Question

A website advertises job openings on its website, but job seekers have to pay to access the list of job openings. The website recently completed a survey to estimate the number of days it takes to find a new job using its service. It took the last 30 customers an average of 60 days to find a job. Assume the population standard deviation is 10 days. Construct a 90% confidence interval of the population mean number of days it takes to find a job.

Answer 1

Solution :

Given that,

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (10 / 30 )

= 3.00

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

- E < < + E

60 - 3.00 < < 60 + 3.00

57.00 < < 63.00

(57.00 , 63.00)