Chapter 7. Hypothesis Testing with One Sample

Chapter 7: Hypothesis Testing with One Sample 21116. Birds and Aircraft CrashesFirst, we check the requirements. The sample does not appear to be a simple random sample, but the subjectsmay comprise a random sample that is representative of the population. The conditions for a binomial aresatisfied. The sample size is 74 with the claim that 10% of aborted takeoffs leading to aircraft going off therunway are due to bird strikes, making p = 0.10, so np = 74 × 0.10 = 7.4 and nq = 74 × 0.90 = 66.6. Therequirements are satisfied.The claim is that 10% of aborted takeoffs leading to aircraft going off the runway are due to bird strikes, sothis is a two-tailed test. In the sample, 5 aborted takeoffs that led to aircraft going off the runway were due tobird strikes. The sample proportion is p ˆ = x / n = 5/74 = 0.0676.H 0 : p = 0.10H 1 : p 0.10 ≠The test statistic is z = p ˆ − p 0.0676 − 0.10= =−0.929pq 0.10 × 0.90n 74In a two-tailed test at the 0.05 significance level, the critical values are ±z α /2=±z .025=±1.96.Since this is a two-tailed test and the test statistic is to the left of center, the P-value is twice the area to the leftof the test statistic. Using Table A-2, we find that the P − value = 2 × 0.1762 = 0.3524 .We fail to reject the null hypothesis.There is not sufficient sample evidence to warrant rejection of the claim that 10% of aborted takeoffs that leadto aircraft going off the runway are due to bird strikes.17. Females Underrepresented in TextbooksFirst, we check the requirements. The sample does not appear to be a simple random sample, but theillustrations may comprise a random sample that is representative of the population. The conditions for abinomial are satisfied. The sample size is 142 + 554 = 696 with the claim that the majority of nonreproductiveillustrations are of males, making p = 0.50, so np = 696 × 0.50 = 348 and nq = 696 × 0.50 = 348. Therequirements are satisfied.The claim is that the majority of nonreproductive illustrations are of males, so this is a right-tailed test. In thesample, 554 nonreproductive illustrations depicted males. The sample proportion isp ˆ = x / n = 554 /696 = 0.796.H 0 : p = 0.50H 1 : p > 0.50The test statistic is z = p ˆ − p 0.796 − 0.50= = 15.618pq 0.50 × 0.50n 696In a right-tailed test at the 0.05 significance level, the critical value is z α= z .05= 1.645.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 − 0.9999) = 0.0001.We reject the null hypothesis.There is sufficient sample evidence to support the claim that the majority of nonreproductive illustrations areof males.