Question

Based on a survey of 160 persons employed by the state of North Carolina, the mean and standard deviation of their ages are found to be 36 and 10 years, respectively. Determine a 90% confidence interval for the mean age of the state of North Carolina employees. Round your answer to the nearest integer.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 36

Population standard deviation =    = 10
Sample size = n =160

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

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

Margin of error = E = Z/2    * ( /n)

=1.645 * ( 10 /  160 )

= 1.3005
At 90% confidence interval estimate of the population mean
is,

- E < < + E

36- 1.3005 <   < 36 + 1.3005

34.6995 <   <37.3005

( 35,.37 )