Chapter 7. Hypothesis Testing with One Sample

210 Chapter 7: Hypothesis Testing with One SampleIn a left-tailed test at the 0.01 significance level, the critical value is −z α=−z .01=−2.33.Since this is a left tailed test, the P-value is the area to the left of the test statistic. Using Table A-2, we findthat the P − value = 0.1093.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the percentage of users of Ziac whoexperience dizziness is less than 5%.14. Aviation FatalitiesFirst, we check the requirements. The sample does not appear to be a simple random sample, but the incidentsmay comprise a random sample that is representative of the population. The conditions for a binomial aresatisfied. The sample size is 8411 with the claim that the percentage of crash landings that result in a fatality isgreater than 5%, making p = 0.05, so np = 8411× 0.05 = 420.55 and nq = 8411× 0.95 = 7990.45. Therequirements are satisfied.The claim is that pilots die in more than 5% of all crash landings, so this is a right-tailed test. In the sample,437 pilots died in crash landings. The sample proportion is p ˆ = x / n = 437 /8411 = 0.05196.H 0 : p = 0.05H 1 : p > 0.05The test statistic is z = p ˆ − p 0.05196 − 0.05= = 0.825pq 0.05× 0.95n 8411In a right-tailed test at the 0.06 significance level, the critical value is z α= z .06= 1.555.Since this is a right tailed test, the P-value is the area to the right of the test statistic. Using Table A-2, we findthat the P − value = (1 − 0.7967) = 0.2033.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that pilots die in more than 5% of all crashlandings. However, we cannot conclude that the fatality rate does not exceed 5%, since all we have seen is thatthere is not enough to show the rate exceeds 5%. It still might. We simply don’t have enough evidence in thissample.15. Air-Bag EffectivenessFirst, we check the requirements. The sample does not appear to be a simple random sample, but the incidentsmay comprise a random sample that is representative of the population. The conditions for a binomial aresatisfied. The sample size is 821 with the claim that the air-bag hospitalization rate is lower than 7.8%, makingp = 0.078, so np = 821× 0.078 = 64.038 and nq = 821× 0.922 = 756.962. The requirements are satisfied.The claim is that the air-bag hospitalization rate is lower than 7.8%, so this is a left-tailed test. The sampleproportion is p ˆ = x / n = 46 /821 = 0.0560.H 0 : p = 0.078H 1 : p < 0.078The test statistic is z = p ˆ − p =pqn0.0560 − 0.0780.078 × 0.922821=−2.351In a left-tailed test at the 0.01 significance level, the critical value is −z α=−z .01=−2.33.Since this is a left tailed test, the P-value is the area to the left of the test statistic. Using Table A-2, we findthat the P − value = 0.0094.We reject the null hypothesis.The sample data support the claim that the air-bag hospitalization rate is lower than the 7.8% rate for crashesof midsized cars equipped with automatic safety belts.