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.

[56.9966, 63.0034]

[56.4216, 63.5784]

[40.4000, 79.6000]

[55.9085, 64.0915]

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 60

Population standard deviation =    = 10
Sample size = n =30

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/  30 )

= 3.0034
At 90% confidence interval estimate of the population mean
is,

- E < < + E

60  - 3.0034<   < 60  + 3.0034

56.9966<   < 63.0034

( 56.9966, 63.0034 )