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

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. 
b. The $100(1–\alpha )\mathrm{\%}$ 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: 
$(\bar{x}-t_{\alpha /2,n-1}\frac{s}{\sqrt{n}},\bar{x}+t_{\alpha /2,n-1\frac{s}{\sqrt{n}}})$ 
such that 
$(\bar{x}-t_{\alpha /2,n-1}\frac{s}{\sqrt{n}},\bar{x}+t_{\alpha /2,n-1\frac{s}{\sqrt{n}}})=1-\alpha$ 
Here, 
$E=t_{\alpha /2,n-1}\frac{S}{\sqrt{n}}$ is the margin of error. 
Here, n is the sample size, $\bar{x}$ is the sample mean, 
and $t_{\alpha /2,n-1}$ 
is the critical value of the t-distribution with $(n–1)$, above which, 
$100(\alpha /2)\mathrm{\%}or\alpha /2$ proportion of the observations lie, and below which, 
$100(1-\alpha +\alpha /2)\mathrm{\%}=100(1-\alpha /2)\mathrm{\%}or(1-\alpha /2)$ 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: 
$x9,6,10,15,19,6,23,26,19,16,11,25,16,11,\sum mx=212$ 
$x^{2}81,36,100,225,361,36,529,676,361,256,121,625,256,121,\sum m{x}^2=3,784$ 
$\bar{x}=\frac{1}{n}\sum m_{i=1}^n{x}_i$ 
$=\frac{212}{14}$ 
$\approx 15.143.$ 
$S=\sqrt{\frac{1}{n-1}(\sum m_{i=1}^n{x}_i^2-n{x}^{-2})}$ 
$=\sqrt{\frac{1}{14-1}[3,784-(14)\cdot (15.143{)}^2]}$ 
$\approx 6.643$ 
The desired confidence level is 95%. Thus, 
$100(1-\alpha )=95$ 
$1-\alpha =0.95$ 
$\alpha =0.05$ 
$\frac{\alpha }{2}=0.025$ 
$1-\frac{\alpha }{2}=0.975$ 
This implies that, 
$t_{\frac{\alpha }{2},n-1}=t_{0.025},13$ 
such that: 
$P(t_{n-1}\le t_{\frac{\alpha }{2},n-1})=P(t_{13}\le t_{0.05},11)$ 
$=0.975$ 
$=P(t_{13}\le 2.16)$ 
[Using the t-distribution table, or the Exel formula: $=T.INV(0.975,13),P(t_{13}\le 2.16)=0.975]$ 
$\ge t_{0.025,13}=2.16$ 
First, the margin of error is calculated below: 
$(\bar{x}-t_{\frac{\alpha }{2},n-1}\frac{S}{\sqrt{n}},\bar{x}+t_{\frac{\alpha }{2},n-1\frac{S}{\sqrt{n}}}=(\bar{x}-E,\bar{x}+E)\approx (15.143-3.835,15.143+3.835)$