Basic Statistics

TESTS FOR EQUALITY OFTWO MEANS: UNPAIRED DATA 105

a single population and randomly assign subjects to two treatment groups. This test

has been shown to be insensitive to minor departures from the normal distribution.

To illustrate testing for equality of population means when the population variance

is unknown, we will use the hemoglobin data for children with cyanotic and acyanotic

congenital heart disease given in Table 8.2.

Just as in Section 7.5.3, when confidence intervals were found, we will use the

pooled variance, si, as an estimate of o', the population variance for each population.

The pooled variance is computed as

(n1 - 1,s: + (n, - 1)s;

s; =

121 + 122 - 2

where s: is the variance from the first sample and si is the variance from the second

sample. From Table 8.2, we have s: = 1.0167, s; = 1.9898. n1 = 19, and n2 = 12.

The computed pooled variance is

= 1.3858

s= 18(1.0167) + ll(1.9898) - - 40.1884

P 19 + 12 - (^2 29)
Note that the pooled variance 1.3858 is between the numerical values of sf = 1.0167
and si = 1.9898. It is somewhat closer to s: since n1 - 1 is > nz - 1. To
obtain the pooled standard deviation, sp, we take the square root of 1.3858 and obtain
sP = 1.1772.
The standard deviation of the XI - Xz distribution is
oJl/nl+ 1/n2
It is estimated by
spJl/nl + l/nz = 1.1772J1/19 + 1/12 = 1.1772413596 = ,4341 g/cm3
The test statistic is
(71 - 72) - (P1 - P2)
SpJl/nl+ l/nz
t=
or
(XI - X,) - 0
spJl/n1+ 1/n2
t=
or for the results from Table 8.2,
13.03 - 15.74 -2.71
t= - - = -6.24
A341 .4341
Table A.3 is then consulted to find the chance of obtaining a t value at least as small
as t = -6.24. The d.f.'s to be used in the table are