Question

A poll is taken in which 332 out of 500 randomly selected voters indicated their preference for a certain candidate.

(a) Find a 90% confidence interval for pp.

≤p≤

(b) Find the margin of error for this 90% confidence interval for p.

Answer 1

Solution:

Given: A poll is taken in which 332 out of 500 randomly selected voters indicated their preference for a certain candidate.

That is:

n = 500

x = 332

thus

Part a) Find a 90% confidence interval for p

Formula:

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

Part b) Find the margin of error for this 90% confidence interval for p.