# A sample of 60 students from a large university is taken. The average age in the sample was 22...

## Question:

A sample of 60 students from a large university is taken. The average age in the sample was 22 years with a standard deviation of 6 years.

Construct a 80% confidence interval for the average age of the population.

Show the value of lower bound in your obtained confidence interval with three decimal places.

## Confidence Interval

There are two types of estimation of a population parameter, one is point estimation and the other one is interval estimation. The confidence interval is the method of interval estimation to estimate population parameters. The advantage of interval estimation is that it gives a probability of how likely the population parameter lies in the confidence interval.

Given Information:

The total number of students from a university is, {eq}n = 60{/eq},

The average age of the sample is, {eq}\bar X = 22{/eq},

The standard deviation is, {eq}s = 6{/eq},

The formula for confidence interval is given as,

{eq}\bar x \pm {t_{\alpha /2}} \cdot \dfrac{s}{{\sqrt n }}{/eq}

Here,

{eq}\bar x{/eq} is sample mean.

{eq}{t_{\alpha /2}}{/eq} is the critical value at level of significance {eq}\alpha{/eq} and degree of freedom {eq}n ? 1{/eq}.

The sample size is small, and the population variance is not known, using t-distribution,

The lower limit of 80% confidence interval is given as,

{eq}\begin{align*} \bar x - {t_{0.1,59}} \cdot \dfrac{s}{{\sqrt n }} &= 22 - \left( {1.296} \right) \cdot \dfrac{6}{{\sqrt {59} }}\\ & = 20.987 \end{align*}{/eq}

The upper limit of 80% confidence interval is given as,

{eq}\begin{align*} \bar x - {t_{0.1,59}} \cdot \dfrac{s}{{\sqrt n }} &= 22 - \left( {1.296} \right) \cdot \dfrac{6}{{\sqrt {59} }}\\ & = 23.012 \end{align*}{/eq}

Therefore, the 80% confidence interval is {eq}\left( {20.987,23.012} \right){/eq} and the lower bound of the confidence interval is 20.987.