In a simple random sample of

180

households, the sample mean number of personal computers was2.3

.Assume the population standard deviation is=

0.38

.

Part:0 / 4

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Part 1 of 4

(a) Construct a

99

%

confidence interval for the mean number of personal computers. Round the answer to at least two decimal places.A99

%

confidence interval for the mean number of personal computers is<

<

.