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?

99% confidence interval is given by

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

As we know that

Margin error = Standard error × critical value

where , Standard error =

so, Margin error becomes

So, 99% confidence interval is given by

step-by-step explanation:

b. 1.25 hr

step-by-step explanation:

if you have a constant speed of 60 mph, then you have a ratio between 60 mi. to 1 hr. dividing both sides by 60 we get the ratio

1 mi. to

then mult. both sides by 75 we get the ratio

75 mi. to

simplifying we get

75 mi. to

or 1.25 hr.