TESTS OF HYPOTHESES FOR A SINGLE MEAN 101- State the statistical conclusion either to reject or fail to reject the null hypothesis.
This should be followed with a statement of the results as they apply to the goals
of the research project.
Similar types of procedures apply to the tests given later in this chapter.8.1.4 Test for a Single Mean When v Is Unknown
In the earlier example of 18 children with acyanotic heart disease, the standard de-
viation of the original population was assumed to be 1.75 months, the same standard
deviation as for age of walking for well children. The assumption that the variability
of a population under study is the same as the known variability of a similar popu-
lation is sometimes reasonable. More often, however, the standard deviation must
be estimated from the sample data, either because one does not know the standard
deviation of the comparable population (one does not know that 1.75 months is the
standard deviation for age of walking for normal children) or because one is not sure
that the variability of the population being studied is close to that of the comparable
population (a small proportion of children with heart disease may walk very late, so
that the age of walking for acyanotic children may be more variable than that for
normal children). Just as in obtaining confidence intervals in Section 7.4, s must be
calculated from the sample and used in place of 0, and the t distribution must be
used instead of the normal distribution. See Section 7.3 for an explanation of the t
distribution. Here, f? = 13.38 and s = 2.10 (see Table 8.1). The standard error of
the mean is sy = s/& or 2.10/m = 0.4944. The test statistic is then
or13.38 - 12.0
Q .4YUt= = 2.79
If the test is being made to decide whether p is > 12.0, a one-sided test is called for.
The null hypothesis is HO : p 5 12.0, so if it is rejected, the conclusion is that the
population mean is > 12.0. In Table A.3 we enter the table with 17 d.f. and look for
2.79; we find 2.567 in the column headed .99 and 2.898 in the column headed .995.
Thus the area to the left oft = 2.79 is between .99 and ,995, and the area to the right
of 2.79 is between .01 and .005. Thus ,005 < P < .01. Usually, this is stated simply
as P < .01 (see Figure 8.3).
If we have chosen a: = .05 as the level of significance, we reject the null hypothesis
and decide that the mean age of walking is > 12.Omonths; acyanotic children walk
later on the average than normal children. If we wish to test the null hypothesis at the
(Y = .05 level without computing P, we compare the calculated t = 2.79 with the
tabled value t[.95] = 1.74 with 17 d.f. and reject the null hypothesis because 2.79 is
1.74.