# 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

foass77W 2020-12-30 Answered
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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## Expert Answer

i1ziZ
Answered 2020-12-31 Author has 92 answers

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:
$\left(\stackrel{―}{x}-{t}_{\frac{\alpha }{2},n-1}\frac{S}{\sqrt{n}},\stackrel{―}{x}+{t}_{\frac{\alpha }{2},n-1\frac{S}{\sqrt{n}}}=\left(\stackrel{―}{x}-E,\stackrel{―}{x}+E\right)\approx \left(15.143-3.835,15.143+3.835\right)$

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