Question

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. n Treatment Placebo H2 34 40 2.39 2.62 0.56 0.85 X S UU. 110 H1 H2 H:H1 H2 UU. 110-11-12 HH1 H2 The 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. 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. OB. 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. O C. 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. OD. 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. <H4-12<0 (Round to three decimal places as needed.) Click to select your answer(s). ?

Answer 1

(a)

Let x̄ and ȳ be the sample means and s_x and s_y be the sample standard deviations of two sets of data of size n_x and n_y respectively then the t-test for difference of means with unequal variance is given by :

(7-7) – (uz – uy) t= Na пу

$t = \frac{2.39-2.62}{\sqrt{\frac{0.56^{2}}{34} +\frac{0.85^{2}}{40}} }$

$t = \frac{-0.23}{0.1651}$

Degree of freedom DF = (s₁²/n₁ + s₂²/n₂)² / { [ (s₁² / n₁)² / (n₁ - 1) ] + [ (s₂² / n₂)² / (n₂ - 1) ] }

$DF = \frac{0.02726^{2}}{\frac{0.0092^{2}}{34-1}+\frac{0.0181^{2}}{40-1}}$

= 68

The p-value is .084229.

The result is significant at p < .10.

Option A : Reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the two samples come from the same popuation with same mean.

(b)

The confidence interval is given by :

$(\overline{X_{1}} - \overline{X_{2}}) \pm t* \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}$

The T-Value at 0.1 % level of significance and two tailed is 1.667572

CI = $(2.39-2.62) \pm 1.6675* 0.1652$

CI = $(-0.23) \pm 0.275471$

CI = (-0.505471, 0.045471)