Question # You are interested in finding a 95% confidence interval for the mean number of visits for physical therapy patients.

Confidence intervals
ANSWERED You are interested in finding a $$\displaystyle{95}\%$$ confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 14 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.
$$\begin{array}{|c|c|}\hline 9 & 6 & 10 & 15 & 19 & 6 & 23 & 26 & 19 & 16 & 11 & 25 & 16 & 11 \\ \hline \end{array}$$
a) To compute the confidence interval use a t or z distribution.
b) With $$\displaystyle{95}\%$$ confidence the population mean number of visits per physical therapy patient is between "?" and "?" visits.
c) If many groups of 14 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About "?" percent of these confidence intervals will contain the true population mean number of visits per patient and about "?" percent will not contain the true population mean number of visits per patient. 2021-08-15

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.”