Question

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Entered Answer Preview Result -0.05 -0.05 incorrect -1.9599 -1.9599 correct -3.0619 -3.0619 correct B B correct At least one of the answers above is NOT correct. (1 point) it is necessary for an automobile producer to estimate the number of miles per gallon achieved by its cars. Suppose that the sample mean for a random sample of 150 cars is 29.1 miles and assume the standard deviation is 2.4 miles. Now suppose the car producer wants to test the hypothesis that, the mean number of miles per gallon, is 29.7 against the alternative hypothesis that it is not 29.7. Conduct a test using a = .05 by giving the following: (a) positive critical z score -0.05 (b) negative critical z score -1.9599 (c) test statistic -3.0619 The final conclustion is A. There is not sufficient evidence to reject the null hypothesis that = 29.7. B. We can reject the null hypothesis that = 29.7 and accept that 29.7.
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Answer 1

Solution:

The provided sample mean is X = 29.1 and the known population standard deviation is o = 2.4, and the sample size is n = 150. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 29.7 Hau # 29.7 This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a two-tailed test is ze = 1.96. The rejection region for this two-tailed test is R= {z: 1:| > 1.96} (3) Test Statistics The Z-statistic is computed as follows: X - μo 29.1 - 29.7 2.4/ 150 3.062 o/vn (4) Decision about the null hypothesis Since it is observed that = 3.062 > % = 1.96, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0022, and since p = 0.0022 <0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is different than 29.7, at the 0.05 significance level.

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The provided sample mean is X = 1.5 and the known population standard deviation is o = 0.1, and the sample size is n = 15. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 1.4 Hau 71.4 This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided the significance level is a = 0.01, and the critical value for a two-tailed test is ze = 2.58. The rejection region for this two-tailed test is R= {z : :| > 2.58} (3) Test Statistics The Z-statistic is computed as follows: x - до 0/vn 1.5 1.4 0.1/15 3.873 (4) Decision about the null hypothesis Since it is observed that 2 = 3.873 > c = 2.58, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0001, and since p= 0.0001 <0.01, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is different than 1.4, at the 0.01 significance level.