Question

Short answer

a. There is insufficient evidence that the magnet’s treatment is better than sham’s treatment in reducing pain.

b. The 90% confidence interval for this test is (-0.61,0.71).

c. It appears that the magnet’s treatment is not effective enough in treating the back pain. If the sample size increases, then it might be helpful to use the magnet’s treatment in treating back pain.

Step-by-step solution

Given information

The effectiveness of magnet treatment at a 0.05 level of significance is to be tested. The statistics are stated below.

\(\begin{array}{l}{{\rm{1}}^{{\rm{st}}}}\;{\rm{sample}}\;{\rm{:magnet's treatment}}\\{n_1}\; = 20,\\{s_1} = 0.96\;\\{{\bar x}_1} = 0.49\end{array}\)

\(\begin{array}{l}{{\rm{2}}^{{\rm{nd}}}}\;{\rm{sample}}\;{\rm{:sham}}\left( {{\rm{placebo}}} \right){\rm{ treatment}}\\{n_2} = 20\;\\{s_2} = 1.4\;\\{{\bar x}_2} = 0.44\end{array}\)

State the hypothesis

To test the claim that the magnet treatment reduces pain better, the hypotheses are stated as follows:

\(\begin{array}{l}{H_{0\;}}:{\mu _1} = {\mu _2}\\{H_1}\;:{\mu _1} > {\mu _2}\end{array}\)

Here,\({\mu _1},{\mu _2}\)are the population means of pain reduction after magnet treatment and sham treatment, respectively.

The test is right tailed.

State the test statistic

This is an example of two independent sample t-test concerning means.

The formula for t-statistics is given below.

\({t_{stat}} = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\)

Find degrees of freedom, critical value

For t-distribution, find the degrees of freedom as follows:

\(\begin{array}{c}df = \min \left( {\left( {{n_1} - 1} \right),\left( {{n_2} - 1} \right)} \right)\\ = \min \left( {\left( {20 - 1} \right),\left( {20 - 1} \right)} \right)\\ = 19\end{array}\)

The critical values are obtained as follows:

\(\begin{array}{c}P\left( {t > {t_\alpha }} \right) = \alpha \\P\left( {t > {t_{0.05}}} \right) = 0.05\end{array}\)

Thus, the critical value obtained from the t-table for 19 degrees of freedom is 1.729.

Compute test statistic

The test statistic of the means of populations is as follows:

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ = \frac{{\left( {0.49 - 0.44} \right)}}{{\sqrt {\frac{{{{\left( {0.96} \right)}^2}}}{{20}} + \frac{{{{\left( {1.4} \right)}^2}}}{{20}}} }}\\ = 0.132\end{array}\)

The test statistic is \(t = 0.132\).

Decision rule using the critical value

The decision criterion for this problem statement is given below.

If thetest statistic is greater than the critical value, reject the null hypothesis at \(\alpha \)level of significance.

If the test statistic is lesser than critical value, indicates failure to accept the null hypothesis at \(\alpha \)level of significance.

Here\(\left| {{t_{stat}} = 0.132} \right|\; < \;{t_{crit}} = 1.729\). Thus, there is a failure to reject the null hypothesis.

It indicates that the magnet’s treatment is ineffective relative to sham’s treatment.

Confidence interval for the difference of means of population

b.

For the 0.05 level of significance, the appropriate confidence level is 90%.

The formula for the confidence interval of the means of population is given by

\(\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\)

E is the margin of error, and the formula for the margin of error is as follows:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}} \times \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \\ = {t_{\frac{{0.10}}{2}}} \times \sqrt {\frac{{{{0.96}^2}}}{{20}} + \frac{{{{1.4}^2}}}{{20}}} \\ = 1.729 \times 0.3795\\ = 0.656\end{array}\)

Substitute the values in the formula to find the confidence interval.

\(\begin{array}{c}{\rm{C}}{\rm{.I}} = \left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\\ = \left( {0.49 - 0.44} \right) - 0.656 < \left( {{\mu _1} - {\mu _2}} \right) < \left( {0.49 - 0.44} \right) + 0.656\\ = - 0.61 < \left( {{\mu _1} - {\mu _2}} \right) < 0.71\end{array}\)

Since 0 is included in the confidence interval, the result is the same as part(a).

Thus, there is insignificant difference in the reduction of pain using magnet treatment.

Analyze the results

c.

The magnets are not effective in treating back pain.

In this case, the sample sizes are 20. There appears to be a possibility that larger sample sizes may result in better results.