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

The point estimate for the population mean μ is the sample mean, x ¯ . In this case, to estimate the mean number of weekly hours of home-computer use among the population of U.S. adults, we used the sample mean obtained from the sample, therefore x ¯ = 8.5.

Explanation for b)

The 95% confidence interval for the mean, μ, is x ¯ ± 2 ⋅ σ n = 8.5 ± 2 ⋅ 3.6 81 = 8.5 ± . 8 = ( 7.7 , 9.3 ) .

Explanation for c)

In general, a more concise (narrower) confidence interval can be achieved in one of two ways: sacrificing on the level of confidence (i.e. selecting a lower level of confidence) or increasing the sample size

Explanation for d)

We would like our confidence interval to be a 95% confidence interval (implying that z* = 2) and the confidence interval length should be 1.2, therefore the margin of error (m) = 1.2 / 2 = .6. The sample size we need in order to obtain this is: 144.