Question 1178386: a) A poll of 2,295 likely voters was conducted on the president’s performance. Approximately what margin of error would the approval rating estimate have if the confidence level is 95%? (Round your answer to 4 decimal places.)
Margin of error
.0172
(b) The poll showed that 46 percent approved the president’s performance. Construct a 90 percent confidence interval for the true proportion. (Round your answers to 4 decimal places.)
The 90% confidence interval is from
to
(c) Would you agree that the percentage of all voters opposed is likely to be 50 percent?
multiple choice 2
Yes, the confidence interval contains .50.
No, the confidence interval does not contain .50.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Std error is sqrt(0.5*0.5/2295), with no other data being given, this is the most conservative result.
it is 0.0104
the margin of error for a 95%CI is z(0.95)*SE=0.020; the 0.0172 would work for using 50% probability and a 90% CI
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Now given that it was 46%, one can recompute the SE as 0.01780 for a 90% Ci.
90%CI is (0.4422, 0.4778)
I would agree that the percentage of voters opposed is likely to be 50%, because the 90% CI for approval does not contain 50%, so therefore the 90% CI for opposition WILL contain 50%.
Two comments: the first part does not give a point estimate, so one has to use 50%. If it did give a point estimate, then the value would change for the CI.
Second, the value algebraically and on the calculator will be different, because the algebraic proportion can have a fraction in the numerator (half of 2295 is 1147.5.) But the calculator will require an integer, and to four decimal places, the result will be different.
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