Chapter 7. Hypothesis Testing with One Sample

Chapter 7: Hypothesis Testing with One Sample 215H 0 : µ = 200H 1 : µ < 200The test statistic is z = x − µ 182.9 − 200=σ / n 121.8 / 54 =−1.032In a left-tailed test at the 0.10 significance level, the critical value is −z α=−z .05=−1.28 .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 = 0.1515.We fail to reject the null hypothesis.There is not sufficient sample evidence to support the claim that the mean weight is less than 200 lb.11. Cotinine Levels of SmokersFirst, we check the conditions. The sample size (n = 40) is greater than 30, we are assuming a value of σ, andthe sample is a random sample. The claim is that the mean cotinine level of all smokers is 200, so this is a twotailedtest. The sample mean is 172.5 and the assumed population standard deviation is 119.5. The significancelevel is 0.01.H 0 : µ = 200H 1 : µ 200 ≠The test statistic is z = x − µ 172.5− 200=σ / n 119.5/ 40 =−1.455In a two-tailed test at the 0.01 significance level, the critical values are ±z α /2=±z .005=±2.575.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 × (0.0735) = 0.1470.We fail to reject the null hypothesis.There is not sufficient evidence to warrant the rejection of the claim that the mean cotinine levels of smokers is200.0.12. Head CircumferencesFirst, we check the conditions. The sample size (n = 100) is greater than 30, we are assuming a value of σ, andthe sample is a random sample. The claim is that the mean head circumference for two month old babies is40.0 cm, so this is a two-tailed test. The sample mean is 40.6 cm and the assumed population standarddeviation is 1.6 cm. The significance level is 0.05.H 0 : µ = 40.0H 1 : µ ≠ 40.0The test statistic is z = x − µ 40.6 − 40.0=σ / n 1.6 / 100 = 3.75In 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 right of center, the P-value is twice the area to theright of the test statistic. Using Table A-2, we find that the P − value = 2 × (1 − 0.9999) = 0.0002.We reject the null hypothesis.There is sufficient evidence to warrant the rejection of the claim that the mean head circumference for twomonth old babies is 40.0 cm.