Chapter 7. Hypothesis Testing with One Sample

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216 Chapter 7: Hypothesis Testing with One Sample13.Testing the Assumed σa. That the population standard deviation is actually known.b. The significance level of the test was 0.05, it was a two tailed test, so any test statistic at least as small as−z .025=−1.96 would result in the reject of the null hypothesis. Solving the test statistic formula for σ yieldsσ = x − µz / n .Substituting n = 106, x = 98.2, µ = 98.6, and z = -1.96, we find the largest value for σ that results in arejection of the null hypothesis.σ = x − µ 98.2 − 98.6=z / n −1.96 / 106 = 2.1011c. Yes, there is a reasonable chance that the population standard deviation is greater than 0.62. Since it wasbased on a sample, we know that it is highly unlikely that the true value of σ is 0.62. The assumption that σis 0.62 is not realistic, but that value may be very close.14. Power and probability of Type II ErrorThe significance level is 0.05, this is a two-tailed test, so the critical values are ±z 0.025=±1.96 .Solving for x , and substituting the values µ=98.6, n = 106, and σ = 0.62, we find two values for xσ0.62x = µ ± z .025= 98.6 ±1.96n 106 = 98.718,98.482We now find the area to the right of each value of x , using the z-score based on a population mean µ = 98.4.z = x − µ 98.482 − 98.4x − µ 98.718 − 98.4= = 1.362 and z = =σ / n 0.62 / 106 σ / n 0.62 / 106 = 5.281For the first value, the power of the test is the area to the left of 1.36, which is 0.9131. For the second, thepower of the test is the area to the right of 5.281, which is 0.9999.b. The value of β = (1 – 0.9131) = 0.0869 and is also β = (1 – 0.9999) = 0.0001.15. Power of TestThe significance is 0.05 and this is a left-tailed test, so the critical value is −z 0.05=−1.645. Solving for x , andsubstituting the values µ=98.6, n = 106, and σ = 0.62, we find the value for xσ0.62x = µ − z .05= 98.6 −1.645n 106 = 98.501.Since the power is to be 0.8, this means that,under the assumption that the mean is really the alternate value,the z-score for 98.501 has area to its left equal to 0.8. Using table A-2 we find this value, 0.84. Solving the teststatistic for µ and substituting in the values gives the followingσµ = x + z .2n= 98.501+ 0.840.62106 = 98.55.7-5 Testing a Claim About a Mean: σ Not KnownIn Exercises 1-4, determine whether the hypothesis test involves a sampling distributions of means that is anormal distribution, Student t distribution, or neither.Note: It is not known how data was collected, so we assume that the sample is a simple random sample.1. The data appear to come from a normally distributed population, and σ is unknown, so this involves theStudent t distribution.2. The data appear to come from a distribution far from normal, the sample size, n = 25, is less than 30, and σ isunknown, so this involves neither.3. The data appear to come from a normally distributed population, and σ is known, so this involves the normaldistribution.
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.
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