Answer to Question #257022 in Statistics and Probability for norn

Question

in a locality of 18000 families a sample of 840 families was selected, of these 840 families, 206 families were found to have a monthly income of 7000 or less. It is desired to estimate how many out of the 18000 families have a monthly income of 3000 or less. Within what limits would you place your estimates

Answer 1

The proportion of families that have income as 7000 or less: $p=\frac{206}{840}=0.245.$

Confidence interval for the the proportion of families that have income as 7000 or less:

$95\%CI=(p-z_{0.025}\sqrt{\frac{p(1-p)}{n}},p+z_{0.025}\sqrt{\frac{p(1-p)}{n}})=$

$=(0.245-1.96\sqrt{\frac{0.245(1-0.245)}{840}},0.245+1.96\sqrt{\frac{0.245(1-0.245)}{840}})=$

$=(0.216,0.274).$

We are 95% confident that the the proportion of families that have income as 7000 or less in a locality

lies between 21.6% and 27.4%.