Question

A survey was run by a high school student in order to determine what proportion of mortgage-holders in his town expect to own their house within 10 years. He surveyed 141 mortgage holders and found that the proportion of these that did expect to own their house within 10 years is 0.70. The student decides to construct a 95% confidence interval for the population proportion. a) Calculate the margin of error that the high school student will have. Give your answer as a decimal to 3 decimal places. Margin of error = A different high school student sees this survey and decides to try and repeat it. This second high school student also surveys 141 mortgage holders, but this student finds that the proportion of these that do expect to own their house within 10 years is 0.88. This student also constructs a 95% confidence interval for the population proportion. b) Calculate the margin of error that the second student will have. Give your answer as a decimal to 3 decimal places. Margin of error =

Answer 1

a. z value for 95% CI is 1.96 as P(-1.96<z<1.96)=0.95

Margin of Error is E=0.70 and n=141

So we will find n using formula of E

PG E = 2* 1.96 * 0.70 * 0.30 141 = 0.076 n

b. Here p=0.88 and n=141

$E=z*\sqrt{\frac{pq}{n}}=1.96*\sqrt{\frac{0.88*0.12}{141}}=0.054$