State the conclusion for the test. A. RejectReject the null hypothesis. There is

Question

State the conclusion for the test.

A.

RejectReject

the null hypothesis. There

is notis not

sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean.

B.

Fail to rejectFail to reject

the null hypothesis. There

isis

sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean.

C.

RejectReject

the null hypothesis. There

isis

sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean.

D.

Fail to rejectFail to reject

the null hypothesis. There

is notis 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.

TreatmentPlacebo 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 | | | 30 X2.35 0.57 40 2.63 0.88 a. Test the claim that the two samples are from populations with the same mean. What are the null and alternative hypotheses? The test statistic, t, is (Round to two decimal places as needed) The P-value isound to three decimal places as needed.) Click to select your answer(s).

Explanation / Answer

Solution:

Part a

What are the null and alternative hypotheses?

Correct Answer: B.

H0: µ1 = µ2

H1: µ1 µ2

Now, we have to find the test statistic value t.

The formula for test statistic is given as below:

Test statistic = t = (X1bar – X2bar) / sqrt[(S1^2/N1)+(S2^2/N2)]

We are given

X1bar = 2.35

X2bar = 2.63

S1 = 0.57

S2 = 0.88

N1 = 30

N2 = 40

Level of significance = = 0.10

Degrees of freedom = 66 (Assuming unequal variances)

Now, plug all values in the above formula given as below:

Test statistic = t = (2.35 – 2.63)/sqrt[(0.57^2/30)+(0.88^2/40)]

Test statistic = t = -0.28/sqrt[0.01083 + 0.01936]

Test statistic = t = -0.28 / sqrt[0.03019]

Test statistic = t = -0.28/0.173753

Test statistic = t = -1.61

Lower critical value = -1.6683

Upper critical value = 1.6683

(By using t-table)

P-value = 0.112

(By using t-table)

= 0.10

P-value >

So, we do not reject the null hypotheses

We do not reject the null hypothesis that two samples are from populations with the same mean.

Correct Answer:

D.

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.

Part b

Construct a confidence interval suitable for testing the claim that the two samples are from populations with the same mean.

Here, we have to construct the confidence interval for difference between two population means and if we find the value zero in the confidence interval then we would not be reject the null hypothesis.

The formula for confidence interval is given as below:

Confidence interval = (X1bar – X2bar) -/+ t* sqrt[(S1^2/N1)+(S2^2/N2)]

Level of significance = = 0.10

Critical t value = 1.6683

Confidence interval = (2.35 – 2.63) -/+ 1.6683*sqrt[(0.57^2/30)+(0.88^2/40)]

Confidence interval = -0.28 -/+ 1.6683*0.173753

Confidence interval = -0.28 -/+ 0.28987213

Lower limit = -0.28 - 0.28987213 = -0.56987213

Upper limit = -0.28 + 0.28987213 = 0.00987213

Confidence interval = (-0.56987213, 0.00987213)

We find the value zero lies in the confidence interval, so we do not reject the null hypothesis that two samples are from populations with the same mean.

There is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean.

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