Chapter 7. Hypothesis Testing with One Sample

214 Chapter 7: Hypothesis Testing with One Sample6. The test statistic is z = x − µ 98.20 − 98.6=σ / n 0.62 / 106 =−6.642The claim is that the mean body temperature of healthy adults is less than 98.6°F, making this a left tailed test.In a left-tailed test at the 0.01 significance level, the critical value is −z α=−z .01=−2.33.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 = 0.0001.There is sufficient evidence to support the claim that mean body temperature of healthy adults is less than98.6°F.7. The test statistic is z = x − µ 5.25− 5.00=σ / n 2.50 / 80 = 0.894The claim is that the mean time between uses of a TV remote control by males during commercials is 5.00seconds, making this a two-tailed test. In 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 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 − 0.8133) = 0.3734.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 is 5.00 seconds.8. The test statistic is z = x − µ 45,678 − 46,000= =−0.262σ / n 9900 / 65The claim is that the mean salary for college graduates who have taken a statistics course is equal to $46,000,making this a two-tailed test. In 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 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 × (0.3974) = 0.7948.There is not sufficient evidence to warrant the rejection of the claim that the mean salary for college graduateswho have taken a statistics course is equal to $46,000In Exercises 9-12, test the given claim. Identify the null hypothesis, alternate hypothesis, test statistic, P-value orcritical value(s), conclusion about the null hypothesis, and final conclusion that addresses the original claim. Usethe P-value method unless your instructor specifies otherwise.9. Everglades TemperaturesFirst, we check the conditions. The sample size (n = 61) is greater than 30, and we are assuming a value of σ.However, the sample may not be a simple random sample. However, we will continue as if it were. The claimis that the mean temperature is greater than 30.0°C, so this is a right tailed test. The sample mean is 30.4°C andthe assumed population standard deviation is 1.7°C. The significance level is 0.05.H 0 : µ = 30.0H 1 : µ > 30.0The test statistic is z = x − µ 30.4 − 30.0=σ / n 1.7 / 61 = 1.838In a right-tailed test at the 0.05 significance level, the critical value is z α= z .05= 1.645.Since this is a right tailed test, the P-value is the area to the right of the test statistic. Using Table A-2, we findthat the P − value = 1− 0.9671 = 0.0329.We reject the null hypothesis.The sample data support the claim that the mean temperature is greater than 30.0°C.10. Weights of BearsFirst, we check the conditions. The sample size (n = 54) is greater than 30, and we are assuming a value of σ.However, the sample may not be a simple random sample. However, we will continue as if it were. The claimis that the mean weight of the bears is less than 200 lb, so this is a left tailed test. The sample mean is 182.9 lband the assumed population standard deviation is 121.8 lb. The significance level is 0.10.