Question

Construct a confidence interval for p 1 minus p 2 at the given level of confidence. x 1 equals29​, n 1 equals241​, x 2 equals31​, n 2 equals312​, 90​% confidence

Answer 1

TRADITIONAL METHOD
given that,
sample one, x1 =29, n1 =241, p1= x1/n1=0.1203
sample two, x2 =31, n2 =312, p2= x2/n2=0.0994
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( (0.1203*0.8797/241) +(0.0994 * 0.9006/312))
=0.0269
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.64
margin of error = 1.64 * 0.0269
=0.0442
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.1203-0.0994) ±0.0442]
= [ -0.0232 , 0.0652]
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DIRECT METHOD
given that,
sample one, x1 =29, n1 =241, p1= x1/n1=0.1203
sample two, x2 =31, n2 =312, p2= x2/n2=0.0994
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 = [ (0.1203-0.0994) ± 1.64 * 0.0269]
= [ -0.0232 , 0.0652 ]
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interpretations:
1) we are 90% sure that the interval [ -0.0232 , 0.0652] 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