# In a random sample of 50 students at college, the average IQ was bar x = 115. Assume that the IQ...

## Question:

In a random sample of 50 students at college, the average IQ was {eq}\bar x = 115 {/eq}. Assume that the IQ of a student in college follows a normal distribution with unknown mean and standard deviation = {eq}15 {/eq}.

a. Give a {eq}95\% {/eq} confidence interval for {eq}\mu {/eq}.

b. Give a {eq}98\% {/eq} confidence interval for {eq}\mu {/eq}.

c. Which confidence interval has a larger margin of error?

## Confidence interval

The confidence interval is the interval estimation for the population parameters which provide the lower limit and upper limits. The true value of population parameters lies between the interval at the specified level of confidence level.

Given Information

Sample size; 50

Sample mean; 11.5

Population standard deviation; 15

a)

The 95% confidence interval for \mu is calculated as follows;

{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( {115 - 1.96\dfrac{{15}}{{\sqrt {50} }} < \mu < 115 - 1.96\dfrac{{15}}{{\sqrt {50} }}} \right) &= 0.95\\ P\left( {110.84 < \mu < 119.16} \right) &= 0.95 \end{align*}{/eq}

Therefore, the required confidence interval is (110.84, 119.16)

b)

The calculation for 98% confidence interval is given 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) &= 0.98\\ P\left( {115 - 2.33\dfrac{{15}}{{\sqrt {50} }} < \mu < 115 + 2.33\dfrac{{15}}{{\sqrt {50} }}} \right) &= 0.98\\ P\left( {110.06 < \mu < 119.93} \right) &= 0.98 \end{align*}{/eq}

Therefore, the required confidence interval is (110.06, 119.93).

c)

The length of the confidence interval is larger when the confidence level is 98%.