Question

Salaries of 47 college graduates who took a statistics course in college have a mean, x, of $62,500. Assuming a standard deviation, o, of $18,547, construct a 99% confidence interval for estimating the population mean . Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. $ <u<$ (Round to the nearest integer as needed.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = $62500

Population standard deviation =   $\sigma$ = $18547

Sample size = n =47

At 99% confidence level the z is ,

$\alpha$  = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z_$\alpha$/2 = Z_0.005 = 2.576

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

= 2.576* (18547 / $\sqrt$ 47)

= 6969.00

At 99% confidence interval estimate of the population mean is,

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

62500 -6969.00 < $\mu$ < 62500+ 6969.00

55531< $\mu$ < 69469

($55531, $69469)