Question

In: Statistics and Probability

A study was done using a treatment group and a placebo group. The results are shown...

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 0.10 significance level for both parts. Treatment Placebo mu mu 1 mu 2 n 32 37 x overbar 2.36 2.67 s 0.51 0.89 a. Test the claim that the two samples are from populations with the same mean. What are the null and alternative​ hypotheses? A. Upper H 0​: mu 1equalsmu 2 Upper H 1​: mu 1greater thanmu 2 B. Upper H 0​: mu 1less thanmu 2 Upper H 1​: mu 1greater than or equalsmu 2 C. Upper H 0​: mu 1not equalsmu 2 Upper H 1​: mu 1less thanmu 2 D. Upper H 0​: mu 1equalsmu 2 Upper H 1​: mu 1not equalsmu 2 The test​ statistic, t, is nothing. ​(Round to two decimal places as​ needed.) The​ P-value is nothing. ​(Round to three decimal 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. 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. D. Fail to 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. b. Construct a confidence interval suitable for testing the claim that the two samples are from populations with the same mean. nothingless thanmu 1 minus mu 2less than nothing ​(Round to three decimal places as​ needed.)

Solutions

Expert Solution

Ho:μ12
Ha: μ1≠μ2
std error =√(S21/n1+S22/n2)= 0.172
test stat t =(x1-x2-Δo)/Se = -1.80
p value : = 0.081

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.

b)

Point estimate of differnce =x1-x2     = -0.310
for 90 % CI & 31 df value of t= 1.696
margin of error E=t*std error                   = 0.291
lower bound=mean difference-E     = -0.60
Upper bound=mean differnce +E      = -0.02
from above 90% confidence interval for population mean =(-0.601,-0.019)

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