# Use this information to construct the 90% and 95% confidence intervals for the population mean....

## Question:

Use this information to construct the {eq}90\% {/eq} and {eq}95\% {/eq} confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.

From a random sample of {eq}61 {/eq} dates, the mean record high daily temperature in a certain city has a mean of {eq}82.37 ^\circ F {/eq}. Assume the population standard deviation is {eq}13.86 ^\circ F {/eq}.

## Confidence interval

The confidence interval includes lower and upper limits which are obtained from sample statistics. If the sample size is greater than 30 and the population standard deviation is known than confidence interval is calculated from a normal distribution.

## Answer and Explanation:

**Given information**

Sample size: 61

Sample mean: 82.37

Population standard deviation: 13.86

The 90% confidence interval for the population mean is calculated as follow.

{eq}\begin{align*} P\left( {\bar X - {Z_{\alpha /2}}\dfrac{\sigma }{{\sqrt n }} < \mu < \bar X + {Z_{\alpha /2}}\dfrac{\sigma }{{\sqrt n }}} \right) &= 0.90\\ P\left( {82.37 - 1.645\dfrac{{13.86}}{{\sqrt {61} }} < \mu < 82.37 + 1.645\dfrac{{13.86}}{{\sqrt {61} }}} \right) &= 0.90\\ P\left( {79.45 < \mu < 85.29} \right)& = 0.90 \end{align*}{/eq}

There are 90% chance that the parameter of population mean is lies between **(79.45 to 85.29)**

The 95% confidence interval for the population mean is calculated as follow.

{eq}\begin{align*} P\left( {\bar X - {Z_{\alpha /2}}\dfrac{\sigma }{{\sqrt n }} < \mu < \bar X + {Z_{\alpha /2}}\dfrac{\sigma }{{\sqrt n }}} \right) &= 0.95\\ P\left( {82.37 - 1.96\dfrac{{13.86}}{{\sqrt {61} }} < \mu < 82.37 + 1.96\dfrac{{13.86}}{{\sqrt {61} }}} \right)& = 0.95\\ P\left( {78.89 < \mu < 85.84} \right)& = 0.95 \end{align*}{/eq}

There are 95% chance that the parameter of population mean is lies between **(78.89 to 85.84)**

We know that if level of confidence is increased than the width of confidence interval becomes larger.

Therefore, the width of the 95% confidence interval is larger than the 90% confidence interval.

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from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 3