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 0.01 significance level for both parts.
\begin{tabular}{|c|c|c|}
\hline & Treatment & Placebo \\
\hline $\boldsymbol{\mu}$ & $\boldsymbol{\mu}_{1}$ & $\boldsymbol{\mu}_{2}$ \\
\hline $\mathbf{n}$ & 31 & 38 \\
\hline $\bar{x}$ & 2.38 & 2.61 \\
\hline $\mathbf{s}$ & 0.91 & 0.64 \\
\hline
\end{tabular}
The P-value is .243 . (Round to three decimal places as needed.)
Question 4
State the conclusion for the test.
A. 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. 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.
C. 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.
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.
\[
< \mu_{1}-\mu_{2}<
\]
(Round to three decimal places as needed.)
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Final Answer: The conclusion for the test is '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.' The confidence interval for the difference in means is \(\boxed{(-0.729, 0.269)}\).
Step 1 ：Given that the P-value is 0.243 and the significance level is 0.01, we compare these two values. Since the P-value is greater than the significance level, we fail to reject the null hypothesis. Therefore, there is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean.
Step 2 ：To construct a confidence interval for the difference in means, we use the formula \(\bar{x}_1 - \bar{x}_2 \pm z \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\), where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, \(n_1\) and \(n_2\) are the sample sizes, and \(z\) is the z-score corresponding to the desired level of confidence.
Step 3 ：Substituting the given values into the formula, we get the confidence interval for the difference in means as \((-0.729, 0.269)\).
Step 4 ：Final Answer: The conclusion for the test is '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.' The confidence interval for the difference in means is \(\boxed{(-0.729, 0.269)}\).
