Basic Statistics

TESTS OF HYPOTHESES FOR A SINGLE MEAN 97

lower than 12.0 months. What must be decided is whether x is significantly different
from 12.0.
We reason this way: There is a whole population of acyanotic children; its mean
age of walking is unknown and we assume that its standard deviation is known to equal
1.75 months. (Here we assume that the variability in the age of walking is the same
in acyanotic children as in well children.) We pretend temporarily that the population
mean age of walking for acyanotic children is 12 months and decide whether the
sample of size 18 could easily have been drawn from such a population. A simple
random sample of children with acyanotic congenital heart disease is assumed.
Under the temporary assumption, then, that 1-1 = 12.0, the means of repeated
samples of size 18 would form a population whose mean is also 12.0 months and
whose standard deviation is aj?- = 1.75/J18 = .412 month. The population of x’s
is approximately normally distributed provided that the original population is not far
from normal.
We compute x = 13.38 months (Table 8.1). The question is: How unusual is this
sample mean, assuming that the sampling was done from a population whose mean
was 12.0 months? To answer this, we compute the chance that a sample will be drawn
whose sample mean is at least as far from ,LL = 12.0 as is x = 13.38. In other words,
we compute the probability of sample means occurring that are as unusual or more
unusual than this one. Since the null hypothesis is Ho : p = 12.0, we reject the null
hypothesis if the x computed is either much smaller than 12.0 or much larger.
This probability, called P, can be obtained by finding the area lying above x =
13.38 under the normal curve whose mean is 12.0 and whose population standard
deviation of the mean is .412, and also the area lying below 10.62. (The point 10.62
is as far below 12.0 as the point 13.38 is above 12.0. Since the normal curve is
symmetric, all sample means that are at least as unusual as 13.38 lie either below
10.62 or above 13.38.) To find the z value forX = 13.38, p = 12.0, and ay = .412,
we compute -
x - Po
a/&

z=-

or

z = (13.38 - 12.0)/.412 = 1.38/.412 = 3.35

By looking in Table A.2, we find for z = 3.35 that the probability of obtaining a
sample mean corresponding to z < 3.35 is .9996, so that the chance of a sample
mean > 13.38 is only .0004. The probability of obtaining a sample mean such that
z is less than -3.35 is also equal to .0004 (see Figure 8.1). We will add these two
probabilities to obtain .0008. So, if 12.0 is the population mean, a sample mean as
unusual as 13.38 would be obtained on the average, in repeated sampling, only 8
times in 10,000 (in other words, P is merely .0008 or .08%). Thus, either a very
unusual sample has been drawn, or else the population mean is not 12.0. It is of
course possible that the population mean is really 12.0 and that an unusual sample
was drawn, but it seems more likely that the population mean is simply not 12.0. We
decide to reject the null hypothesis, That is, we conclude that the mean age at walking
for acyanotic children differs from 12.0 months, the mean for well children.