Math  /  Data & Statistics

QuestionALEKS - Ava Drexler-Amey 3xe/10_u-lgNslkasNW8D8A9PVVfu7nMe56bnd_Vyg51BuSaxDUY2OFMoF5b1Zv10r8-H1SkOK3yuqYtwBSWbk4e-rnDxN-aPljMa5N8wOnqAN-BMdqzxt1H40q?10Bw7QYjlbavbSPXtx-Y Confidence Intervals and Hypothesis Testing Confidence interval for a population proportion 1/51 / 5
A researcher wants to estimate the proportion of depressed individuals taking a new anti-depressant drug who find relief. A random sample of 200 individuals who had been taking the drug is questioned; 162 of them found relief from depression. Based upon this, compute a 90%90 \% confidence interval for the proportion of all depressed individuals taking the drug who find relief. Then find the lower limit and upper limit of the 90%90 \% confidence interval.
Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.)
Lower limit: \square Upper limit: \square

Studdy Solution
Find the lower and upper limits of the confidence interval:
Lower limit:
0.810.0456=0.76440.760.81 - 0.0456 = 0.7644 \approx 0.76
Upper limit:
0.81+0.0456=0.85560.860.81 + 0.0456 = 0.8556 \approx 0.86
The 90% 90\% confidence interval for the proportion is approximately:
Lower limit: 0.76 \boxed{0.76}
Upper limit: 0.86 \boxed{0.86}

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord