3) please help A study was done using a treatment group and a placebo group. The

Question

3) please help

A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal, Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. C. H_0: mu_1 = mu_2 H_1: mu_1 > mu_2 D. H_0: mu_1 = mu_2 H_1: mu_1 notequalto mu_2 The test statistic, t is. (Round to two decimal places as needed.) The P-value is. (Round to three decima places as needed.) State the conclusion for the test. A. Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean. B. Reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean C. Fail to reject the null hypothesis. There is sufficient evidence to warrant reject on of the claim that the two samples are from populations with the same mean D. Reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean. b. Construct a confidence interval suitable for testing the claim that the two samples are from populations with the same mean.

Explanation / Answer

Below are the null and alternate hypothesis
H0: mu1 = mu2
H1: mu1 not equals to mu2

Test statistics, t = 1.39

p-value = 0.1685

As p-value is greater than significance level, we fail to reject null hypothesis. This means there is not significant evidence that two means are different. Hence Option A is correct.

(B)

lets consider CI = 95%, t-value = 2

lower limit = (x1bar - x2bar) - t*SE = -0.28 - 2*0.2 = -0.68

upper limit = (x1bar - x2bar) + t*SE = -0.28 + 2*0.2 = 0.12

Hence, -0.68 < mu1-mu2 < 0.12

x1(bar) 2.37 x2(bar) 2.65 s1 0.68 s2 0.94 n1 28 n2 37 SE = sqrt[ (s12/n1) + (s22/n2) ] (s12/n1) 0.0165 (s22/n2) 0.0239 SE 0.20 DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] } [ (s12 / n1)2 / (n1 - 1) ] 0.000010 [ (s22 / n2)2 / (n2 - 1) ] 0.000016 (s12/n1 + s22/n2)2 0.001632 DF = 63

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