In 2003, an organization surveyed 1,508 adult Americans and asked about a certai
ID: 3319610 • Letter: I
Question
In 2003, an organization surveyed 1,508 adult Americans and asked about a certain war, "Do you believe the United States made the right or wrong decision to use military force?" Of the 1 comma 5081,508 adult Americans surveyed, 1,085 stated the United States made the right decision. In 2008, the organization asked the same question of 1,508 adult Americans and found that 573 believed the United States made the right decision. Construct and interpret a 90% confidence interval for the difference between the two population proportions, p2003p2008. The lower bound of a 90% confidence interval is nothing. (Round to three decimal places as needed.) The upper bound of a 90% confidence interval is nothing. (Round to three decimal places as needed.) Interpret the 90% confidence interval for the difference between the two population proportions, p2003p2008. Choose the correct answer below. A. There is 90% confidence that the difference in the proportion of adult Americans from 2003 to 2008 who believe the United States made the right decision to use military force in the country is greater than the lower bound. B. There is 10% confidence that the difference in the proportion of adult Americans from 2003 to 2008 who believe the United States made the right decision to use military force in the country is between the lower and upper bounds of the interval. C. There is 90% confidence that the difference in the proportion of adult Americans from 2003 to 2008 who believe the United States made the right decision to use military force in the country is between the lower and upper bounds of the interval.
Explanation / Answer
TRADITIONAL METHOD
given that,
sample one, x1 =1085, n1 =1508, p1= x1/n1=0.719
sample two, x2 =573, n2 =1508, p2= x2/n2=0.38
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.719*0.281/1508) +(0.38 * 0.62/1508))
=0.017
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
margin of error = 1.645 * 0.017
=0.028
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.719-0.38) ±0.028]
= [ 0.312 , 0.368]
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DIRECT METHOD
given that,
sample one, x1 =1085, n1 =1508, p1= x1/n1=0.719
sample two, x2 =573, n2 =1508, p2= x2/n2=0.38
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.719-0.38) ± 1.645 * 0.017]
= [ 0.312 , 0.368 ]
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interpretations:
1) we are 90% sure that the interval [ 0.312 , 0.368] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the difference between
true population mean P1-P2
option:C
There is 90% confidence that the difference in the proportion of adult Americans from 2003 to 2008
who believe the United States made the right decision to use military force in the country is between the lower and upper bounds of the interval.
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