Question

Random samples that are drawn independently from two normally distributed populations yielded the following statistics. Group 1 Group 2 - 10 ny = 15 *, -276.3 72 - 2628 2745.76 3 - 625 (The first row gives the sample sizes, the second row gives the sample means, and the third row gives the sample variances.) Can we conclude, at the 0.01 significance level, that the two population variances, o and a differ? Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table

Answer 1

no, we can not conclude that population variances differ.

The provided sample means are shown below: X1 = 276.3 X2 = 262.8 Also, the provided sample standard deviations are: Si = 52.4 S2 = 25 and the sample sizes are ni = 10 and n2 = 15. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = M2 Ha: M1 = 42 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. Testing for Equality of Variances A F-test is used to test for the equality of variances. The following F-ratio is obtained: F= Solo 52.42 252 = 4.393 The critical values are FL = 0.164 and Fu = 4.717, and since F = 4.393, then the null hypothesis of equal variances is not rejected. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the degrees of freedom are df = 23. In fact, the degrees of freedom are computed as follows, assu ming that the population variances are equal: = 2.807, for Hence, it is found that the critical value for this two-tailed test is tc 0.01 and df = 23. a = The rejection region for this two-tailed test is R = {t: 1t| > 2.807}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t X-X (n + m) (n1-1) sí +(n2-1)s ni+n2-2 276.3 – 262.8 0.023 (10-1)52.42 +(15-1)252 10+15-2 Go + 1). (4) Decision about the null hypothesis Since it is observed that t| = 0.023<tc = 2.807, it is then concluded that the null hypothesis is not rejected. 0.9821 Using the P-value approach: The p-value is p = 0.9821, and since p 0.01, it is concluded that the null hypothesis is not rejected.

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