# A researcher wants to find the average IQ of Missourians. She randomly samples 100 people who...

## Question:

A researcher wants to find the average IQ of Missourians. She randomly samples 100 people who live in the state, and obtains the following results: Mean = 106.5 SD = 12.0

Calculate the 95% confidence interval for the mean IQ of Missourians.

Explain what the 95% confidence interval obtained really means.

## Confidence Interval

The two-sided confidence interval for the estimation of the population mean is calculated by solving the lower limit and upper limit. The margin of error for the population mean is calculated by the product of critical value and standard error (standard deviation divided by the square root of sample size).

Given information

Sample size: 100

Sample standard deviation: 12

Sample mean: 106.5

The 95% confidence interval for the mean IQ of Missourians is calculated as follows.

{eq}\begin{align*} P\left( {\bar X - {t_{\alpha /2}}\dfrac{s}{{\sqrt n }} < \mu < \bar X + {t_{\alpha /2}}\dfrac{s}{{\sqrt n }}} \right) &= 0.95\\ P\left( {106.5 - 1.984*\dfrac{{12}}{{\sqrt {100} }} < \mu < 106.5 - 1.984*\dfrac{{12}}{{\sqrt {100} }}} \right) &= 0.95\\ P\left( {104.12 < \mu < 108.88} \right) &= 0.95 \end{align*} {/eq}

Explanation: The 95% confidence interval states that there is a 95% chance the true value of population within the interval (104.12, 108.88). 