Testing a Single Mean
Prerequisites
Logic of Hypothesis Testing, Areas
Under Normal Distributions, Sampling
Distribution of the Mean, Introduction
to Sampling Distributions, t
Distribution
Learning Objectives
 Compute the probability of a sample mean being at least as high as
a specified value when σ is known
 Compute a twotailed probability
 Compute the probability of a sample mean being at least as high as
a specified value when σ is estimated
 State the assumptions required for item 3 above.
This section shows how to test the null hypothesis that the population mean is equal to some hypothesized value.
For example, suppose an experimenter wanted to know if people
are influenced by a subliminal message and performed the following
experiment. Each of nine subjects is presented with a series of
100 pairs of pictures. As a pair of pictures is presented, a subliminal
message is presented suggesting the picture that the subject should
choose. The question is whether the (population) mean number of
times the suggested picture is chosen is equal to 50. In other
words, the null hypothesis is that the population mean (μ)
is 50. The (hypothetical) data are shown in Table 1. The data
in Table 1 have a sample mean (M) of 51. Thus the sample mean
differed from the hypothesized population mean by 1.
The significance test consists of computing the probability of
a sample mean differing from μ by one (the difference between
the hypothesized population mean and the sample mean) or more.
The first step is to determine the sampling distribution of
the mean. As shown in a previous
section, the mean and standard
deviation of the sampling distribution of the mean are
μ_{M} = μ
and
respectively. It is clear that μ_{M}
= 50. In order to compute the standard deviation of the sampling
distribution of the mean, we have to know the population standard
deviation (σ). In practice, it is
very unlikely that you would know σ
and therefore you would use s, the sample
estimate of σ. However, it is instructive
to see how the probability is computed if σ
is known before proceeding to see how it is calculated when it
is estimated.
In this example, if the null hypothesis is true,
then σ = 5(it
is not important for the present discussion to see how this was
calculated, but the calculation is based on the binomial
distribution). For a σ of
5 and an N of 9, the standard deviation of the sampling distribution
of the mean is 5/3 = 1.667. Recall that the standard deviation
of a sampling distribution is called the standard error.
To recap, we wish to know the probability of obtaining
a sample mean of 51 or more when the sampling distribution of
the mean has a mean of 50 and a standard deviation of 1.667. To
compute this probability, we will make the assumption that the
sampling distribution of the mean is normally distributed. We
can then use the normal
distribution calculator as shown in Figure 1.
Notice that the mean is set to 50, the standard
deviation to 1.667, and the area above 51 is requested and
shown to be 0.274.
Therefore, the probability of the obtaining a sample
mean of 51 or larger is 0.274. Since a mean of 51 or higher is
not unlikely under the assumption that the subliminal message
has no effect, the effect is not significant and the null hypothesis
is not rejected.
The test conducted above was a onetailed test
because it computed the probability of a sample mean being one
or more points higher than the hypothesized mean of 50 and
the area computed was the area above 51. To test the two tailed hypothesis,
you would compute the probability of a sample mean differing
by one or more in either direction from the hypothesized mean
of 50. You would do so by computing the probability of a mean
being less than or equal to 49 or greater than or equal to 51.
The results of the normal
distribution calculator are shown in Figure 2.
As you can see, the probability is 0.54 which, as expected, is
twice the probability of 0.27 shown in Figure 1.
Before normal calculators such as the one illustrated
above were widely available, probability calculations were made
based on the standard normal distribution. This was done by computing
Z based on the formula
where
Z is the value on the standardized normal distribution, M is
the sample mean, μ is the hypothesized value of the mean,
and σ_{M} is the standard error of
the mean. For this example, Z = (5150)/1.667 = 0.60. Use the
normal calculator with a mean of 0 and a standard deviation
of 1 as shown below.
Notice that the probability (the shaded area)
is the same as previously calculated (for the onetailed test).
As noted, in realworld data analysis it is very
rare that you would know σ and wish
to estimate μ. Typically σ is not
known and is estimated in a sample by s and σ_{M}
is estimated by s_{M}. For our example,
we will consider the data in the ADHD
treatment case study. These data consist of the scores
of 24 children with ADHD on a delay of gratification (DOG)
task. Each child was tested under four dosage levels. Table
2 shows the data for the placebo (0mg) and highest dosage
level (0.6mg) of methylphenidate. Of particular interest here
is the column labeled "Diff" that shows the difference
in performance between the 0.6mg (D60) and the 0mg (D0) conditions.
These scores are positive for children who performed better
in the 0.6 mg condition than in the control condition and
negative for those who scored better in the control condition.
If methylphenidate has a positive effect, then the mean difference
score in the population will be positive. The null hypothesis
is that the mean difference score in the population is 0.
To test this null hypothesis, we compute t using
a special case of the following formula
The special case of this formula applicable to testing a single
mean is:
where t is the value of t we compute for the
significance test, M is the sample mean, μ is the hypothesized
value of the population mean, and s_{M} is the estimated
standard error of the mean. Notice the similarity of this formula
to the formula
for Z.
In the previous example, we assumed the distribution
of scores was normally distributed. In this case,
it is the population of difference scores that we assume to
be normally distributed.
The mean (M) of the N = 24 difference scores is
4.9583. the hypothesized value of μ is 0, and the standard
deviation (s) is 7.5382. The estimate of the standard error of
the mean is computed as:
Therefore, t = 4.96/1.54 = 3.22. The probability
value for t depend on the degrees of freedom. The degrees of
freedom is equal to N  1 = 23. As shown below, the t
distribution calculator shows that the probability of a
t less than 3.22 or greater than 3.22 is only 0.0038. Therefore,
if the drug had no effect, the probability of finding a difference
between means as large or larger (in either direction) as the
difference found is very low. Therefore the null hypothesis
that the population mean difference score is zero can be rejected.
The conclusion is that the population mean for the drug condition
is higher than the population mean for the placebo condition.
Review of Assumptions
 Each value is sampled independently from each
other value.
 The values are sampled from a normal distribution.
