Question

In: Statistics and Probability

(a) A poll of 2,291 likely voters was conducted on the president’s performance. Approximately what margin...

(a) A poll of 2,291 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.

Expert Solution

Solution:

Part a)

Given: n = Sample Size = 2291

c = confidence level = 95%

We have to find margin of error for the proportion of the approval rating estimate using 95% confidence level.

Formula:

Since estimate of sample proportion is unknown , we use

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

Part b) The poll showed that 46 percent approved the president’s performance. Construct a 90 percent confidence interval for the true proportion.

We have

Formula for confidence interval is:

where

Z_c is z critical value for c = 90% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.90) / 2 = 1.90 / 2 = 0.9500

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_c = 1.645

Thus

Thus a 90% confidence interval is from 0.4528 to 0.4872

Part c) Would you agree that the percentage of all voters opposed is likely to be 50 percent?

No, the confidence interval does not contain 0.50

orchestra answered 2 years ago

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