Question

Consider two independent random samples with the following results: n1=561 pˆ1=0.38   n2=642 pˆ2=0.26 Use this data to find the 98% confidence interval for the true difference between the population proportions. Step 2 of 3 : Find the margin of error. Round your answer to six decimal places.

Can you explain how I get the critical value and then how to find the endpoints.

Answer 1

TRADITIONAL METHOD
given that,
sample one, x1 =213.18, n1 =561, p1= x1/n1=0.38
sample two, x2 =166.92, n2 =642, p2= x2/n2=0.26
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( (0.38*0.62/561) +(0.26 * 0.74/642))
=0.026826
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.33
margin of error = 2.33 * 0.026826
=0.062505
critical value is 2.33
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.38-0.26) ±0.062505]
= [ 0.057495 , 0.182505]
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DIRECT METHOD
given that,
sample one, x1 =213.18, n1 =561, p1= x1/n1=0.38
sample two, x2 =166.92, n2 =642, p2= x2/n2=0.26
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 = [ (0.38-0.26) ± 2.33 * 0.026826]
= [ 0.057495 , 0.182505 ]
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interpretations:
1) we are 98% sure that the interval [ 0.057495 , 0.182505] 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, 98% of these intervals will contains the difference between
true population mean P1-P2