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

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

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

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.

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

The formula for t-statistics is given below.

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

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.

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 \(t = 0.132\).

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.

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.

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.