Question

4)

Assume that IQ scores of college graduates are normally distributed. A researcher collects a random sample of 25 college graduates and tests their IQ. The sample had a mean IQ of 108 with a standard deviation of 15.6.

a) Find a 95% confidence interval for the true mean IQ for college graduates.

b) Provide the margin of error of the interval as your answer.

Round your answer to 1 decimal place.

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 108

sample standard deviation = s = 15.6

sample size = n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,24 = 2.064

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.064 * (15.6 / $\sqrt$ 25)

= 6.4

The 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

108 - 6.4 < $\mu$ < 108 + 6.4

101.6 < $\mu$ < 114.4

(101.6 , 114.4)