Chapter 7. Hypothesis Testing with One Sample

220 Chapter 7: Hypothesis Testing with One Sample19. Olympic WinnersThe claim is that the mean winning time in the Olympic 100 m dash is less than 10.5 sec, so this is a left-tailedtest. The sample size is n = 23 making the degrees of freedom df = 22. The significance level is not given sowe take the significance level to be 0.05. Calculating x from the sample data givesx = Σx /n = 241.27 / 23 = 10.49Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 223⋅ 241.27 − (2536.576)2= = 0.507n(n −1)23⋅ 22We move on to the test.H 0 : µ = 10.5H 1 : µ < 10.5The test statistic is t = x − µ 10.49 −10.5=s / n 0.507 / 23 =−0.095In a left-tailed test at the 0.05 significance level with df = 22, the critical value is −t α=−t .05=−1.717.In the row for 22 degrees of freedom, the absolute value for the test statistic falls to the right of 1.321 so the P-value is greater than 0.10.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the mean winning time in the Olympic 100 mdash is less than 10.5 sec.Notice that the precision of the numbers increases in later years.One important characteristic of the data is that it, generally, decreases in later Olympics. Although the resultsare that there is no evidence that times will be lower than 10.5, it is an invalid conclusion, as they are clearlygoing to continue being lower than 10.5 sec.20. Nicotine in CigarettesThe claim is that the mean nicotine content of cigarettes is greater than 40 mg, so this is a right-tailed test. Thesample size is n = 10 making the degrees of freedom df = 9. The significance level is 0.01. Calculating x fromthe sample data givesx = Σx /n = 433/10 = 43.3Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 210 ⋅18878.9 − (433)2= = 3.801n(n −1)10 ⋅ 9We move on to the test.H 0 : µ = 40H 1 : µ > 40The test statistic is t = x − µs / n=43.3− 40.03.801/ 10 = 2.745In a right-tailed test at the 0.01 significance level with df = 9, the critical value is t α= t .01= 2.821.In the row for 9 degrees of freedom, the test statistic lies between 2.821 and 2.262, so the P-value is between.01 and .025.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the editor’s belief that the mean nicotine content ofcigarettes is greater than 40 mg.21. Norwegian Babies Weigh More?The claim is that Norwegian newborn babies, on average, weigh more than 3420g, which is the mean weightof American newborn babies, so this is a right-tailed test. The sample size is n = 90 making the degrees offreedom df = 89, the sample mean is x = 3570, and the sample standard deviation is s = 498g. Thesignificance level is 0.05.H 0 : µ = 3420H 1 : µ > 3420The test statistic is t = x − µ 3570 − 3420=s / n 498 / 90 = 2.857