Question

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

Answer 1

20.96-22.0378

Explanation:

For a 95% confidence interval the z value is 1.96

The confidence interval will be given by

Mean±z×б/√n

21.5-1.96×2.5/√83 =20.96

21.5+1.96×2.5/√83 =22.0378

With 95% confidence the average hours spent by employees on social media per hour is between 20.96 to 22.0378 based on this sample data

Answer 2

95% of the population spent between 16.8 hours to 26.2 hours on social media per month

Explanation:

1. subtracting 1 from sample size: 83-1= 82
2. Note Mean=21.5 and standard deviation=2.5
3. Subtracting confidence interval from 1 and dividing by 2: (1-0.95)/2=0.025
4. At α=0.025 and dF=82, the t-distribution table gives 1.984
5. Divide standard deviation by square root of sample size: 21.5/√83= 2.36
6. Multiplying answer in step 4 by answer in step 5: 2.36×1.984= 4.68
7. For lower end: 21.5-4.68= 16.82
8. for upper end: 21.5+4.68=26.18