# You are given the sample mean and standard deviation of the population. Use this information to construct the displaystyle{90}%{quadtext{and}quad} {95

You are given the sample mean and standard deviation of the population. Use this information to construct the​ $90\mathrm{%}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}95\mathrm{%}$ confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.
From a random sample of 66 ​dates, the mean record high daily temperature in a certain city has a mean of ${85.69}^{\circ }F$. Assume the population standard deviation is ${13.60}^{\circ }F.$
The​ $90\mathrm{%}$ confidence interval is
The​ $95\mathrm{%}$ confidence interval is
Which interval is​ wider?
Interpret the results.
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Derrick
Step 1
Given:
This mean recorded temperature is: $\stackrel{―}{x}=85.69{F}^{\circ }$
The population's standard deviation is: $\sigma =13.60{F}^{\circ }$
The number of random samples is: $n=66$ dates.
Confidence level $-90\mathrm{%}$:
The expression for the confidence level is,
$1-2\alpha =0.9$
$\alpha =\frac{1-0.9}{2}$
$=0.05$
Step 2
From the graph of the normal distribution, the confidence level is symmetric about the P axis. The expression to calculate the corresponding value of P is,
$P=1-\alpha$
$=1-0.05$
$=0.95$
From the z-table corresponding to $P=0.95$, the value of z is 1.65
The expression to calculate the interval is,
$c.I=\stackrel{―}{x}±z\frac{\sigma }{\sqrt{n}}$
Substitute the values in the above expression.
$c.l=85.69±1.65×\frac{13.60}{\sqrt{66}}$
$=85.69±2.762$
$=\left(85.69-2.762,85.69-2.762\right)$
$=\left(82.928,88.452\right)$
Therefore, the interval for confidence leven $90\mathrm{%}$ is from 82.928 to 88.452
Step 3
Confidence level $-95\mathrm{%}:$
The expression for the confidence level is,
$1-2\alpha =0.95$
$\alpha =\frac{1-0.95}{2}$
$=0.025$
From the graph of the normal distribution, the confidence level is symmetric about the P axis. The expression to calculate the corresponding value of P is,
$P=1-\alpha$
$=1-0.025$
$=0.975$
From the z-table corresponding to $P=0.975$, the value of z is 1.96.
The expression to calculate the interval is,
$c.l=\stackrel{―}{x}±z\frac{\sigma }{\sqrt{n}}$
Substitute the values in the above expression.
$c.l-85.69±1.96×\frac{13.60}{\sqrt{66}}$
$=85.69±3.281$
$=\left(85.69-3.281,85.69-3.281\right)$
$=\left(82.409,88.971\right)$
Hence, the interval for confidence level $95\mathrm{%}$ is from 82.409 to 88.971, and the interval for confidence level $95\mathrm{%}$ is the widest one.