Chapter 7. Hypothesis Testing with One Sample

224 Chapter 7: Hypothesis Testing with One Sample6. Supermodel WeightsThe claim is that the population standard deviation of supermodel body weights is less than 29lb, so this is aleft-tailed test. The sample size is n = 9 making the degrees of freedom df = 8. The significance level is 0.01.Calculating the sample standard deviation from the data givess = nΣ(x 2 ) − (Σx) 29 ⋅132,223− (1,089)2= = 7.533n(n −1)9 ⋅8We move on to the test.H 0 : σ = 29H 1 : σ < 29(nThe test statistic is χ 2 −1)s2=σ 2 = 8 ⋅ 7.533229 2 = 0.540In a left-tailed test at the 0.01 significance level with df = 8, the critical value is χ 2 = 2.733.In the row for 8 degrees of freedom, the test statistic lies below 1.344, so the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that the population standard deviation of supermodel body weights is lessthan 29lb.7. Supermodel HeightsThe claim is that the population standard deviation of supermodel heights vary less than the heights of womenin general, who have a standard deviation of 2.5 in, so this is a left-tailed test. The sample size is n = 18making the degrees of freedom df = 17. The significance level is 0.05. Calculating the sample standarddeviation from the data givess = nΣ(x 2 ) − (Σx) 218 ⋅88,084 − (1,259)2= = 1.187n(n −1)18 ⋅17We move on to the test.H 0 : σ = 2.5H 1 : σ < 2.52(n −1)s 17The test statistic is χ 2 ⋅1.1872= = = 3.832σ 2 2.5 2In a left-tailed test at the 0.05 significance level with df = 17, the critical value is χ 2 = 8.672.In the row for 17 degrees of freedom, the test statistic lies below 5.697, so the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that the population standard deviation of supermodel heights vary less thanthe heights of women in general.8. Using Birth Weight DataThe claim is that the population standard deviation of birth weights of male babies born to mothers taking aspecial vitamin supplement is 0.470 kg, so this is a two-tailed test. The sample size is n = 16 making thedegrees of freedom df = 15. The significance level is not mentioned, so we use 0.05. Calculating the samplestandard deviation from the data givess = nΣ(x 2 ) − (Σx) 2216 ⋅ 222.571− (58.8)= = 0.657n(n −1)16 ⋅15We move on to the test.H 0 : σ = 0.470H 1 : σ 0.470 ≠(nThe test statistic is χ 2 −1)s2 15⋅ 0.6572= = = 29.311σ 2 0.470 2In a two-tailed test at the 0.05 significance level with df = 15, the critical values are χ 2 = 6.262 andχ 2 = 27.488.In the row for 15 degrees of freedom, the test statistic lies between 27.488 and 30.578, so the P-value isbetween 0.05 and 0.02.We reject the null hypothesis.