Answer to Question #324901 in Statistics and Probability for kayel

Question

A.     Using the t-table, give the confidence coefficients for each of the following:

a. _n=21, 95% confidence

b. _n=26, 99% confidence

B.     Compute the population proportion interval estimate given , and the confidence level.

a. _{n=420;p-hat=0.61}, 95% confidence

b. _{n=960;p-hat=0.17}, 99% confidence

C.     Estimate the interval for the population proportion from each of the following. Then, interpret the results.

a. x=_{610, n=1050}, 95% confidence

b. x=_{734, n=1540}, 99% confidence

Answer 1

A. a.1.721

b.2.479

B. $Confidence \ level=p\plusmn z\sqrt{(pq)/n}$

a.n=420, p=0.61, q=1-p=1-0.61=0.39, z=1.96

$Confidence \ level=0.61 \plusmn 1.96\sqrt{0.61x0.39/420}=0.61 \plusmn 0.047$

b.n=960,,p=0.17,q=1-0.17=0.83, z=2.58

$Confidence\ level=0.17 \plusmn 2.58 \sqrt{0.17x0.83/960}=0.17 \plusmn 0.031$

C. a. p=x/n=610/1050=0.58, q=0.42, n=1050, z=1.96

$Interval= 0.58 \plusmn 1.96 \sqrt{0.58x0.42/1050}=0.58 \plusmn 0.03$

b.p=734/1540=0.48, q=1-0.48=0.52, n=1540, z=2.58

$Interval=0.48 \plusmn 2.58\sqrt{0.48x0.52/1540}=0.48 \plusmn 0.032$