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

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 $100\left(1-\alpha \right)\%$ confidence interval for the population mean, $\mu$ when the population standard deviation, $\sigma$ is unknown and the sample standard deviation, s is used, is the t-confidence interval as follows:
$\left(\overline{x}-t_{{}^{\alpha }{/}_{2;n-1}}\frac{s}{\sqrt{n}}, \overline{x}+t_{{}^{\alpha }{/}_{2;n-1}}\frac{s}{\sqrt{n}}\right)$
such that:
$P\left(\overline{x}-t_{{}^{\alpha }{/}_{2;n-1}}\frac{s}{\sqrt{n}}\le \mu \le \overline{x}+t_{{}^{\alpha }{/}_{2;n-1}}\frac{s}{\sqrt{n}}\right)=1-\alpha$
Here,
$E=t_{{}^{\alpha }{/}_{2;n-1}}\frac{s}{\sqrt{n}}$ is the margin of error.
Here, n is the sample size, $\overline{x}$ is the sample mean, and $t_{{}^{\alpha }{/}_{2;n-1}}$ is the critical value of the t-distribution with $\left(n-1\right)$, above which, $100\left(\frac{\alpha }{2}\right)\%$ or $\frac{\alpha }{2}$ proportion of the observations lie, and below which, $100\left(1-\alpha +\frac{\alpha }{2}\right)\%=100\left(1-\frac{\alpha }{2}\right)\%$ or $\left(1-\frac{\alpha }{2}\right)$ proportion of the observations lie.
The sample size is, $n=14,$ so that the degrees of freedom is, $n-1=13.$
The means and standard deviations are obtained as follows:
$\begin{array}{|cc|}\hline x& x^2\\ 9& 81\\ 6& 36\\ 10& 100\\ 15& 225\\ 19& 361\\ 6& 36\\ 23& 529\\ 26& 676\\ 19& 361\\ 16& 256\\ 11& 121\\ 25& 625\\ 16& 256\\ 11& 121\\ \sum x=212& \sum x^2=3.784\\ \hline\end{array}$
$\overline{x}=\frac{1}{n}\sum _{i=1}^n{x}_i$
$=\frac{212}{14}$
$\approx 15.143$
$s=\sqrt{\frac{1}{n-1}\left(\sum _{i=1}^n{x}_i^2-n{\overline{x}}^2\right)}$
$=\sqrt{\frac{1}{14-1}\left[3,784-\left(14\right)·{\left(15.143\right)}^2\right]}$
$\approx 6.643.$
The desired confidence level is $95\%$. Thus,
$100\left(1-\alpha \right)=95$
$1-\alpha =0.95$
$\alpha =0.05$
$\frac{\alpha }{2}=0.025$
$1-\frac{\alpha }{2}=0.975.$
This implies that,
$t_{{ }^{\left\{\alpha \right\}}\mathrm{\angle }\{2.n-1\}}=t_{0.025.13},$
such that:
$P(t_{n-1}\le t_{{ }^{\left\{\alpha \right\}}\mathrm{\angle }\{2.n-1\}})=P(t_{13}\le t_{0.05.11})$
$=0.975$
$=P\left(t_{13}\le 2.16\right)$
$\left[\text{Using the t-distribution table, or the Excel formula:}=T.INV\left(0.975,13\right),P\left(t_{13}\le 2.16\right)=0.975\right]$
$\Rightarrow t_{0.025.13}=2.16$
First, the margin of error is calculated below:
$E=t_{{ }^{\left\{\alpha \right\}}\mathrm{\angle }\{2.n-1\}}\frac{s}{\sqrt{n}}$
$=t_{0.025.13}\frac{s}{\sqrt{n}}$
$=\left(2.16\right)·\frac{6.643}{\sqrt{14}}$
$\approx 3.835$
The confidence interval can be obtained as follows:
$(\overline{x}-t_{{ }^{\left\{\alpha \right\}}\mathrm{\angle }\{2.n-1\}}\frac{s}{\sqrt{n}},\overline{x}+t_{{ }^{\left\{\alpha \right\}}\mathrm{\angle }\{2.n-1\}}\frac{s}{\sqrt{n}})=(\overline{x}-E,\overline{x}+E)$
$\approx \left(15.143-3.835,15.143+3.835\right)$
$=\left(11.308,18.978\right).$
Hence, with “$95\%$ 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 $95\%$ of the confidence intervals calculated on from several samples of the same size will contain the true population mean, therefore, the remaining $5\%\left(=100\%-95\%\right)$ 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.”