Question

In order to estimate the average electric usage per month, a sample of 47 houses were selected and the electric usage determined. The sample mean is 2,000 KWH. Assume a population standard deviation of 142 kilowatt hours. At 90% confidence, compute the margin of error?

In order to estimate the average electric usage per month, a sample of 46 houses were selected and the electric usage determined. The sample mean is 2,000 KWH. Assume a population standard deviation of 127 kilowatt hours. At 99% confidence, compute the upper bound of the interval estimate for the population mean.

Consider the population of electric usage per month for houses. The standard deviation of this population is 118 kilowatt-hours. What is the smallest sample size to provide a 90% confidence interval for the population mean with a margin of error of 22 or less? (Enter an integer number.)

Answer 1

1 ) Given ; n=47 , , ,

Now , ; From standard normal dsitribution table

Therefore , the margin of error is ,

2 ) Given : n=46 , , ,

Now , ; From standard normal dsitribution table

Therefore , the upper bound of the 99% confidence interval estimate for the population mean is ,

Upper bound=

3) Given : Margin of error=E=22 , ,

Now , ; From standard normal dsitribution table

Therefore , the required sample size is ,