Question

Two placement exams are available that students can take to determine which math class they should begin with in their freshman ye. It is believed that there is no difference in the mean scores that would be received for the two tests. To test this using a 0.05 level of significance, a randomly selected group of students took both tests and had their scores recorded the following data were obtained Student Test A Test B 1 78 2 74 3 79 4 72 5 80 6 68 82 77 99 Dutermine 1 Point estimate 2. Pooled standard deviation Sp. 3. Calculate test statistics 4. State the decision rule 5 Decision 6. Conclusion

Answer 1

The provided sample means are shown below: X1 = 75.71 X2 = 84.14 Also, the provided sample standard deviations are: 81 = 5.62 82 = 8.71 and the sample sizes are ni = 7 and ng = 7. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 2 Ha Al + H2 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 = 12. 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 te = 2.179, for a = 0.05 and df = 12. The rejection region for this two-tailed test is R = {t: It > 2.179}. (3) Test Statistics Since it is assumed that the population variances are equal, the t- statistic is computed as follows: t = X1 - X2 (n-1) s +(ne - 1) ( 1 + 2 + 75.71 - 84.14 (7-15.62% -47- 118.71 (1 + 1) -2.152 (4) Decision about the null hypothesis Since it is observed that t = 2.152 <tc = 2.179, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.0525, and since p= 0.0525 > 0.05, 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 piis different than pe, at the 0.05 significance level. Confidence Interval The 95% confidence interval is -16.966<Mi - M2 <0.106.

Point estimate = -8. 43

Pooled estimate = 7.3297

All parts given in above test result.

Dear student,
I am waiting for your feedback. I have given my 100% to solve your queries. If you satisfied with my answer then please please like this.
Thank You