SOLUTION: A recent poll of 1500 U.S. adults of voting age found that 930 of them voted in a recent election. Make a 99% confidence interval for the proportion of all U.S. adults of voting a

Question 1189888: A recent poll of 1500 U.S. adults of voting age found that 930 of them voted in a recent election. Make a 99% confidence interval for the proportion of all U.S. adults of voting age population who voted. Write endpoints as percents rounded totthe nearest tenth of a percent. Write a sentence interpreting the meaning of the interval you find.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
p hat mss 930/1500=0.62
99%CI half-interval is z(0.995)*sqrt (p*(1-p)/n)=2.576*sqrt (0.62*0.38/1500)=0.0323 or 3.2%
interval is (58.8%, 65.2%)
We are highly confident that the true proportion of US adults of voting age who vote lies in this interval.
If we ran 100 different samples and intervals, 95% of intervals would be expected to contain the true value.
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