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.

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

different from the national average

Step-by-step 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

the answer for this question is 38

to plot points   for x just start at the origin (middle) and go to the left however many times it say(if the number is a positive) if the number is a negative then go right however many times it says if it is 0 go nowhere

now keep you pencil there and go up however many times it says on y(if it is positive) if it is negative then go down however many times it says. if it says 0 go nowhere.

you have just plotted a point! now when you finish plotting all of them connect them with a ruler

plot each point as (-3, 7), (0,4), (3,1), (8,-4)   in the parenthesis on the left of the comma is x and the right is y.