Questions: Researchers conducted a study to determine whether magnets are effective in treating back pain. Pain was measured using the visual analog scale, and the results shown below are among the results obtained in the study. Higher scores correspond to greater pain levels. 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) to (c) below. Reduction in Pain Level After Magnet Treatment (μ1): n=16, x̄=0.41, s=0.85 Reduction in Pain Level After Sham Treatment (μ2): n=16, x̄=0.36, s=1.39 a. Use a 0.01 significance level to test the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment (similar to a placebo). What are the null and alternative hypotheses? A. H0: μ1<μ2 B. H0: μ1=μ2 H1: μ1 ≥ μ2 H1: μ1 ≠ μ2 C. H0: μ1=μ2 D. H0: μ1 ≠ μ2 H1: μ1>μ2 H1: μ1<μ2 The test statistic, t, is .12. (Round to two decimal places as needed.) The P-value is . (Round to three decimal places as needed.)

Solution Steps

We are testing the effectiveness of magnets in reducing pain compared to a sham treatment. The null and alternative hypotheses are defined as follows:

• Null Hypothesis \( H_0: \mu_1 \leq \mu_2 \) (the mean reduction in pain for the magnet treatment is less than or equal to that for the sham treatment)
• Alternative Hypothesis \( H_1: \mu_1 > \mu_2 \) (the mean reduction in pain for the magnet treatment is greater than that for the sham treatment)

The standard error \( SE \) is calculated using the formula:

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{0.85^2}{16} + \frac{1.39^2}{16}} = \sqrt{\frac{0.7225}{16} + \frac{1.9321}{16}} = \sqrt{0.04515625 + 0.12076375} = \sqrt{0.16592} \approx 0.4073 \]

The test statistic \( t \) is calculated using the formula:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{0.41 - 0.36}{0.4073} \approx \frac{0.05}{0.4073} \approx 0.1227 \]

The degrees of freedom \( df \) are calculated using the formula:

\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} = \frac{(0.04515625 + 0.12076375)^2}{\frac{(0.04515625)^2}{15} + \frac{(0.12076375)^2}{15}} \approx \frac{0.0274}{0.000135 + 0.000968} \approx \frac{0.0274}{0.001103} \approx 24.83 \]

The P-value is calculated based on the test statistic. Since \( t \approx 0.1227 \) and we are conducting a right-tailed test, we find:

\[ P = 1 - T(t) \approx 1 - T(0.1227) \approx 0.450 \]

At a significance level of \( \alpha = 0.01 \), we compare the P-value with \( \alpha \):

• Since \( P \approx 0.450 > 0.01 \), we fail to reject the null hypothesis.

Final Answer