MATH SOLVE

4 months ago

Q:
# A study was conducted in order to estimate ?, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be ? = 3.6 hours.Based on this information, what would be the point estimate for ??(a) 81(b) 8.5(c) 3.6(d) None of the above.We are 95% confident that the mean number of weekly hours that U.S. adults use computers at home is:(a) between 8.1 and 8.9.(b) between 7.8 and 9.2.(c) between 7.7 and 9.3.(d) between 7.5 and 9.5.(e) between 7.3 and 9.7.Which of the following will provide a more informative (i.e., narrower) confidence interval than the one in problem 3?(a) Using a sample of size 400 (instead of 81).(b) Using a sample of size 36 (instead of 81).(c) Using a different sample of size 81.(d) Using a 90% level of confidence (instead of 95%).(e) Using a 99% level of confidence (instead of 95%).(f) Both (a) and (d) are correct.(g) Both (a) and (e) are correct.How large a sample of U.S. adults is needed in order to estimate ? with a 95% confidence interval of length 1.2 hours?(a) 6(b) 12(c) 20(d) 36(e) 144

Accepted Solution

A:

Answer:d. the current study does not provide significant evidence that the mean number of weekly hours has changed over the past year, since 8 falls inside the confidence interval.Step-by-step explanation:Mean was of 8 hours, test if it has changed:At the null hypothesis, we test if it has not changed, that is, the mean is still of 8, so:[tex]H_0: \mu = 8[/tex]At the alternative hypothesis, we test if it has changed, that is, the mean is different of 8, so:[tex]H_1: \mu \neq 8[/tex]Using a 95% confidence interval of (7.7, 9.3), our conclusion is that:8 is part of the confidence interval, which means that the study does not provide evidence that the mean has changed, and the correct answer is given by option d.