A survey of households in a small town showed that in 850 of 1,200 sampled households,at least one member attended a town meeting during the year. Using the 99% level of confidence,what is the confidence interval for the proportion of households represented at a town meeting?
A)[0.674,0.742]
B)[0.655,0.705]
C)[0.665,0.694]
D)[0.679,0.680]

Question

Quizplus · Accepted Answer

The proportion of households represented at a town meeting is 850/1200 = 0.708. At the 99% confidence level, the Z-score is approximately 2.576. The standard error (SE) is calculated using the formula SE = sqrt[p(1-p)/n], where p is the sample proportion and n is the sample size. So, SE = sqrt[0.708(1-0.708)/1200] ≈ 0.013. The margin of error (ME) is Z*SE, so ME = 2.576 * 0.013 ≈ 0.034. Therefore, the confidence interval is 0.708 ± 0.034, which gives us [0.674, 0.742].

A Survey of Households in a Small Town Showed That