# A sample of the ages of 100 males was taken and the results were as follows. The mean of the...

## Question:

A sample of the ages of 100 males was taken and the results were as follows. The mean of the sample was {eq}\bar x = 45 {/eq} years and the standard deviation of the population was known to be {eq}\sigma = 5 {/eq}.

(a) Find a {eq}95\% {/eq} confidence interval for the true mean of the population.

(b) What is the margin of error in the above situation?

## Confidence interval

The confidence interval is a type of interval which is described as the population parameter. A confidence interval contains a lower and upper limit which is determined the width of the interval and the most commonly used confidence level is 90%,95% and 99%.

Given information

{eq}\begin{align*} {\rm{sample \ size}}\left( n \right) &= 100;\\ {\rm{sample \ mean}}\left( {\bar X} \right) &= 45;\\ {\rm{population \ standard \ deviation}}\left( \sigma \right) &= 5; \end{align*} {/eq}

a)

The calculation for 95% confidence interval for the true mean of the population is calculated as;

{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( {45 - 1.645\dfrac{5}{{\sqrt {100} }} < \mu < 45 + 1.645\dfrac{5}{{\sqrt {100} }}} \right) &= 0.95\\ P\left( {44.1775 < \mu < 45.8225} \right) &= 0.95 \end{align*}{/eq}

Therefore, the required confidence interval is (44.1775 to 45.8225).

b)

The margin of error is given as;

{eq}\begin{align*} M.O.E &= {Z_{\alpha /2}}*\dfrac{\sigma }{{\sqrt n }}\\ &= 1.645*\dfrac{5}{{\sqrt {100} }}\\ &= 0.8225 \end{align*}{/eq}

The required margin of error is 0.8225. 