Researchers conducted a study to determine whether magnets are effective in treating back pain. The results are shown in the table for the treatment (with magnets) group and the sham (or placebo) group. The results are a measure of reduction in back pain. 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.
\begin{tabular}{|c|c|c|}
\hline & Treatment & Sham \\
\hline$\mu$ & $\mu_{1}$ & $\mu_{2}$ \\
\hline $\mathrm{n}$ & 16 & 16 \\
\hline $\bar{x}$ & 0.55 & 0.37 \\
\hline $\mathrm{s}$ & 0.78 & 1.19 \\
\hline
\end{tabular}
a. Use a 0.05 significance level to test the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment
What are the null and alternative hypotheses?
\[
\begin{array}{l}
H_{0}: \mu_{1} \neq \mu_{2} \\
H_{1}: \mu_{1}< \mu_{2} \\
H_{0}: \mu_{1}< \mu_{2} \\
H_{1}: \mu_{1} \geq \mu_{2}
\end{array}
\]
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1} \neq \mu_{2} \\
H_{0}: \mu_{1}=\mu_{2} \\
H_{1} \cdot \mu_{1}> \mu_{2}
\end{array}
\]
The test statistic, $t$, is $\square$. (Round to two decimal places as needed)
Answer
Final Answer: The null and alternative hypotheses are \(H_{0}: \mu_{1}=\mu_{2}\) and \(H_{1}: \mu_{1}>\mu_{2}\). The test statistic, t, is \(\boxed{0.51}\).
Step 1 ：The null and alternative hypotheses for a statistical test comparing the means of two groups are typically a statement of no effect or no difference, and what we are testing for, respectively. In this case, we are testing whether the mean reduction in pain is greater for those treated with magnets than those given a sham treatment.
Step 2 ：The null hypothesis is \(H_{0}: \mu_{1}=\mu_{2}\), which states that there is no difference in the mean reduction in pain between the two treatments.
Step 3 ：The alternative hypothesis is \(H_{1}: \mu_{1}>\mu_{2}\), which states that the mean reduction in pain is greater for those treated with magnets.
Step 4 ：The test statistic, t, can be calculated using the formula for the t statistic in a two-sample t test. This involves the sample means, sample sizes, and sample standard deviations of the two groups.
Step 5 ：Given that the sample sizes (n1 and n2) are both 16, the sample means (\(x_{\bar{1}}\) and \(x_{\bar{2}}\)) are 0.55 and 0.37 respectively, and the sample standard deviations (s1 and s2) are 0.78 and 1.19 respectively, we can substitute these values into the formula.
Step 6 ：The calculated t statistic is approximately 0.51 when rounded to two decimal places.
Step 7 ：Final Answer: The null and alternative hypotheses are \(H_{0}: \mu_{1}=\mu_{2}\) and \(H_{1}: \mu_{1}>\mu_{2}\). The test statistic, t, is \(\boxed{0.51}\).
