34 in. < mu1 – mu2 < -0.03 in |

19. Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (sigma1 = sigma2), so that the standard error of the difference between means is obtained by pooling the sample variances.A paint manufacturer wanted to compare the drying times of two different types of paint. Independent simple random samples of 11 cans of type A and 9 cans of type B were selected and applied to similar surfaces. The drying times, in hours, were recorded. The summary statistics are as follows. Construct a 99% confidence interval for mu1 – mu2, the difference between the mean drying time for paint type A and the mean drying time for paint type B. (Points : 5) |

20. The two data sets are dependent. Find d-Bar to the nearest tenth.WIT20(Points : 5) 44.1 20.3 33.9 203.4 |