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

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b