Statistics for the Behavioral Sciences by Frederick J. Gravetter, Larry B. Wallnau ISBN 10: 1305504917 ISBN 13: 9781305504912

262 CHAPTER 8 | Introduction to Hypothesis Testing
Most statistical tests are now done with computer programs that provide an exact probability
(p value) for a Type I error. Because an exact value is available, most researchers
simply report the p value from the computer printout rather than setting an alpha level at
the beginning of the test. However, the same criterion still applies: A result is not significant
unless the p value is less than .05.
3. Take time to consider the implications of your decision about the null hypothesis. The
null hypothesis states that there is no effect. Therefore, if your decision is to reject H 0
, you
should conclude that the sample data provide evidence for a treatment effect. However, it is
an entirely different matter if your decision is to fail to reject H 0
. Remember that when you
fail to reject the null hypothesis, the results are inconclusive. It is impossible to prove that
H 0
is correct; therefore, you cannot state with certainty that “there is no effect” when H 0
is
not rejected. At best, all you can state is that “there is insufficient evidence for an effect.”
4. It is very important that you understand the structure of the z-score formula (page 234). It
will help you understand many of the other hypothesis tests that are covered later.
5. When you are doing a directional hypothesis test, read the problem carefully, and watch
for key words (such as increase or decrease, raise or lower, and more or less) that tell you
which direction the researcher is predicting. The predicted direction will determine the
alternative hypothesis (H 1
) and the critical region. For example, if a treatment is expected
to increase scores, H 1
would contain a greater than symbol, and the critical region would
be in the tail associated with high scores.
DEMONSTRATION 8.1
HYPOTHESIS TEST WITH Z
A researcher begins with a known population—in this case, scores on a standardized test that
are normally distributed with μ = 65 and σ = 15. The researcher suspects that special training
in reading skills will produce a change in the scores for the individuals in the population.
Because it is not feasible to administer the treatment (the special training) to everyone in
the population, a sample of n = 25 individuals is selected, and the treatment is given to this
sample. Following treatment, the average score for this sample is M = 70. Is there evidence
that the training has an effect on test scores?
STEP 1
State the hypothesis and select an alpha level The null hypothesis states that the special
training has no effect. In symbols,
H 0
: μ = 65 (After special training, the mean is still 65.)
The alternative hypothesis states that the treatment does have an effect.
H 1
: μ ≠ 65 (After training, the mean is different from 65.)
At this time you also select the alpha level. For this demonstration, we will use α = .05.
Thus, there is a 5% risk of committing a Type I error if we reject H 0
.
STEP 2
STEP 3
Locate the critical region With α = .05, the critical region consists of sample means that
correspond to z-scores beyond the critical boundaries of z = ±1.96.
Obtain the sample data, and compute the test statistic For this example, the distribution
of sample means, according to the null hypothesis, will be normal with an expected value of
μ = 65 and a standard error of
s M
5 s Ïn 5 15
Ï25 5 15 5 5 3