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 0.05 significance level for both parts.
\begin{tabular}{|c|c|c|}
\hline & Treatment & Placebo \\
\hline $\boldsymbol{\mu}$ & $\mu_{1}^{\prime}$ & $\mu_{2}$ \\
\hline $\mathrm{n}$ & 31 & 38 \\
\hline $\bar{x}$ & 2.38 & 2.64 \\
\hline $\mathrm{s}$ & 0.91 & 0.63 \\
\hline
\end{tabular}
a. Test the claim that the two samples are from populations with the same mean.
What are the null and alternative hypotheses?
A. $\mathrm{H}_{0} \cdot \mu_{1} \neq \mu_{2}$
B.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1} \neq \mu_{2}
\end{array}
\]
\[
H_{1}: \mu_{1}< \mu_{2}
\]
C. $\mathrm{H}_{0}: \mu_{1} \leqslant \mu_{2}$
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1}> \mu_{2}
\end{array}
\]
\[
\mathrm{H}_{1}: \mu_{1} \geq \mu_{2}
\]
The test statistic, $t$, is -1.35 . (Round to two decimal places as needed.)
The P-value is $\square$. (Round to three decimal places as needed.)
Answer
Final Answer: The null and alternative hypotheses are given by option B: \[H_{0}: \mu_{1}=\mu_{2}\] \[H_{1}: \mu_{1} \neq \mu_{2}\] The P-value is \(\boxed{0.182}\).
Step 1 ：The null and alternative hypotheses for a two-sample t-test are typically that the two population means are equal, while the alternative hypothesis is that they are not equal. This corresponds to option B in the question: \[H_{0}: \mu_{1}=\mu_{2}\] \[H_{1}: \mu_{1} \neq \mu_{2}\]
Step 2 ：To calculate the P-value, we need to use the cumulative distribution function (CDF) for the t-distribution. The CDF gives the probability that a random variable drawn from the distribution is less than or equal to a given value.
Step 3 ：Since we are doing a two-tailed test (because the alternative hypothesis is that the means are not equal), we need to find the probability that the absolute value of the test statistic is greater than the observed value, which is equivalent to 1 minus the probability that the absolute value of the test statistic is less than the observed value.
Step 4 ：Given that the test statistic, t, is -1.35 and the degrees of freedom is 67, we can calculate the P-value.
Step 5 ：The P-value is calculated to be approximately 0.182.
Step 6 ：Final Answer: The null and alternative hypotheses are given by option B: \[H_{0}: \mu_{1}=\mu_{2}\] \[H_{1}: \mu_{1} \neq \mu_{2}\] The P-value is \(\boxed{0.182}\).
