Question

In a study of red/green color blindness, 800 men and 3000 women are randomly selected and tested. Among the men, 76 have red/green color blindness. Among the women, 8 have red/green color blindness. Test the claim that men have a higher rate of red/green color blindness.

The test statistic is

Construct the 99% confidence interval for the difference between the color blindness rates of men and women. <(pm−pw)<

Answer 1

TRADITIONAL METHOD
given that,
sample one, x1 =76, n1 =800, p1= x1/n1=0.095
sample two, x2 =8, n2 =3000, p2= x2/n2=0.003
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( (0.095*0.905/800) +(0.003 * 0.997/3000))
=0.01
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.58
margin of error = 2.58 * 0.01
=0.027
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.095-0.003) ±0.027]
= [ 0.065 , 0.119]
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DIRECT METHOD
given that,
sample one, x1 =76, n1 =800, p1= x1/n1=0.095
sample two, x2 =8, n2 =3000, p2= x2/n2=0.003
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 = [ (0.095-0.003) ± 2.58 * 0.01]
= [ 0.065 , 0.119 ]
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interpretations:
1) we are 99% sure that the interval [ 0.065 , 0.119] 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, 99% of these intervals will contains the difference between
true population mean P1-P2
Answer:

99% confidence interval for the difference between the color blindness rates of men and women l [ 0.065 , 0.119]