Question

In a sample of 100 households, the mean number of hours spent on social networking sites during the month of January was 50 hours. In a much larger study, the standard deviation was determined to be 6 hours. Assume the population standard deviation is the same. What is the 98% confidence interval for the mean hours devoted to social networking in January?

 A.The 98% confidence interval ranges from 6 to 50 hours.

  B.The 98% confidence interval ranges from 48.60 to 51.40 hours.  
C.The 98% confidence interval ranges from 49.40 to 50.60 hours.

 
D.The 98% confidence interval ranges from 50 to 52 hours.

Answer 1

The 98% confidence interval ranges from 48.60 to 51.40 hours.
Please research the included formula below, as it is crucial to understanding this topic.

better late than never :P

Answer 2

The correct option is B

Explanation:

The formula for confidence interval for the mean is

`Interval=\mu\pm z** (\sigma)/(√(n))`

Where, μ is population mean, σ is standard deviation, n is sample size and z* is z-score at given confidence interval.

From the z-table the value of z-score at 98% confidence interval is 2.33.

From the given information it is clear that

`\mu=50`

`\sigma=6`

`n=100`

The 98% confidence interval for the mean hours devoted to social networking in January is

`Interval=50\pm 2.33* (6)/(√(100))`

`Interval=50\pm 2.33* 0.6`

`Interval=50\pm 1.398`

`Interval=[50-1.398,50+1.398]`

`Interval=[48.602,51.398]`

`Interval=[48.60,51.40]`

Therefore the 98% confidence interval ranges from 48.60 to 51.40 hours. Option B is correct.