Question

An Internet user wants to know the average download speed for his Internet connection. Suppose we know that the standard deviation is 7 mbps. He runs a speed test everyday at noon for 30 days. The average speed of the tests was 13.68 mbps. Construct a 95% confidence interval for the average speed of this Internet connection at noon.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 13.68

Population standard deviation =    = 7

Sample size n =30

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2 * ( /n)

=1.96 * ( 7 /  30 )

E= 2.5049
At 95% confidence interval estimate of the population mean
is,

- E < < + E

13.68 - 2.5049 <   < 13.68 + 2.5049

11.1751 <   < 16.1849

( 11.1751,16.1849, )

At 95% confidence interval estimate of the population mean
is,( 11.1751,16.1849, )