$K$
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.
\begin{tabular}{|c|c|c|}
\hline & Treatment & Placebo \\
\hline $\boldsymbol{\mu}$ & $\mu_{1}$ & $\mu_{2}$ \\
\hline $\mathbf{n}$ & 27 & 34 \\
\hline$\overline{\mathbf{x}}$ & 2.33 & 2.65 \\
\hline $\mathbf{s}$ & 0.69 & 0.96 \\
\hline
\end{tabular}
a. Test the claim that the two samples are from populations with the same mean. What are the null and altemative hypotheses?
A.
\[
\begin{array}{l}
H_{0}: H_{1} \neq H_{2} \\
H_{1}: H_{1}\mu_{2}
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1} \neq \mu_{2}
\end{array}
\]
The test statistic, $t$ is $\square$. (Round to two decimal places as needed.)
The P-value is $\square$. (Round to three decimal places as needed.)
State the conclusion for the test.
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.
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 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 nul 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.
(1) Time Remaining: 03:59:41
Next
Solution
Step 1 ：Define the null and alternative hypotheses. The null hypothesis is that the two population means are equal, and the alternative hypothesis is that the two population means are not equal. So, we have: \[\begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array}\]
Step 2 ：Calculate the test statistic, t, using the formula for the two-sample t-test: \[t = \frac{(\overline{x}_{1} - \overline{x}_{2})}{\sqrt{\left(\frac{s_{1}^{2}}{n_{1}}\right) + \left(\frac{s_{2}^{2}}{n_{2}}\right)}}\] where \(\overline{x}_{1}\) and \(\overline{x}_{2}\) are the sample means, \(s_{1}\) and \(s_{2}\) are the sample standard deviations, and \(n_{1}\) and \(n_{2}\) are the sample sizes.
Step 3 ：Substitute the given values into the formula: \[t = \frac{(2.33 - 2.65)}{\sqrt{\left(\frac{0.69^{2}}{27}\right) + \left(\frac{0.96^{2}}{34}\right)}}\]
Step 4 ：Solve the equation to find the test statistic, t. The calculated t statistic is approximately -1.51. So, \(t = \boxed{-1.51}\)
Step 5 ：Calculate the P-value using the t-distribution table or a statistical software. The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true.
Step 6 ：The calculated P-value is approximately 0.136. So, \(P-value = \boxed{0.136}\)
Step 7 ：Compare the P-value with the significance level (0.10 in this case). If the P-value is less than the significance level, we reject the null hypothesis. If the P-value is greater than the significance level, we fail to reject the null hypothesis.
Step 8 ：Since the P-value (0.136) is greater than the significance level (0.10), we 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.
