Question

A recent survey of 9 social networking sites has a mean of 12.51 million visitors for a specific month. The standard deviation was 4.8 million. Find the 98% confidence interval of the true mean. Assume the variable is normally distributed. Round your answers to at least two decimal places.

Answer 1

Solution :

Given that,

t /2,df = 2.896

Margin of error = E = t/2,df * (s /n)

= 2.896 * (4.8 / 9)

Margin of error = E = 4.63

The 98% confidence interval estimate of the population mean is,

- E < < + E

12.51 - 4.63 < < 12.51 + 4.63

7.88 < < 17.14

(7.88 , 17.14)