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(1 point) The time needed for college students to complete a certain paper-and-pencil maze follows a normal distribution with a mean of 30 seconds and a standard deviation of 2.3 seconds. You wish to see if the mean time is changed by vigorous exercise, so you have a group of 15 college students randomly selected exercise vigorously for 30 minutes and then complete the maze. It takes them an average of 3 = 30.7 seconds to complete the maze. Use this information to test the hypotheses H:N = 30 Hu#30 Conduct a test using a significance level of a = 0.05. Round your answers to two decimal places where appropriate. (a) The test statistic (b) The positive critical value, z* = (c) The final conclusion is A. We can reject the null hypothesis that u = 30 and accept that 30. B. There is not sufficient evidence to reject the null hypothesis that w = 30.
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Answer 1

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The provided sample mean is X = 30.7 and the known population standard deviation is 0 = 2.3, and the sample size is n = 15. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 30 Ha: u + 30 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 2c 1.96. = The rejection region for this two-tailed test is R = {z: 121 > 1.96} (3) Test Statistics The z-statistic is computed as follows: z X Mo oln 30.7 – 30 2.3/V15 = 1.179

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(4) Decision about the null hypothesis = Since it is observed that |2| 1.179 < Zc 1.96, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.2385, and since p = 0.2385 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 30, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 29.536 < u < 31.864.

The provided sample mean is X = 1.6 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 = 1.5 Ha: u > 1.5 This corresponds to a right-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 right-tailed test is ze = 1.64. The rejection region for this right-tailed test is R= {z :z > 1.64} (3) Test Statistics The z-statistic is computed as follows: 1.6 - 1.5 z = X Ho olſn = = 3.873 0.1/V15

(4) Decision about the null hypothesis Since it is observed that z = 3.873 > 2c = 1.64, 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 < 0.05, 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 greater than 1.5, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 1.549 < u < 1.651.

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The provided sample mean is X = 29.7 and the known population standard deviation is o= 1, and the sample size is n = 27. .(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 30 Ha: u < 30 This corresponds to a left-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 left-tailed test is ze = -2.33. The rejection region for this left-tailed test is R= {z: 2 < -2.33} (3) Test Statistics The Z-statistic is computed as follows: 2 = X μο o/n 29.7 – 30 1/v27 = = -1.559

(4) Decision about the null hypothesis Since it is observed that z = -1.559 > 2c -2.33, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.0595, and since p 0.0595 > 0.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is less than 30, at the 0.01 significance level. Confidence Interval The 99% confidence interval is 29.204 <ll< 30.196.

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known population standard deviation is o = 1.2, and the sample size is n = 17. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 3 Ha: u > 3 This corresponds to a right-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 right-tailed test is ze = 1.64. The rejection region for this right-tailed test is R={z: 2 > 1.64} (3) Test Statistics The Z-statistic is computed as follows: 2 = X Mo o/vn = 3.4 - 3 1.2/717 = = 1.374

(4) Decision about the null hypothesis Since it is observed that z = 1.374 < ze = 1.64, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.0847, and since p 0.0847 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is greater than 3, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 2.83 < u < 3.97.
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