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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 :)

Respuesta :

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

Answer:

Step-by-step 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 = [tex]\frac{7}{\sqrt{n} } =0.495[/tex]

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

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