We wish to estimate what percent of adult residents in a certain county are pare

Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 500 adult residents sampled, 195 had kids. Based on this, construct a 99% confidence interval for the proportion, p, of adult residents who are parents in this county.

Give your answers as decimals, to three places.

Giving a test to a group of students, the grades and gender are summarized below

Let represent the percentage of all male students who would receive a grade of A on this test. Use a 99.5% confidence interval to estimate p to three decimal places.
Enter your answer as a tri-linear inequality using decimals (not percents).

Explanation / Answer

1)estimated proportion p=195/500=0.39

n=500

therefore std error =(p(1-p)/n)1/2 =0.0218

for 99% CI; z=2.5758

hence 99% confidence interval =p -/+ z*std error =0.3338 ; 0.4462

2)estimated proportion for all male students who would receive a grade of A p=14/21 =0.6667

n=21

therefore std error =(p(1-p)/n)1/2 =0.1029

for 99.5% CI; z=2.8070

hence 99.5% confidence interval =p -/+ z*std error =0.3779 ; 0.9554

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