Question #324901

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



1
Expert's answer
2022-04-07T13:14:54-0400

A. a.1.721

b.2.479

B. Confidence level=p±z(pq)/nConfidence \ 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±1.960.61x0.39/420=0.61±0.047Confidence \ 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±2.580.17x0.83/960=0.17±0.031Confidence\ 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±1.960.58x0.42/1050=0.58±0.03Interval= 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±2.580.48x0.52/1540=0.48±0.032Interval=0.48 \plusmn 2.58\sqrt{0.48x0.52/1540}=0.48 \plusmn 0.032







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