(22.27) Exercise 22.25 describes a Harris Poll survey of smokers in which 848 of
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Question
(22.27) Exercise 22.25 describes a Harris Poll survey of smokers in which 848 of a sample of 1010 smokers agreed that smoking would probably shorten their lives. Harris announces a margin of error of ±3 percentage points for all samples of about this size. Opinion polls announce the margin of error for 95% confidence.
Step 1:
What is the actual margin of error (in percent) for the large-sample confidence interval from this sample?
1.15%
Step 2:
The margin of error is largest when pˆ = 0.5.
What would the margin of error (in percent) be if the sample had resulted in pˆ = 0.5?
Give your answer to 2 decimal places.
Fill in the blank:
Step 3:
Why do you think that Harris announces a ±3% margin of error for all samples of about this size?
1.15%
Step 2:
The margin of error is largest when pˆ = 0.5.
What would the margin of error (in percent) be if the sample had resulted in pˆ = 0.5?
Give your answer to 2 decimal places.
Fill in the blank:
Step 3:
Why do you think that Harris announces a ±3% margin of error for all samples of about this size?
Explanation / Answer
Solution:
Step 1: b) 2.26%
Given x = 848, n = 1010
so p = x/n = 848/1010 = 0.8396
here = 95% or 0.05
Critical z-value = Z/2 = Z0.025 = 1.96
Standard error SE = (p * (1-p)/n)
= ((424/505 * 81/505) /1010) = 0.0115
Margin of Error E = Z/2 * SE
= Z/2 * (p * (1-p)/n) = 1.96 * 0.0115
= 0.02254 or 2.26%
Step 2:
Standard error SE = (p * (1-p)/n)
= (0.5* 0.5 /1010)
= 0.0157
Margin of Error E = Z/2 * SE
= Z/2 * (p * (1-p)/n)
= 1.96 * 0.0157
= 0.030772
a) Because the margin of error for samples of about this size is no more than the announced margin of error.
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