Question

Determine the 90% confidence interval estimate for the population mean of a normal distribution given n=64, 6 = 112, and x= 1,500. The 90% confidence interval for the population mean is from to (Round to two decimal places as needed. Use ascending order.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 1500

T

Population standard deviation =   $\sigma$ = 112
Sample size = n =64

At 90% confidence level the z is

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.1

$\alpha$ / 2 = 0.1 / 2 = 0.05

Z$\alpha$/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z$\alpha$/2    * ( $\sigma$ /$\sqrt$n)

= 1.645 * ( 112 /  $\sqrt$64 )

= 23.03
At 90% confidence interval estimate of the population mean
is,

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

1500- 23.03 < $\mu$    < 1500 + 23.03

1476.97 <  $\mu$ < 1523.03

1476.97 to 1523.03