Question 1076339: In a simple random sample of 1300 full-time students at a large university, it is found that 850 study an average of more than 65 minutes per day. Use this sample to estimate the true percentage, pp, of full-time students at this university who study an average of more than 65 minutes per day.
Find a 90% confidence interval for p
Find a 95% confidence interval for p
Find a 98% confidence interval for p
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! A CI has 0.654 +/- z sqrt {(0.654*0.346)1300}.sqrt=0.131
for 90%, where z=1.645, the half interval is 0.021--(0.633,0.675)
for 95%, where z=1.96, the half interval is 0.026--(0.0628, 0.680)
for 98%, where z=2.326, the half interval is 0.30-(0.624, 0.684)
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