Step 1

a) Here, the population standard deviation is unknown, and the sample standard deviation must be used for the calculation of the confidence interval.

Thus, to compute the confidence interval, the t-distribution must be used.

Step 2

b) The \(\displaystyle{100}{\left({1}-\alpha\right)}\%\) confidence interval for the population mean, \(\displaystyle\mu\) when the population standard deviation, \(\displaystyle\sigma\) is unknown and the sample standard deviation, s is used, is the t-confidence interval as follows:

\(\left(\bar{x}-t_{^{\alpha}/_{2;n-1}}\frac{s}{\sqrt{n}},\ \bar{x}+t_{^{\alpha}/_{2;n-1}}\frac{s}{\sqrt{n}}\right)\)

such that:

\(P\left(\bar{x}-t_{^{\alpha}/_{2;n-1}}\frac{s}{\sqrt{n}}\leq\mu\leq\bar{x}+t_{^{\alpha}/_{2;n-1}}\frac{s}{\sqrt{n}}\right)=1-\alpha\)

Here,

\(E=t_{^{\alpha}/_{2;n-1}}\frac{s}{\sqrt{n}}\) is the margin of error.

Here, n is the sample size, \(\displaystyle\overline{{{x}}}\) is the sample mean, and \(t_{^{\alpha}/_{2;n-1}}\) is the critical value of the t-distribution with \(\displaystyle{\left({n}-{1}\right)}\), above which, \(\displaystyle{100}{\left(\frac{\alpha}{{2}}\right)}\%\) or \(\displaystyle\frac{\alpha}{{2}}\) proportion of the observations lie, and below which, \(\displaystyle{100}{\left({1}-\alpha+\frac{\alpha}{{2}}\right)}\%={100}{\left({1}-\frac{\alpha}{{2}}\right)}\%\) or \(\displaystyle{\left({1}-\frac{\alpha}{{2}}\right)}\) proportion of the observations lie.

The sample size is, \(\displaystyle{n}={14},\) so that the degrees of freedom is, \(\displaystyle{n}-{1}={13}.\)

The means and standard deviations are obtained as follows:

\(\begin{array}{|c|c|}\hline x & x^{2} \\ \hline 9 & 81 \\ \hline 6 & 36 \\ \hline 10 & 100 \\ \hline 15 & 225 \\ \hline 19 & 361 \\ \hline 6 & 36 \\ \hline 23 & 529 \\ \hline 26 & 676 \\ \hline 19 & 361 \\ \hline 16 & 256 \\ \hline 11 & 121 \\ \hline 25 & 625 \\ \hline 16 & 256 \\ \hline 11 & 121 \\ \hline \sum\ x=212 & \sum\ x^{2}=3.784 \\ \hline \end{array}\)

\(\displaystyle\overline{{{x}}}={\frac{{{1}}}{{{n}}}}{\sum_{{{i}={1}}}^{{{n}}}}{x}_{{{i}}}\)

\(\displaystyle={\frac{{{212}}}{{{14}}}}\)

\(\displaystyle\approx{15.143}\)

\(\displaystyle{s}=\sqrt{{{\frac{{{1}}}{{{n}-{1}}}}{\left({\sum_{{{i}={1}}}^{{{n}}}}{{x}_{{{i}}}^{{{2}}}}-{n}\overline{{{x}}}^{{{2}}}\right)}}}\)

\(\displaystyle=\sqrt{{{\frac{{{1}}}{{{14}-{1}}}}{\left[{3},{784}-{\left({14}\right)}\cdot{\left({15.143}\right)}^{{{2}}}\right]}}}\)

\(\displaystyle\approx{6.643}.\)

The desired confidence level is \(\displaystyle{95}\%\). Thus,

\(\displaystyle{100}{\left({1}-\alpha\right)}={95}\)

\(\displaystyle{1}-\alpha={0.95}\)

\(\displaystyle\alpha={0.05}\)

\(\displaystyle\frac{\alpha}{{2}}={0.025}\)

\(\displaystyle{1}-\frac{\alpha}{{2}}={0.975}.\)

This implies that,

\(\displaystyle{t}_{{^{\left\lbrace\alpha\right\rbrace}\angle{\left\lbrace{2}.{n}-{1}\right\rbrace}}}={t}_{{{0.025}{.13}}},\)

such that:

\(\displaystyle{P}{\left({t}_{{{n}-{1}}}\leq\ {t}_{{^{\left\lbrace\alpha\right\rbrace}\angle{\left\lbrace{2}.{n}-{1}\right\rbrace}}}\right)}={P}{\left({t}_{{{13}}}\leq\ {t}_{{{0.05}{.11}}}\right)}\)

\(\displaystyle={0.975}\)

\(\displaystyle={P}{\left({t}_{{{13}}}\leq{2.16}\right)}\)

\(\displaystyle{\left[\text{Using the t-distribution table, or the Excel formula:}={T}.{I}{N}{V}{\left({0.975},{13}\right)},{P}{\left({t}_{{{13}}}\leq{2.16}\right)}={0.975}\right]}\)

\(\displaystyle\Rightarrow\ {t}_{{{0.025}{.13}}}={2.16}\)

First, the margin of error is calculated below:

\(\displaystyle{E}={t}_{{^{\left\lbrace\alpha\right\rbrace}\angle{\left\lbrace{2}.{n}-{1}\right\rbrace}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

\(\displaystyle={t}_{{{0.025}{.13}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

\(\displaystyle={\left({2.16}\right)}\cdot{\frac{{{6.643}}}{{\sqrt{{{14}}}}}}\)

\(\displaystyle\approx{3.835}\)

The confidence interval can be obtained as follows:

\(\displaystyle{\left(\overline{{{x}}}-{t}_{{^{\left\lbrace\alpha\right\rbrace}\angle{\left\lbrace{2}.{n}-{1}\right\rbrace}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}},\overline{{{x}}}+{t}_{{^{\left\lbrace\alpha\right\rbrace}\angle{\left\lbrace{2}.{n}-{1}\right\rbrace}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\right)}={\left(\overline{{{x}}}-{E},\overline{{{x}}}+{E}\right)}\)

\(\displaystyle\approx{\left({15.143}-{3.835},{15.143}+{3.835}\right)}\)

\(\displaystyle={\left({11.308},{18.978}\right)}.\)

Hence, with “\(\displaystyle{95}\%\) confidence, the population mean number of visits per physical therapy patient is between 11.308 (or approximately 11) and 18.978 (or approximately 19) visits.”

Step 3

c) As it can be said that \(\displaystyle{95}\%\) of the confidence intervals calculated on from several samples of the same size will contain the true population mean, therefore, the remaining \(\displaystyle{5}\%{\left(={100}\%-{95}\%\right)}\) of the confidence intervals will not contain the true population mean.

Hence, if “many groups of 14 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About 95 percent of these confidence intervals will contain the true population mean number of visits per patient and about 5 percent will not contain the true population mean number of visits per patient.”