Question

Men Problem II (30pts) A sociologist believes that for people under the age of 30, women sleep longer per night on the weekend than men. 30 people were surveyed. Test at alpha =1%. The following summary statistics were determined from the sample. Women Mean hrs 10.45 11.18 St dev hrs 3.95 2.81 n 16 14 a) State the Null and Alternate Hypothesis in symbols and in words. (3pts) b) Calculate the expected results for the hypothesis test sampling distribution, assuming the Null Hypothesis is true. (i.e., name and graph of the sampling distribution, mean and standard error, ASSUME POOLED STANDARD DEVIATIONS) (3pts) c) Identify the standard distribution that best approximates the sampling distribution. (1pts) d) Formulate the decision rule(you can use either critical test scores or p-values) (3pts) e) Determine the statistical sample results, i.e., the differences between the sample means, the pooled standard deviations, the standard error, the test statistic, degrees of freedom and p-value. (16pts) f) Determine the conclusion in terms of the Null and Alternate Hypotheses. Do the sample results indicate that the sociologist's claim is supported at alpha= 1%? (4pts)

Answer 1

The provided sample means are shown below: Xi 10.45 X2 = 11.18 Also, the provided sample standard deviations are: $1 = 3.95 S2 = 2.81 and the sample sizes are ni = 16 and n2 = 14. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = M2 Ha: Mi < 2 This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the degrees of freedom are df = 28. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this left-tailed test is te = -2.467, for a = 0.01 and df = 28. The rejection region for this left-tailed test is R= = {t:t< -2.467} (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X1 - X2 t= (na1)+(12-1)5% (na + m) ni+n2-2 10.45 – 11.18 -0.575 |(16–1)3:062-114–1)2.81 ( 16 + D) (4) Decision about the null hypothesis Since it is observed that t = -0.575 > tc = -2.467, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.2849, and since p= 0.2849 > 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 mi is less than 42, at the 0.01 significance level.

Z-Test Results: t-stats = -1.709, p-value =0.0437 0.40 0.35 0.30 0.25 - 0.20 - 0.15 0.10 - 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

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