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?

Question

contestada

Respuesta :

nobillionaireNobley nobillionaireNobley · Answer 1 · 2016-11-01T19:29:15+01:00

99% Confidence Interval = 55 + or - 2.58 x 7/sqrt(200) = 55 + or - 2.58 x 0.495 = 55 + or - 1.277 = 53.773 or 56.227

Answer Link

RenatoMattice RenatoMattice · Answer 2 · 2019-04-03T09:30:13+02:00

Answer: 99% confidence interval is given by [tex](53.72,56.27)[/tex]

Step-by-step explanation:

Since we have given that

Number of households in a sample = 200

Mean number of hours spent on social networking sites during the month of January = 55 hours

Standard deviation = 7 hours

Since we have given that the population standard deviation is the same.

and there is 99% confidence interval for the mean hours devoted to social networking in January.

According to the critical value table we get that

α = tail area = 0.05

central area = 1-2α = 0.99 (99% confidence interval)

So, critical value will be [tex]z_{0.05}=2.58[/tex]

As we know that

Margin error = Standard error × critical value

where , Standard error = [tex]\frac{\sigma}{\sqrt{n}}=\frac{7}{\sqrt{200}}=0.495[/tex]

so, Margin error becomes

[tex]0.495\times 2.58\\\\\approx \pm 1.28[/tex]

So, 99% confidence interval is given by

[tex](53.72,56.27)[/tex]