106 TESTS OF HYPOTHESES ON POPULATION MEANS(as usual the d.f.’s are the numbers used in the denominator of the estimate of the
variance sg). Looking at the row in Table A.3 that corresponds to d.f. = 29, we
see that t[.9995] = 3.659, so that > .9995 of the t distribution lies below 3.659 and
< .0005 lies above 3.659. The area below -3.659 is by symmetry also < .0005.
Summing the area in both tails we have P < .001.
The null hypothesis of equal means is rejected; the conclusion is that the mean
hemoglobin level for acyanotic children is different from that for cyanotic children.
The mean level for the acyanotic children is XI = 13.03 and that of the cyanotic
children X2 = 15.74. As before, 511 - x2 = -2.71, so if we reject the null
hypothesis of equal means we can also say that acyanotic children have lower mean
hemoglobin levels with P < .001.
A one-sided test could also be used. Because acyanotic children are less severely
handicapped than cyanotic children, we may believe that their mean hemoglobin level
(1-11) is lower than 1-12, The question then asked is: Is the mean hemoglobin level for
acyanotic children lower than for cyanotic children? In symbols, the question can be
written as: Is 1-11 < 1-12? The null hypothesis is stated in the opposite direction, that
is, the mean level for acyanotic children is greater than or equal to that of cyanotic
children, or HO : 1-11 2 1-12. If we are able to reject this null hypothesis, we can state
that 1-11 is < 1-12, and then from the value of P we know our chance of making an error.
Now a one-sided test is appropriate, and we calculate the chance of obtaining an
XI - 512 as small as -2.71 g/cm3 or smaller. From the results obtain previously,
P < .0005, and the conclusion is that the hemoglobin level of acyanotic children is
lower on average than that of cyanotic children.
Note the difference in the conclusion between two-sided and one-sided tests. In
a two-sided test, the question asked is whether 1-11 and 1-12 are different, and the
conclusion is either that they are equal or that they are unequal; in a one-sided test
the question asked is whether 1-11 is < p2, and the conclusion is either that 1-11 is < 1-12
or p1 is 2 1-12.
Sometimes the question asked is whether 1-11 is > 1-12. Then, the rejection region
in the right side of the distribution of XI - X2 is used, and the conclusion is either
that p1 is > 1-12 or 1-11 is 5 1-12.
In Chapter 9 a test for equal variance is given. There the test for equal variance
is made using the data above on the acynotic children as an example, and the null
hypothesis of equal variances was not rejected. If the assumption for equal variances
in the two groups cannot be made, one possible procedure is given in that chapter. An
analysis is given for testing the equality of two means where the results are adjusted
for the inequality of the variances. In Chapter 13, additional tests are given that do
not assume equal variances.
The t test for independent groups (unpaired) is one of the most commonly used
statistical tests. It is widely available both in statistical programs and even in some
spreadsheet programs. In most programs, the results are given for a two-sided test.
Most programs print out t and the actual level of P instead of simply saying that it is
less than a given value. The user then compares the computed P value to a prechosen
LY level or simply reports the P value. Minitab, SAS, SPSS, and Stata will all perform
a two-sample t test.