Question

Algebra 2

In a sample of 200 households, the mean number of hours spent on social networking sites during the month of January was 55 hours. In a much larger study, the standard deviation was determined to be 7 hours. Assume the population standard deviation is the same. What is the 99% confidence interval for the mean hours devoted to social networking in January?

A) The 99% confidence interval ranges from 53.73 to 56.27 hours.
B) The 99% confidence interval ranges from 7 to 55 hours.
C) The 99% confidence interval ranges from 54.51 to 55.49 hours.
D) The 99% confidence interval ranges from 55 to 60 hours.

Can i get an explanation?
All i know is that this question has something to do with the Margin of error.
The Margin of error has two different formulas
Please help :)

Answer 1

MOE = +\- 1.27
Confidence Interval is 53.73 to 56.27

Answer 2

Explanation:

n=sample size =200

Mean = x bar = 55 hours

Std deviation (Population) = 7 hours

For 99% confidence level

Z critical value = 2.58

Margin of error = ±2.58(std error)

Std error =
`(7)/(√(n) ) =0.495`

Hence margin of error = ±0.495(258)=1.28

Confidence interval = (55-1.28, 55+1.28) = (53.72, 56.28)

~(53.73, 56.27)

Hence option A