Treatment PlaceboA 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 normallydistributed 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.H1H225342.372.67S0.920.65a. Test the claim that the two samples are from populations with the same mean.What are the null and alternative hypotheses?O A. Ho: H1< H2O B. Ho: H1 = H2O D. Ho: H1 =H2%3DO C. Ho: H1 # H2The test statistic, t, is(Round to two decimal places as needed.)The P-value is |. (Round to three decimal places as needed.)State the conclusion for the test.O A. Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two samplesare from populations with the same mean.O B. Reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the two samples are frompopulations with the same mean.O C. Reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two samples arefrom populations with the same mean.O D. Fail to reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the two samples arefrom 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.

Name: Treatment PlaceboA study was done using a treatment group and a place
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Description: Treatment PlaceboA 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 normallydistributed populations, and do not assume that

The claim of the hypotheses test is that the two samples are from population with μ2the same mean or not. Let μ1 be the first population mean and be the second population mean. Null hypothesis: H0:μ1=μ2 Alternative hypothesis: H1:μ1≠μ2 Correct answer: Option B. The test statistic is, t=x¯1-x¯2s12n1+s22n2=2.37-2.670.92225+0.65234=-0.300.2151=-1.39 Thus, the test statistic is -1.39. The degrees of freedom is, df=s12n1+s22n22s12n12n1-1+s22n22n2-1=0.92225+0.6523420.92225225-1+0.65234234-1=40 The degrees of freedom is 40.The hypothesis is two tailed. The probability of t not equal to 1.39 with 40 degrees of freedom can be obtained using the excel formula “=T.DIST.2T(1.39,40)”. The probability value is 0.172. Thus, the P-value is 0.172. Decision rule: If P-value≤α, then reject the null hypothesis. Otherwise, do not reject the null hypothesis. Conclusion: The P-value (0.172) is greater than the level of significance (0.10). Based on decision rule, fail to reject the null hypothesis. Thus, there is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean. Correct answer: Option A.b. Computation of critical value: The critical value of t-distribution for 0.10 at 40 degrees of freedom can be obtained using the excel formula “=T.INV.2T(0.10,40)”. The critical value is 1.68385. The 90% confidence interval for μ1-μ2 is, x¯1-x¯2-tα2s12n1+s22n2<μ1-μ2<x¯1-x¯2+tα2s12n1+s22n22.37-2.67-1.683850.92225+0.65234<μ1-μ2<2.37-2.67+1.683850.92225+0.65234-0.30-0.362<μ1-μ2<-0.30+0.362-0.662<μ1-μ2<0.062 Thus, the 90% confidence interval for μ1-μ2 is -0.662<μ1-μ2<0.062. Since the confidence interval contains 0, the null hypothesis is failed to be rejected. This shows that, there is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean. That is, the populations have the same mean.

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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.

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.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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