Question 1:

A 95% confidence interval for u is computed to be (2.10, 3.15). For each of the following hypotheses, state whether H0 will be rejected at 0.05 level. Part: 0/4 Part 10f4 H0: u=4 versus H1: \"9E4 Since the 95% confidence interval (Choose one) V 11.0, then H0 (Choose one) V be rejected at X f) the 0.05 level. contains . does not contain Part:1/4 _ Part20f4 H0: u=1versusH1: uql Since the 95% confidence interval (Choose one) V \"0! then H0 (Choose one) V be rejected at X 5) the 0.05 level. Part:2/4 Part 3 Of4 H0: u=3.6 versus H1: \"$3.6 Since the 95% confidence interval (Choose one) V 11.0, then H0 (Choose one) V be rejected at X 53 the 0.05 level. Part 4 of 4 Ho: M= 1.3 versus H1 : u# 1.3 Since the 95% confidence interval (Choose one) Ho, then Ho (Choose one) V be rejected at X the 0.05 level.A test of H0: u=4 versus H]: u4 is performed using a significance level of a: 0.01. The Pvalue is 0.12. Part: 0/3 Part 1 of3 (a) 15 H0 rejected? SinceP (Choose one) V 0., we Choose one) V H0 at the (1:0.01 level. Pal't:1/3 Part 2 of 3 (b) If the true value of u is 4, is the result a Type I error, a Type II error, or a correct decision? The result is a (Choose one) V . correct decision X \\(3 Type I error Type II error Part:2/3 Part 3 of 3 (c) If the true value of u is 3, is the result a Type I error, a Type II error, or a correct decision? The result is a (Choose one) V - X 6) Measuring lung function: One of the measurements used to determine the health of a person's lungs is the amount of air a person can exhale under force in one second. This is called forced expiratory volume in one second, and is abbreviated FEV1. Assume the mean FEV1 for 10-year-old boys is 2.1 liters and that the population standard deviation is o= 0.4. A random sample of 90 10-year-old boys who live in a community with high levels of ozone pollution are found to have a sample mean FEV1 of 1.96 liters. Can you conclude that the mean FEV1 in the highpollution community is less than 2.1 liters? Use the a: 0.05 level of significance and the P-value method with the TI84 Plus calculator. Part: 0 / 5 Part 1 of 5 (a) State the appropriate null and alternate hypotheses. H : = 0 H 2'1 I:II:I El=l2l H1: 11 0 0=0 H This hypothesis test is a (Choose one) V test. X 5 Part: 3 / 7 Part 4 of 7 (d) Compute the value of the test statistic. Round the answer to two decimal places. Z = X 5 Part: 4 / 7 Part 5 of 7 (e) Compute the P-value. Round the answer to four decimal places. P-value = X Part: 5 / 7 Part 6 of 7 (f) Determine whether to reject Ho. Use the a = 0.01 level of significance. (Choose one) |the null hypothesis Ho. X 5Part7of7 (9) State a conclusion. There (Choose one) V enough evidence to conclude that the mean lead amount is less than 15 micrograms per liter. Credit card debt: Following are outstanding credit card balances for a sample of 16 college seniors at a large university. 1309 2000 529 1531 1040 509 1874 1176 381 1529 363 431 2238 967 604 511 Send data to Excel According to a report, the mean outstanding balance for college seniors in a recent year was $520. Perform a hypothesis test to determine whether the mean debt for seniors at this university is greater than $520. Use the 01: 0.10 level of significance and the Pvalue method with the TI-84 Plus calculator: Part: 0 / 5 Part 1 of 5 (a) State the appropriate null and alternate hypotheses. H0; l l I:I|:| El=El H1: ll I:II:I M This hypothesis test is a (Choose one) V test. X 3 Part:1/5 - Part20f5 (b) Compute the value of the test statistic. Round the answer to at least three decimal places. t: '_| X \\(D Part30f5 (c) Compute the P-value of the test statistic. Round the answer to at least four decimal places. P-value= '_I X \\(D Part:3/5 Part4of5 (d) Determine whether to reject H0. (Choose one) V the null hypothesis H0. X 8 Re'ect Do not reject Part:4/5 Part 50f5 (e) State a conclusion. There (Choose one) V enough evidence to conclude that the mean debt for seniors at this university is greater than $520