Question

A random sample of 50 respondents is selected, and each respondent is instructed to keep a detailed record of time spent engaged viewing content across all screens (Internet, video on a computer video on mobile phone) in a particular week. The sample mean content viewing time per week is 42 hours with a standard deviation of 3.2 hours.

Construct a 99% confidence interval estimate for the mean content viewing time per week in this area. What is the margin of error?

Answers should be accurate up to 4 decimal places.

Lower limit:

Upper limit:

Margin of error:

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 42

Population standard deviation =    = 3.2

Sample size = n =50

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

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

=2.576 * (3.2 / 50)

E= 1.1658

At 99% confidence interval estimate of the population mean is,

- E < < + E

42 -1.1658 < < 42+ 1.1658

40.8342< < 43.1658

(40.8342, 43.1658 )

Lower limit:40.8342

Upper limit: 43.1658

Margin of error=1.1658