# If n={13}, (x)=31, and s=17, construct a confidence interval at a 99% confidence level.

If $n=13,\left(\stackrel{―}{x}\right)=31,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}s=17,$ construct a confidence interval at a $99\mathrm{%}$ confidence level. Assume the data came from a normally distributed population.

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Step 1
Given:
Sample size, $n=13$
Sample mean, $\stackrel{―}{x}=31$
Sample standard deviation, $s=17$
The degree of confidence, $c=99\mathrm{%}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}0.99$
Level of significance,
$\alpha =1-c$
$=1-0.99$
$=0.01$
Let the population mean to be estimated using confidence intervals be $\mu$
Step 2
The $99\mathrm{%}$ confidence interval for the population mean is:
$\mu =\stackrel{―}{x}±{t}_{\frac{a}{2},df}\frac{s}{\sqrt{n}}$
$=31±{t}_{\frac{0.01}{2},12}\frac{17}{\sqrt{13}}$
$=31±3.055×\frac{17}{\sqrt{13}}$
$=31±14.4$
$=\left(16.6,45.4\right)$
Thus, the required $99\mathrm{%}$ confidence interval for the population mean is:
$16.6<\mu <45.4$