Question

The national average time spent using the internet at public libraries is 6 hours

per week with a standard deviation of 2 hours. In a local public library a random
sample of 64 people revealed an average time spent using the internet to be 6
1/2 hours. Is there reason to believe that the local average time on the internet is
different from the national average? Conduct a hypothesis test at the .05 level of
significance.

Answer 1

We reject H0 and conclude that local average time on the internet is

different from the national average

Explanation:

H0 : μ = 6

H0 : μ ≠ 6

Test statistic :

(xbar - μ) ÷ (s/√(n))

(6.5 - 6) / (2 / √64)

0.5 / (2/8)

0.5 / 0.25

Test statistic = 0.5 / 0.25

Test statistic = 2

We obtain the Pvalue :

Using the Pvalue from test statistic (Z) score calculator :

Pvalue = 0.0455

α = 0.05

Since Pvalue < α ; We reject H0 and conclude that local average time on the internet is

different from the national average