Question

A multinational firm wants to estimate the average number of hours in a month that their employees spend on social media while on the job. A random sample of 83 employees showed that they spent an average of 21.5 hours per month on social media, with a standard deviation of 2.5. Construct and interpret a 95% confidence interval for the population mean hours spent on social media per month.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 21.5

sample standard deviation = s = 2.5

sample size = n = 83

Degrees of freedom = df = n - 1 = 82

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,82 = 1.989

Margin of error = E = t_/2,df * (s /n)

= 1.989 * (2.5 / 83)

= 0.546

The 95% confidence interval estimate of the population mean is,

- E < < + E

21.5 - 0.546 < < 21.5 + 0.546

20.954 < < 22.046

A 95% confidence interval for the population mean hours spent on social media per month is (20.954 , 22.046)