Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. Suppose that the mean income is found to be $24.3$⁢24.3 for a random sample of 758758 people. Assume the population standard deviation is known to be $9.2$⁢9.2. Construct the 90%90% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = $ 24.3

Population standard deviation =   $\sigma$ = $ 9.2

Sample size = n = 758

At 90% confidence level

$\alpha$ = 1 - 90%  

$\alpha$ = 1 - 0.90 =0.10

$\alpha$/2 = 0.05

Z$\alpha$/2 = Z0.05  = 1.645

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

= 1.645 * ( 9.2 /  $\sqrt$758 )

= 0.5

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

$\bar x$  ± E

24.3 ± 0.5

( 23.8, 24.8 )