Question

If, for a sample in which the subjects are randomly chosen, the mean income is $45,000, the sample size is 1600, the standard deviation is $4,000, and the standard error of the mean is $______, what, approximately, is the 95% confidence interval for the population mean? (show all work)

(a) $41,000 to $49,000.

(b) $37,000 to $53,000.

(c) $44,800 to $45,200.

(d) It is impossible to know, because we do not know if the incomes are normally distributed.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $45000

Population standard deviation =    = $4000

Sample size = n =1600

standard error ( /n) =(4000 /  1600 ) =100

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2 * ( /n)

= 1.96 * (4000 /  1600 )

E= 196
At 95% confidence interval estimate of the population mean
is,

- E < < + E

45000 - 196 <   < 45000+ 196

44804 <   < 45196

answer is closes to

$44800 to $45200