Basic Statistics

TESTS FOR EQUALITY OFTWO MEANS: UNPAIRED DATA 103

8.2 TESTS FOR EQUALITY OF TWO MEANS: UNPAIRED DATA

As a second example of testing a hypothesis, let us suppose that we are interested in
comparing the hemoglobin level of children with acyanotic heart disease with that of
children with cyanotic heart disease. The question is: Is the mean hemoglobin level
for acyanotic children different from that for cyanotic children? The null hypothesis
is that the mean levels for acyanotic and cyanotic are the same.

8.2.1 Testing for Equality of Means When c Is Known

From experience with hemoglobin levels, researchers expect the population standard
deviation to be about 1 .O g/cm3. The available data consists of two samples, one of
19 acyanotic children, the other of 12 cyanotic children (see Table 8.2). The mean
for each sample is computed as XI = 13.03 g/cm3 for the acyanotic children and
as 7, = 15.74 for the cyanotic children. We calculate the difference between them,
X1 - X2 = -2.71 g/cm3. The problem then is to decide whether there really is a
difference between the mean hemoglobin levels of the two populations of children
with congenital heart disease or whether the difference 5?1-5f2 = -2.71 was caused
simply by sampling variation. (One way of stating this question is to ask whether the
difference between ;j?l and
We have two populations: A population of hemoglobin levels of acyanotic children
(XI’S), with mean PI, and a population of hemoglobin levels of cyanotic children
(X~’S), with mean p2, and it is assumed that the two populations have the same
population standard deviations of 1 .O g/cm3.
From Section 7.5, the population ofxl -xz’s obtained taking all possible samples
of size n1 and 122, respectively, from the two populations, has a mean 1-11 - 1-12 and a
standard deviation
aJl/nl + l/nz = 1.OJm = 1.Odm = .3687 g/cm3.

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is signijicant.)

The hypothesis that will be tested is Ho : 1-11 = 1-12; it may also be written as

HO : p1- ,u2 = 0, which is equivalent to saying that PI- 1-12, the mean of the -x2

population, is 0. If the original populations are not far from normal and simple

random sampling can be assumed, the -fTz population is approximately normally

distributed. Under the null hypothesis, then, we have a population of - xz’s,

which is normally distributed, whose mean is 0 and whose standard deviation is .3687.

This hypothesis will be rejected if the numerical value of xl - xz is either too large

or too small.

• xz as unusual as -2.71 g/cm3 if
  actually, 1-11 - p2 = 0, the z value corresponding to xl - xz = -2.71 is computed
  using the test statistic:

To find P, the probability of obtaining an

(71 - Zz) - (Pl - Pz)

aJI/nl+ l/nz

z=

or

-2.71 - 0

z= = -7.35

.3687