Question

A simple random sample of size n = 20 is drawn from a population that is normally distributed. The sample mean is found to be x = 66 and the sample standard deviation is found to be s = 10. Construct a 90​% confidence interval about the population mean.

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 66

sample standard deviation = s = 10

sample size = n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 90% confidence level

$\alpha$ = 1 - 90%

$\alpha$ =1 - 0.90 =0.10

$\alpha$/2 = 0.05

t$\alpha$/2,df = t_0.05,19 = 1.729

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

= 1.729 * (10 / $\sqrt$ 20)

Margin of error = E =3.87

The 90% confidence interval estimate of the population mean is,

$\bar x$  ± E  

= 66  ± 3.87

= ( 62.13, 69.87 )