Question

The 2000 census "long form" asked the total 1999 income of the householder, the person in whose name the dwelling unit was owned or rented. This census form was sent to a random sample of 17% of the nation's households. Suppose that the households that returned the long form are an SRS of the population of all households in each district. In Middletown, a city of 40,000 persons, 2621 householders reported their income. The mean of the responses was x = $33,453, and the standard deviation was s = $8721. The sample standard deviation for so large a sample will be very close to the population sta

Answer 1

ndard deviation s. Use these facts to give an approximate 99% confidence interval for the 1999 mean income of Middletown householders who reported income.

Answer 2

Could any one explain or give me an answer on how to solve this question or how to come to the answer?

Answer 3

you want to look at the normal distribution table and find where the confidence interval is .99. It's around 2.3 I believe. Then you want to use the equation: sample proportion ± zscore x sqrt(sample proportion * (1-sample proportion)) / full population) using Middletown's numbers. So the sample proportion will be 2621/40000. I'm not sure how to take into account the 17% of nation's households given at the beginning. If they're only talking about the 17% who reported income then that equation should give the confidence interval which would be ( ( sample proportion - zscore x sqrt(sample proportion * (1-sample proportion)) / full population), (sample proportion + zscore x sqrt(sample proportion * (1-sample proportion)) / full population)).

Answer 4

Thank you : ) I have another question if you could help with it.

Answer 5

no problem, and sure what is it?