Z-STATS 8: TWO-SAMPLE AND DIRECTIONAL HYPOTHESIS TESTING
Z-STATS 8: This lesson describes some refinements to the hypothesis testing approach that was introduced in the previous lesson. The truth of the matter is that the previous lesson was somewhat oversimplified in order to focus on the concept and general steps in the hypothesis testing procedure. With that background, we can now get into some of the finer points of hypothesis testing.
The "two sample case" is a special case in which the difference between two sample means is examined. This sounds like what we did in the last lesson, but we actually looked at the difference between an observed or sample group mean and a control group mean, which was treated as if it were a population mean (rather than an observed or sample mean). There are some fine points that need to be considered about the calculations and the procedure.
Mathematically, there are some differences in the formula that should be used. Recall the form of the equation for calculating a t-value:
where Xbar is the mean of the experimental group, and µ is a population parameter or the mean of the population. In the last lesson, we substituted the control group mean XbarB for µ. However, the control group was actually a sample from the population of all such mice (in this example) and following suit, the experimental group was also just a sample from its population. Phrasing the situation this way, there are now two sets of differences that must be considered: the difference between sample means or XbarA-XbarB and the difference between population means from which these samples came or µA-µB. Those expressions should be substituted into the tcalc formula to give the proper mathematical form:
Remember the steps for testing a hypothesis are: (1) State the hypotheses; (2) Set the criterion for rejection of Ho; (3) Compute the test statistic; (4) Decide about Ho.
Remember the example of testing the effect of antibiotics on mice in Lesson 7. The point of the study was to find out if the mice who were treated with the antibiotic would outlive those who were not treated (i.e., the control group). Are you surprised that the researcher did not hypothesize that the control group might outlive the treatment group? Would it make any difference in how the hypothesis testing were carried out? These questions raise the issue of directional testing, or one-tailed vs two-tailed tests.
Now don't get confused - we're
not testing to see if our mice have two tails! We're testing to
see if the mean of the sample group is either less than or greater
than the mean of the control group, which - in statistical terms
- is considered to be a two-direction or two-tailed test. Remember
that the hypotheses were Ho:XbarA = XbarB and Ha: XbarA is not
equal to XbarB. In this alternate hypothesis, all that has been
said is that the two means are not the same, which would be true
(a) if the mean of the sample group is higher than that of the
control group or (b) if the mean of the sample group is lower
than that of the control group. There is nothing in the phrasing
of the hypothesis that stipulates the group A animals (treated)
must actually have longer life spans as compared to the group
B animals.
The issue of two-side vs one-side tests becomes important when
selecting the critical t-value. In the earlier discussion of this
example, the alpha level was set to 0.05, but that 0.05 was actually
divided equally between the left and right tails of the distribution
curve. The condition being tested is that group A has a "different"
life span as compared to group B, which represents a two-tailed
test as illustrated in Figure 8-2.
If the conclusion were to support
the claim that the antibiotic prolonged group A life spans, then
the researchers should use a directional alternate hypothesis,
such as Ha: XbarA > XbarB. Here group A's life span is hypothesized
to be greater (longer) than group B's (the control group). In
this case, an alpha level of 0.05 implies that all 0.05 would
have to appear in the right or high tail of the curve, which then
is a one-tailed or directional test, as shown in Figure 8-3. This
figure shows that the critical t-value will actually be smaller
for the one-tail test, that is, +1.65 instead of 1.96 or 2.00
from the two-tail test. This happens because 95% of the area under
the curve begins to accumulate from the left-most side of the
curve (including that tail) and includes less of the right side
of the curve. The result is that tcalc can be smaller
(1.65 instead of 1.96) and still cause Ho to be rejected.
This can be confusing and it should be helpful to think about having one or two gates on the curve. For example, for an alpha level of 0.05 and a two-tail test, there are two gates - one at -1.96 and one at +1.96. If you "walk" out of either of those gates, then you have demonstrated significance at p=0.05 or less. For a one-tail test, there is only one gate at +1.65. If you "walk" out of this gate, you have demonstrated significance at p=0.05 or less.
The t-table provides for one-tailed
versus two-tailed tests listing separate tcrit. As
in all t-tests, in order to reject Ho and accept Ha (that group
A lives longer than group B), the calculated t would have to be
greater than the tcrit found in the table. But with
the one-tail test the tcrit is 1.65 instead of 1.96, so tcalc
may be smaller as well when rejecting Ho. It is actually easier
to reject Ho using a one-tailed t-test. Figure 8-4 superimposes
the two graphs in order to demonstrate the last statement.
For example, given a tabled t value (tcrit) of 1.96 at 0.05 significance
for a 2-tailed t-test and a tcalc of 1.90, the Ho cannot
be rejected. We are 0.06 away from walking out the gate. Using
the one-tail test, this same value for tcalc exceeds
the 1.65 tcrit for 0.05 alpha level, so the Ho can
be rejected. We have walked out the gate and are standing a distance
of 0.25 outside. So tcrit for a one-tail test is less
than that for a two-tail test, at comparable alpha levels. That
is, it takes less of a calculated t to overturn Ho. This relationship
is shown in Figure 8-4. There is another more interesting point
to be made. The figure actually shows that if you used a tcrit
of 1.96 on a one-tailed test, the alpha level would actually drop
to half of what it was before, from 0.05 to 0.025!
To reiterate, if you are standing right at the gate (1.96) for a two-tail test, then you have just barely met the p=0.05 requirement. However, if you are standing at the 1.96 point when running a one-tail test, then you have already exceeded the 1.65 gate and the probability must be even more significant, say p=0.025. It's important to find the critical t-value that is correct for the intended directional nature of the test.
Madelon F. Zady is an Assistant Professor at the University of Louisville, School of Allied Health Sciences Clinical Laboratory Science program and has over 30 years experience in teaching. She holds BS, MAT and EdD degrees from the University of Louisville, has taken other advanced course work from the School of Medicine and School of Education, and also advanced courses in statistics. She is a registered MT(ASCP) and a credentialed CLS(NCA) and has worked part-time as a bench technologist for 14 years. She is a member of the: American Society for Clinical Laboratory Science, Kentucky State Society for Clinical Laboratory Science, American Educational Research Association, and the National Science Teachers Association. Her teaching areas are clinical chemistry and statistics. Her research areas are metacognition and learning theory.
