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.
Answer:
The correct option is B
Step-by-step explanation:
The formula for confidence interval for the mean is
[tex]Interval=\mu\pm z*\times \frac{\sigma}{\sqrt{n}}[/tex]
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
[tex]\mu=50[/tex]
[tex]\sigma=6[/tex]
[tex]n=100[/tex]
The 98% confidence interval for the mean hours devoted to social networking in January is
[tex]Interval=50\pm 2.33\times \frac{6}{\sqrt{100}}[/tex]
[tex]Interval=50\pm 2.33\times 0.6[/tex]
[tex]Interval=50\pm 1.398[/tex]
[tex]Interval=[50-1.398,50+1.398][/tex]
[tex]Interval=[48.602,51.398][/tex]
[tex]Interval=[48.60,51.40][/tex]
Therefore the 98% confidence interval ranges from 48.60 to 51.40 hours. Option B is correct.
