98 TESTS OF HYPOTHESES ON POPULATION MEANS
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10 62 12 0 13 38 X (months)- 3.35 0 3 35 z
Figure 8.1 Two-sided test of HO : p = 12.0 when x = 13.38.
11.6 12.0 12.4 X (months)
- 0.97 0 0.97 z
Figure 8.2 Two-sided test of Ho : p = 12.0 when x = 11.6.
It is possible that our conclusion is incorrect, that the mean age is actually 12.0,
and that this sample is unusual. We cannot know that p is not 12.0, since we have
only one sample from the population on which to base the decision. However, the
conclusion that p is not 12.0 seems more reasonable than that a highly unusual sample
was obtained.
Suppose that we had obtained x = 11.6 months for the sample mean instead of
X = 13.38. We proceed exactly as we did with x = 13.38. This time z is
z = (11.6 - 12.0)/.412 = -.4/.412 = -.97
In Table A.2, the area to the left of z = +.97 is .8340, or 83.4%. Note that even
though the computed z = -.97, we use t = .97. This is possible because the normal
distribution is symmetric. The chance of a sample mean’s being < 11.6 is then
1.00 - 3340 = .1660, and the chance of a sample mean’s being < 11.6 or > 12.4
(or .4 unit from p = 12.0) is twice as large, so that P = 2(.1660) = .3320, or 33%
(see Figure 8.2).
In other words, if the population mean is really 12.0, and if the experiment were
repeated many times, and each time the mean age of walking calculated from a random
sample of 18 acyanotic children, about 33 out of 100 of the sample means would differ
from 12.0 by at least as much as did the sample mean actually obtained. Clearly, 11.6
is close enough to 12.0 that it could have been obtained very easily by chance. We
look at P = .33 and decide to accept the null hypothesis. The acceptance is a negative
type of acceptance; we simply have failed to disprove the null hypothesis. We decide
that as far as we can tell from the sample, the population mean may be 12.0 and that