A city claims that 75% of all low income housing is 1500 sq.ft.The tenants believe the proportion of housing this size is smaller than the claim,and hire an independent firm to test a random sample size of 120 and found 1500 sq.ft.housing at 66%.Create a 95% confidence interval for the proportion of housing 1500 sq.ft.

Question

Quizplus · Accepted Answer

To calculate a 95% confidence interval for the proportion, we use the formula: $$CI = \hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ where $\hat{p} = 0.66$, $n = 120$, and $Z_{\alpha/2} = 1.96$ for a 95% confidence level. Plugging these values into the formula gives us the interval $(0.5752, 0.7448)$ or $57.52\%, 74.48\%$.

A City Claims That 75% of All Low Income Housing