214 <strong>Chapter</strong> 7: <strong>Hypothesis</strong> <strong>Testing</strong> <strong>with</strong> <strong>One</strong> <strong>Sample</strong>6. 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.<strong>7.</strong> 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.
<strong>Chapter</strong> 7: <strong>Hypothesis</strong> <strong>Testing</strong> <strong>with</strong> <strong>One</strong> <strong>Sample</strong> 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.