Question

A poll of 2,256 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 (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 to (c) Would you agree that the percentage of all voters opposed is likely to be 50 percent? No, the confidence interval does not contain .50. Yes, the confidence interval contains .50.

Answer 1

1) Margin of error = Confidence coefficient*Standard error of p
p = 0.5 can be considered
Confidence coefficient is the critical value of z for 95% confidence level = 1.96
Standard error of p = sqrt [p*(1-p)/n]
= sqrt [0.5*0.5/2256]
Therefore,
Margin of error estimated = 1.96 * sqrt [0.5*0.5/2256]
= 0.0206 or 2.06%

2)Confidence interval for the true proportion =
sample proportion +/- margin of error
Sample proportion = p = 46% = 0.46

Critical value of z for 90% confidence level=1.645

Margin of error = 1.645 * sqrt [0.46*0.54/2256] = 0.0173 or 1.73%
The CI required is
0.46+/- 0.0173
lower limit is 0.46 - 0.0173 = 0.4427
upper limit is 0.46 + 0.0173 = 0.4773
The CI is (0.4427, 0.4773) or (44.27%, 47.73%)

C) percentage of all voters opposed is likely

=(100-47.73 %, 100-44.27 %)

=(52.27%, 55.73%)

No, the confidence interval does not contain 0.5 or 50℅