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 $\boldsymbol{\mu}$ & $\mu_{1}$ & $\mu_{2}$ \\ \hline $\mathbf{n}$ & 25 & 25 \\ \hline $\mathbf{x}$ & 0.55 & 0.37 \\ \hline $\mathbf{s}$ & 0.68 & 1.04 \\ \hline \end{tabular} The test statistic, $t$, is 72 . (Round to two decimal places as needed.) The P-value is 0.236 . (Round to three decimal places as needed.) State the conclusion for the test. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment. Is it valid to argue that magnets might appear to be effective if the sample sizes are larger? Since the sample mean for those treated with magnets is greater than the sample mean for those given a sham treatment, it is valid to argue that magnets might appear to be effective if the sample sizes are larger. b. Construct a confidence interval suitable for testing the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment. \[ \square<\mu_{1}-\mu_{2}<\square \] (Round to three decimal places as needed.)
Solution
Step 1 ：Given values are: sample mean for treatment group (\(x_1\)) = 0.55, sample mean for sham group (\(x_2\)) = 0.37, standard deviation for treatment group (\(s_1\)) = 0.68, standard deviation for sham group (\(s_2\)) = 1.04, sample size for both groups (\(n_1 = n_2\)) = 25, and test statistic (\(t\)) = 72.
Step 2 ：Calculate the standard error (\(se\)) using the formula \(se = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\).
Step 3 ：Substitute the given values into the formula to get \(se = \sqrt{\frac{0.68^2}{25} + \frac{1.04^2}{25}} = 0.2485155930721451\).
Step 4 ：Construct a confidence interval for the difference in means using the formula \((x_1 - x_2) \pm t \cdot se\).
Step 5 ：Substitute the given values into the formula to get the lower bound of the confidence interval as \((0.55 - 0.37) - 72 \cdot 0.2485155930721451 = -17.713122701194447\) and the upper bound as \((0.55 - 0.37) + 72 \cdot 0.2485155930721451 = 18.073122701194446\).
Step 6 ：Since the confidence interval includes 0, we cannot conclude that there is a significant difference in means between the two groups. Therefore, we fail to reject the null hypothesis that the means are equal.
Step 7 ：Final Answer: The confidence interval for the difference in means between the treatment and sham groups is \(\boxed{[-17.713, 18.073]}\).
