Chapter 7. Hypothesis Testing with One Sample

Chapter 7: Hypothesis Testing with One Sample 219The test statistic is t = x − µ 1.538 −1.5=s / n 1.914 / 5 = 0.044In a right-tailed test at the 0.01 significance level with df = 5, the critical value is t α= t .01= 3.365.In the row for 5 degrees of freedom, the test statistic lies above 1.476, 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 amount of lead in the air in Building 5 ofthe World Trade Center following the collapse of the two building is greater than 1.5 µg/m 3 .The assumption that the data came from a normally distributed population seems unrealistic, as there is anextreme outlier in the data set.17. Sugar in CerealThe claim is that the mean sugar content for all cereals is les than 0.3g, so this is a left-tailed test. The samplesize is n = 16 making the degrees of freedom df = 15, the sample mean is x = 0.295, and the sample standarddeviation is s = 0.168. The significance level is 0.05.H 0 : µ = 0.3H 1 : µ < 0.3The test statistic is t = x − µ 0.295− 0.3=s / n 0.168 / 16 =−0.119In a left-tailed test at the 0.05 significance level with df = 15, the critical value is −t α=−t .05=−1.753.In the row for 15 degrees of freedom, the absolute value for the test statistic falls to the right of 1.341 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 sugar content for all cereals is lessthan 0.3g.18. Treating Chronic Fatigue SyndromeThe claim is that the mean score on the fatigue scale is positive, or greater than zero, so this is a right-tailedtest. The sample size is n = 21 making the degrees of freedom df = 20. The significance level is 0.01.Calculating x from the sample data givesx = Σx /n = 84 /21 = 4.0Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 2221⋅ 430 − (84)= = 2.168n(n −1)21⋅ 20We move on to the test.H 0 : µ = 0H 1 : µ > 0The test statistic is t = x − µs / n = 4 − 02.168 / 21 = 8.455In a right-tailed test at the 0.01 significance level with df = 20, the critical value is t α= t .01= 2.528.In the row for 20 degrees of freedom, the test statistic lies to the left of 2.845, so the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that the mean score on the fatigue scale is positive. The treatments appear tobe effective.