Question 1155344: In a survey of 650 males ages 18-64, 394 say they have gone to the dentist in the past year.
Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 394/650=0.606, point estimate
SE is sqrt(p*(1-p)/n)=0.0192
half-interval is z*SE, so for 90% it is 1.645*0.0192=0.0315 and the point estimate+/- half-interval is the CI, or (0.575, 0.638)
for 95% CI the half-interval is 1.96*0.0192=0.038 and the CI is (0.568, 0.644).
The higher confidence has wider intervals, because to have more confidence, there has to be more possibilities, basically. This says we don't know the exact percentage of those who have gone to the dentist, but we can be 90% confident (95% confident) it lies in the interval given. It either is or isn't, and we will never know which, but either or is not a probabilistic issue but rather a confidence issue.
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