# 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

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 visits for 14 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.
$9610151962326191611251611$
a. To compute the confidence interval use a t or z distribution.
b. With 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.
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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 $100\left(1–\alpha \right)\mathrm{%}$ 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:
$\left(\stackrel{―}{x}-{t}_{\alpha /2,n-1}\frac{s}{\sqrt{n}},\stackrel{―}{x}+{t}_{\alpha /2,n-1\frac{s}{\sqrt{n}}}\right)$
such that
$\left(\stackrel{―}{x}-{t}_{\alpha /2,n-1}\frac{s}{\sqrt{n}},\stackrel{―}{x}+{t}_{\alpha /2,n-1\frac{s}{\sqrt{n}}}\right)=1-\alpha$
Here,
$E={t}_{\alpha /2,n-1}\frac{S}{\sqrt{n}}$ is the margin of error.
Here, n is the sample size, $\stackrel{―}{x}$ 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,
$100\left(\alpha /2\right)\mathrm{%}or\alpha /2$ proportion of the observations lie, and below which,
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:
$x9,6,10,15,19,6,23,26,19,16,11,25,16,11,\sum mx=212$
${x}^{2}81,36,100,225,361,36,529,676,361,256,121,625,256,121,\sum m{x}^{2}=3,784$
$\stackrel{―}{x}=\frac{1}{n}\sum {m}_{i=1}^{n}{x}_{i}$
$=\frac{212}{14}$
$\approx 15.143.$
$S=\sqrt{\frac{1}{n-1}\left(\sum {m}_{i=1}^{n}{x}_{i}^{2}-n{x}^{-2}\right)}$
$=\sqrt{\frac{1}{14-1}\left[3,784-\left(14\right)\cdot \left(15.143{\right)}^{2}\right]}$
$\approx 6.643$
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,
${t}_{\frac{\alpha }{2},n-1}={t}_{0.025},13$
such that:
$P\left({t}_{n-1}\le {t}_{\frac{\alpha }{2},n-1}\right)=P\left({t}_{13}\le {t}_{0.05},11\right)$
$=0.975$
$=P\left({t}_{13}\le 2.16\right)$
[Using the t-distribution table, or the Exel formula: $=T.INV\left(0.975,13\right),P\left({t}_{13}\le 2.16\right)=0.975\right]$
$\ge {t}_{0.025,13}=2.16$
First, the margin of error is calculated below:
$\left(\stackrel{―}{x}-{t}_{\frac{\alpha }{2},n-1}\frac{S}{\sqrt{n}},\stackrel{―}{x}+{t}_{\frac{\alpha }{2},n-1\frac{S}{\sqrt{n}}}=\left(\stackrel{―}{x}-E,\stackrel{―}{x}+E\right)\approx \left(15.143-3.835,15.143+3.835\right)$