Given : The purpose of the study was to determine whether the mean weekly number of hours that Internet users in that age group spend online differs from the mean for Internet users in general which is 12.50 hours.

The set of hypothesis for the given claim :-

$H_0:\mu=12.50\\\\ H_a:\mu\neq12.50$

A 95% confidence interval for the mean weekly number of hours that Internet users in that age group spend online was calculated to be (11.439 , 12.577).

Since 12.50 contained in the 95% confidence interval (11.439 , 12.577).

Therefore, we support the null hypothesis.

So the final conclusion will be, we have evidence to support the claim that Internet users in that age group spend online differs from the mean for Internet users in general which is 12.50 hours.

7 0

3 years ago

What are the zeros of the function below? Check all that apply

vovangra [49]

Answer:

-7. is the answer so tick it mark me brainlist

3 0

3 years ago

Jason hires groundskeepers. They work an average of 60 hours per month. If seven groundskeepers are hired at $12/hour, what is t

bekas [8.4K]

$5040 because 12*60 = 720 then multiply it by the 7 workers so 5040

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