North South UniversitySchool of Life Sciences

Comparing Proportions between Two Independent Populations

Obaidur Rahman, PhD.NSU

1OrR_NSU_School of life science 1

SE(pAZT)= 0.01924

SE(pPLAC)= 0.0305507

Estimated 95% CI based on a single sample of size 180 for AZT

0.07 + 2 SE(pAZT) = 0.07 + 0.01924 = (0.0390137, 0.1203358)

Estimated 95% CI based on a single sample of size 180 for Placebo

0.22 + 2 SE(pPLAC) = 0.22 + 0.0305507 = (0.160984, 0.2855248)

E(1,1) = (53*180)/363 = 26.3

E(1,2) = (53*183)/363 = 26.72

E(2,1) = (310*180)/363 = 153.7

E(2,2) = (310*183)/363 = 156.3

X2 = (13-26.3)2/26.3 + (40-26.7)2/26.7 + (167-153.7)2/153.7 + (143-156.3)2/156.3

= 6.72 + 6.625 + 1.15 + 1.13

= 15.6

Degree of freedom = (2-1) * (2-1) = 1

Chi square distribution

This statistic at 1 df, indicate a significant correlation (p<.001) relating

AZT and placebo group.

Chi square distribution

In 1979 the U.S. Veterans Administration conducted a health survey of 11,230

veterans. The advantages of this survey are that it includes a large random

sample with a high interview response rate, and it was done before the public

controversy surrounding the issue of the health effects of possible exposure to

Agent Orange. The data shown in Table relate Vietnam service to having sleep

problems among the 1787 veterans who entered the military service between

1965 and 1975.

Chi square distribution