Question

Treatment Sham Researchers conducted a sady to determine whether marts are effective in treating back on the results are shown in the table for the treatment (with magnets) group and the sham for placebo group The results are a measure of reduction in back pain Assume that the two samples are independent simple random sample selected from normally distributed population, and do not assume that the population vandard deviations are equal. Complete parts (a) and (b) below. 16 053 0.73 11 0.45 141 What are the null and alternative hypotheses? ОА на на OCH OD H: H2 The test stat, is 0.17. Round to two decimal places as needed) The P value is 0.436 (Round to three decimal places as needed.) State the conclusion for the best Fail to reject the null hypothesis. There is not suficient evidence to support the claim that those treated with magrets have a greater mean reduction in pain than the given a sham treatment Is it valid to argue that magnets might appear to be effective of the sample sizes are larger? Since the W for those treated with magnets is the sample mean for those given a sham treatment, it valid to argue that magnets might spear to be efective if the sample sizes larger b. Constru sample standard deviation esting the claim that those treated with magnets have a greater moun reduction in pain than those given a shom trostment -0.940 sample mean (Round to free decimal places as needed

Answer 1

Solution: The provided sample means are shown below: X = 0.53 X = 0.45 Also, the provided sample standard deviations are: S1 -0.73 82 = 1.41 and the sample sizes are n = 11 and no = 11. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Hi HP 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 degrees of freedom are df = 20. 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 = 2.086, for a = 0.05 and df = 20. The rejection region for this two-tailed test is R = {t: t > 2.086). (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: ta X-X (n-1)+(12-1) -G 7:12 + The 0.53-0.45 = 0.167 (11–170.73 +(11-1)1.412 ( 1 + 1) V 11-11-2 (4) Decision about the null hypothesis Since it is observed that t = 0.167<te = 2.086, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.869, and since p = 0.869 > 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 ui is different than 2, at the 0.05 significance level.