Question

The average starting salaries for 55 graduates with a certain major was $43949. The standard deviation for all graduates with that major was $8937. Give the 94% confidence interval for the population mean. Round the endpoints to the nearest dollar, and do not put commas in your numbers.

Confidence interval: ( , )

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 43949

sample standard deviation = s = 8937

sample size = n = 55

Degrees of freedom = df = n - 1 = 54

At 94% confidence level the z is ,

$\alpha$ = 1 - 94% = 1 - 0.94= 0.06

$\alpha$ / 2 = 0.06 / 2 = 0.03

t_{$\alpha$ /2,df} = t_0.03,54 = 1.921

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

= 1.921 * ( 8937/ $\sqrt$ 55)

= 2314.930

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

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

43949 - 2314.930 < $\mu$ < 43949 + 2314.930

41634.07 < $\mu$ < 46263.93

(41634 , 46264)