MenWomenH2A study was done on body temperatures of men and women. The results are shown in the table. Assume that the two samples are independent simple random samples selected fromnormally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below.1197.67°F0.91°F5997.35°F0.73°Fa. Use a 0.05 significance level to test the claim that men have a higher mean body temperature than women.What are the null and alternative hypotheses?OB. Ho H =H2O A. Ho 4, 2H2H: 1 State the conclusion for the test.O A. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that men have a higher mean body temperature than women.O B. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that men have a higher mean body temperature than women.OC. Reject the null hypothesis. There is not sufficient evidence to support the claim that men have a higher mean body temperature than women.O D. Reject the null hypothesis. There is sufficient evidence to support the claim that men have a higher mean body temperature than women.b. Construct a confidence interval suitable for testing the claim that men have a higher mean body temperature than women.(Round to three decimal places as needed.)Does the confidence interval support the conclusion found with the hypothesis test?V because the confidence interval contains

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Description: MenWomenH2A study was done on body temperatures of men and women. The results are shown in the table. Assume that the two samples are independent simple random samples selected fromnormally distributed populations, and do not assume that the populat

Given Information : A study was done on body temperature of men and women . The results are provided . Assume that the two sample are normally distributed and do not assume that the population standard deviation are equal . Test the claim : Men have a higher mean body temperature than women . The provided sample means are shown below: Also, the provided sample standard deviations are: and the sample sizes are n1=11 and n2=59. Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: \mu_1μ1 = \mu_2μ2 Ha: \mu_1μ1 > \mu_2μ2 This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. Rejection Region Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the degrees of freedom are df = 12.512df=12.512. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal: Hence, it is found that the critical value for this right-tailed test is t_c = 1.776tc=1.776, for \alpha = 0.05α=0.05 and df = 12.512df=12.512. The rejection region for this right-tailed test is R = \{t: t > 1.776\}R={t:t>1.776}. Test Statistics Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: Decision about the null hypothesis Since it is observed that t=1.102≤tc=1.776, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.1456p=0.1456, and since p = 0.1456 \ge 0.05p=0.1456≥0.05, it is concluded that the null hypothesis is not rejected. Conclusion Fail to reject the null hypothesis. Therefore, there is not enough evidence to claim that the men have a higher mean body temperature than women, at the 0.05 significance level.We need to construct the 95% confidence interval for the difference between the population means \mu_1 - \mu_2μ1−μ2, for the case that the population standard deviations are not known. Based on the information provided, we assume that the population variances are unequal, so then the number of degrees of freedom is computed as follows: The critical value for α=0.05 and df=13.009 degrees of freedom is tc=t1−α/2;n−1=2.16. The corresponding confidence interval is computed as shown below: Since we assume that the population variances are unequal, the standard error is computed as follows: Now, we finally compute the confidence interval: The 95% confidence interval is −0.31<μ1−μ2<0.95.

Math

Statistics

A study was done on body temperatures of men and women. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. 59 97.35 F 0.73 F 11 97.67 F 0.91°F a. Use a 0.05 significance level to test the claim that men have a higher mean body temperature than women. What are the null and alternative hypotheses? O B. Ho H =P H: H2 O A. Ho 2 H2 OC. Ho 2 H <2 OD. Ho P = P2 H > 2 The test statistic, t, is (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed.)

A study was done on body temperatures of men and women. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. 59 97.35 F 0.73 F 11 97.67 F 0.91°F a. Use a 0.05 significance level to test the claim that men have a higher mean body temperature than women. What are the null and alternative hypotheses? O B. Ho H =P H: H2 O A. Ho 2 H2 OC. Ho 2 H <2 OD. Ho P = P2 H > 2 The test statistic, t, is (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed.)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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