Chapter 7. Hypothesis Testing with One Sample

Chapter 7: Hypothesis Testing with One Sample 2174. The data appear to come from a distribution that is not normal, the sample size, n = 150, is larger than 30, andσ is unknown, so this involves the Student t distribution..In Exercises 5-8, use the given information to find a range of numbers for the P-value..5. In the row for 11 degrees of freedom, the test statistic falls between 3.106 and 2.718, so the P-value is between0.005 and 0.01.6. In the row for 11 degrees of freedom, the absolute value of the test statistic falls to the right of 1.363 so the P-value is more than 0.10.7. In the row for 15 degrees of freedom, the test statistic falls to the left of 2.947, so the P-value is less than 0.01.8. In the row for 8 degrees of freedom, the absolute value of the test statistic falls between 1.860 and 1.397, sothe P-value is between 0.10 and 0.20.In Exercises 9-12, assume that a simple random sample has been selected from a normally distributedpopulation. test the given claim. Find the test statistic, P-value or critical value(s), and state the final conclusion.9. The claim is that the mean IQ score of statistics professors is greater than 118, so this is a right-tailed test. Thesample size is n = 20, the sample mean is x = 120, and the sample standard deviation is s = 12. Thesignificance level is 0.05.The test statistic is t = x − µ 120 −118=s / n 12 / 20 = 0.745In a right-tailed test at the 0.05 significance level with 19 degrees of freedom, the critical value ist α= t .05= 1.729.Since the test statistic is smaller than 1.328, 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 IQ of statistics professors is greaterthan 118.10. The claim is that the mean body temperature of healthy adults is less than 98.6°F, so this is a left-tailed test.The sample size is n = 35, the sample mean is x = 98.20, and the sample standard deviation is s = 0.62. Thesignificance level is 0.01.The test statistic is t = x − µ 98.20 − 98.60= =−3.817s / n 0.62 / 35In a left-tailed test at the 0.01 significance level with 34 degrees of freedom, the critical value is−t α=−t .01=−2.441.Since the absolute value of the test statistic is larger than 2.728, the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that the mean body temperature of healthy adults is less than 98.6°F.11. The claim is that the mean time between uses of a TV remote control by males during commercials equals 5.00sec, so this is a two-tailed test. The sample size is n = 81, the sample mean is x = 5.25, and the samplestandard deviation is s = 2.50. The significance level is 0.01.The test statistic is t = x − µ 5.25− 5.00=s / n 2.50 / 81 = 0.90In a two-tailed test at the 0.01 significance level with 80 degrees of freedom, the critical values are±t α /2=±t .005=±2.639.Since the test statistic is smaller than 1.292, the P-value is greater than 0.20.We fail to reject the null hypothesis.There is not sufficient evidence to warrant the rejection of the claim that the mean time between uses of a TVremote control by males during commercials equals 5.00 sec.