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1. Students of the author estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Uses a 0.05 significance level to test the claim that these times are from a population with mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute?Assume times are normally distributed 69 81 39 65 42 21 60 63 66 48 64 70 96 91 65 a. State the null and alternative hypotheses b. Use technology to calculate the test statistic and P- Value. [Put the values in L1 (say). Then go to STAT-一→ TESTS-一→ T-Test → Input: select Data, and complete the rest] c. Use Critical - Value Method to do the test. d. Use P-Value Method to do the test. 2. Use technology to find the P- Values. Would you reject or fail to reject Ho at 0.01 significance level? Why or why not? Do 2ND (VARS) > Select tedf (lower value, upper value, degree of freedom), for lower value, use -99999 for lower tail of the T-distribution. Remember to multiple you answer by two for a two- tailed test] The claim is that for Verizon data speeds at airports, the mean is a. μ-14.00 mbps. The size is -14 and the test statistic is t-1.645. b. The claim is that for 12 AM body temperatures, the mean is μ < 98, F. The size is n = 5 and the test statistic is-2.603.

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dear student please post the question one at a time.

1)sample mean is X 62.67 and the sample standard deviation is s19.48, and the sample size is n 15 a) Null and Aternative Hypotheses The following null and alternative hypotheses need to be tested: Ha: μ#60 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. Rejection Region Based on the information provided, the significance level is α-: 0.05, and the critical value for a two- tailed test is te 2.145 The rejection region for this two-tailed test is R = {t : It! > 2.145) b) Test Statistics The t-statistic is computed as follows t_X - fig 62.67-60 8/Vn 19.48/V15

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