In a simple random sample of130 households, the sample mean number of personal computers was2.87.Assume the population standard deviation is= 0.45.

(a) Construct a99.5%

confidence interval for the mean number of personal computers. Round the answer

to at least two decimal places.

A99.5%confidence

interval for the mean number of personal computers is

?<

b) If the sample size were 160 rather than 130, would the margin of error be larger or smaller than the result in part (a)? Explain. The margin of error would be (larger or smaller) , since (a decrease or an increase) in the sample size will (increase or decrease) the standard error.

(C) If the confidence levels were 80% rather than 99.5%, would the margin of error be larger or smaller than the result inpart (a)?Explain.

The margin of error would be ( larger or smaller), since(an increase or decrease)in the confidence level will (increase or decrease) the critical value

z/2

.

(d) Based on the confidence interval constructed in part (a), is it likely that the mean number of personal computers is less than 2%

It ( is or is not) likely that the mean number of personal computers is less than 2%