Question

I need ---

95% confidence interval - the p-value for testing - descriptive statistics & graphs

30	30
26	18.5
25	30
30	18.5
19	19
18	19
18.5	25
25.5	18.5
30	25.5
27	18.5
26.5	25
25	19
26	26
25	17
28	18.5

Answer 1

Solution: The provided sample means are shown below: X = 25.3 X = 21.87 Also, the provided sample standard deviations are: $1 = 3.95 S2 = 4.53 and the sample sizes are n = 15 and ng = 15. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = He Ha: 4 + 112 This corresponds to a two-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 = 0.05, 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 two-tailed test is t = 2.048, for a = 0.05 and df = 28 The rejection region for this two-tailed test is R = {t: t > 2.048} (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X1 - 2 t = (1,-1): +(12-1); T: +1-2 6+ 1) 25.3 -21.87 2.21 (15-1)3.952+(15-1)4.53 15+15-2 (+1) = V (4) Decision about the null hypothesis Since it is observed that t = 2.21 > t = 2.048, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0354, and since p= 0.0354 <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 population mean ui is different than up, at the 0.05 significance level.

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = n1 + n2 2 = 15+ 15 - 2 = 28. The critical value for a = 0.05 and df = 28 degrees of freedom is te = t1-a/2:1–1 = 2.048. The corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: Sp (nı - 1)si + (n2 - 1)s ni + ng 2 (15 - 1) 3.952 + (15 - 1) x 4.532 15 + 15 - 2 = 4.25 Since we assume that the population variances are equal, the standard error is computed as follows: 1 se SpV ni n2 4.25 x 1 15 15 = 1.552 Now, we finally compute the confidence interval: СІ (X1 - X, - te X se, Xi - X2 + te x se) (25.3 - 21.87 – 2.048 x 1.552, 25.3 -21.87 +2.048 x 1.552) (0.251, 6.609

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