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Lab 20
Independent samples t-tests
Hypothesis tests analyzed with related samples t-tests
Let's
start with a brief review. In the last several
labs we looked at ways to test samples.
We used z-scores to test whether a sample was probably from a
population for which we knew the population mean μ
and the population standard deviation σ
We used one-sample t-tests to examine the same
situation, except we don't know the population σ,
so we need to use an estimate (the sample standard deviation).
The last lab used a different computational formula to
calculate the observed t for two more situations:
- repeated measures, in which there is one sample,
but each individual is tested twice.
- matched pairs, in which there are two samples, but
they are related on a subject by subject basis.
The logic of today's lab should seem similar to the last
several labs. The overall logic is the same, we still use the
t-distribution to
find our critical values. However things get a little more complicated,
because of the formulas we'll use. Now we are going to
look at a situation where we are interested in the potential difference
between two independent populations. And again, we'll deal with
situations in which we don't know the μ or
σ for either of these
populations, so we'll have to use estimates.
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An experiment that uses a separate independent
sample for each treatment condition (or each population) is called an
independent-measures
research design. Often you'll also see it referred to as a
between-subjects
or between-groups design.
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So we'll use the same logic and
steps for hypothesis testing that we used in the previous labs, and
fill in the details of the differences as we go.
Step 1:
State your H0
and H1 and choose your criterion: α
Step 2: Compute the observed t for your
samples
Step 3: Compare observed t to critical t (or p to
α) and make a decision
Let's
start with Step 1:
Figuring
out your criteria is exactly the same process as before, you pick what
your field has decided as being an accepted level of alpha (chance of
making a type I error). For our example, let's assume α = 0.05
The hypotheses are
going to be a bit different, because
the situation is different. Remember, that now we are making hypotheses
about two different populations, not just comparing a treatment to what
is known.
For example,
suppose that you want to compare two
different treatments (e.g., two ways of studying, two different drugs,
etc), or you want to compare two groups of people (e.g., men vs. women,
young vs. old, etc.). So now, the hypotheses are about population A
(men) and population B (women), and how they are different from one
another.
Suppose
that we are interested in how tall men and women are.

Is this going to be
a one-tailed test or a two-tailed
test? In this case, we'll conduct a two-tailed test. We won't make a
directional prediction.
So the H0
hypothesis would be that men and
women are the same height. That is,
H0: μMen
= μWomen
- or -
H0: μMen
- μWomen = 0
Our alternative
hypothesis would be that men and women
have
different mean heights.
That is,
H1: μMen
≠ μWomen
- or -
H1:
μMen
- μWomen ≠0
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Note: What might the hypothesis be for
a one-tailed test? Men are taller than women.
H0: μMen <= μWomen
H1: μMen > μWomen
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Step 2:
We are going to be
using two samples, one to represent
each population.
This is a good time
to look at some sample data:
Men's
heights: 67, 73, 74, 70, 70, 75, 73, 68, 69
Women's heights: 69, 63, 67, 64, 61, 66, 60, 63, 63
Any guesses as to
how we'll compute our df?
Think about it this
way, with one sample we used n
- 1 because all of the values in the sample are free to vary but one,
because we know the value of the sample mean.
Now consider the
current situation. We've got two
samples. How many values are free to vary?
sample 1: nMen
- 1
sample 2: nWomen - 1
so together there are nMen + nWomen
- 2 = df
We need to know how
many individuals we have in our
samples.
nA
= 9
and nB = 9
So the df
for our example is: nA
+ nB - 2 = 9 + 9 - 2 = 16
Remember, that
because we're using samples, we can only
estimate the values of the population parameters and so we're going to
need to take degrees of freedom into account.
So what is our
critical t for a 2-tailed test?
Critical t =
TINV(2α/tails, df) = TINV(2 * .05 / 2, 16) =
TINV(.05, 16) = 2.12
Now comes the big difference. We need to compute
our observed t statistic. At the conceptual level, the
formula is the same. However, at the practical level, it is a much more
complex because we have two samples, which means that we have two
estimates. Let's break the formula below into several parts.

In
other words, we're interested in
the difference between the two populations, so to compute the t
statistic we need to see if the difference between our two samples is
different from the difference between the two populations.
So
the numerator is straightforward:
is the difference between
the sample means
is
the difference between
the population means. Since H0 says the means are the same, the
difference between
them always is 0.
The
denominator is where things become
complex:
is the estimated
standard error of the difference between sample means. That is, when we
estimate the difference in population means using 2 sample means, we
are
typically going to be off somewhat in our estimates of the true
population difference. This error, "on average," is the standard error
of the difference between sample means.
The
formula is going to be a little
tricky but is based on the same idea as other standard errors. Let's
rewrite
the estimated standard error formula so both the numerator and
denominator are under the square root like this:

We
see that there is an estimated
variance (s2)
in the numerator and a sample size n in the denominator. Eventually,
we'll do roughly the same thing with the independent samples
t-test but it will require several steps to get there.
In
order to calculate the standard
error, we are going to need the formulas for the degrees of freedom
first.
dfA
and dfB are
simply the sample sizes of groups A and B minus 1.
The
total degrees of freedom is df = dfA
+ dfB = (nA -1) +
(nB - 1) = nA
- nB - 2
Second,
we need a sort of average
variance in the 2 samples. We call this, "pooled variance." It is a
measure of variance that is a weighted average of both samples'
variance. Here is the
formula for pooled variance:

An
alternate formula for pooled
variance that makes it easier to calculate from SPSS output is:

The
pooled variance is just a step on
the way to calculating the estimated standard
error of
the difference between sample means. Here is the formula for that:

Thus,
you can see that there is a sort
of variance in the numerators and sample sizes in the denominators.
Thus, although the formula looks very weird, at its core, it is very
much like the formula we've already seen.
Example:
Suppose
we have Groups 1 and 2 with 5 and 4 scores,
respectively.
| Person |
Score |
Group |
| 1 |
23 |
1 |
| 2 |
35 |
1 |
| 3 |
45 |
1 |
| 4 |
33 |
1 |
| 5 |
22 |
1 |
| 6 |
16 |
2 |
| 7 |
22 |
2 |
| 8 |
14 |
2 |
| 9 |
18 |
2 |
If we
calculate the means, sums of squares, and
degrees of freedom, we get the following:
|
Group
1 |
Group
2 |
| Mean |
31.6 |
17.5 |
s
|
9.476 |
3.416 |
| df |
4 |
3 |
So pooled variance = (4*9.476*9.476 + 3*3.416*3.416) / (4 + 3) = 56.31
and the estimated standard error = sqrt(56.31 / 5 + 56.31 / 4) = 5.03
Observed t = (31.6 - 17.5) / 5.03 = 2.80
Total
degrees of freedom = 4 + 3 = 7
Using Excel, we see that a 2-tailed critical t = TINV(2 * 0.05 / 2, 7)
= 2.36
Finally,
we see that the observed t of 2.80 is
larger than the critical t of 2.36 and thus we reject the null
hypothesis.
In ordinary language, we would conclude that Group 1 appears to come
from a population with a mean that is significantly larger than that of
Group 2.
Assumptions
of all t-tests
(1)
The observations are independent
(both between and
within groups)
(2)
The two populations are normally
distributed
** New Assumption
** (3) The two
populations
have equal variances. This is referred to as the homogeneity of
variance assumption. Recall that in the formula we pool our
sample variances. This is an
okay thing to do if the variances are about the same. However, it isn't
okay if they are very different. SPSS provides a test for this
assumption. In the output for the Independent Samples Test, you'll see
a box labeled Levene's Test for Equality of Variances. If this test is
significant (the Sig. value is 0.05 or less), then there is evidence
that this assumption has been violated and a corrected formula must be
used. We won't deal directly with this corrected formula but we can
make use of it from SPSS output.
Using
SPSS to compute independent samples t-tests
| Note: To do an independent samples t-test
you'll need to have two variables (columns) in your data file. One
column will contain the data (your dependent measure). The other column
will be an independent variable that specifies which group the subject
belongs to (e.g., 1 for group 1, 2 for group 2). |
| Person |
Score |
Group |
| 1 |
23 |
1 |
| 2 |
35 |
1 |
| 3 |
45 |
1 |
| 4 |
33 |
1 |
| 5 |
22 |
1 |
| 6 |
16 |
2 |
| 7 |
22 |
2 |
| 8 |
14 |
2 |
| 9 |
18 |
2 |
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| Go to the Analyze menu and select the submenu
Compare Means. In this submenu you'll see several tests. The one that
we're interested in today is independent samples t-test. |

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| After selecting Independent samples
t-test, you'll get a window that
looks like this. Here you should select the variables that you are
testing. Your test variable is your dependent variable. Your group
variable is the independent variable that assigns each subject to a
group. |

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| Before you can do the analysis, you must
define the groups. Click the button and then enter the values that you
used to define the groups (e.g., 1 for group 1 and 2 for group 2). |

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| Here is what the output will look like. |

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Notice that SPSS doesn't tell you
to reject or
fail to reject the H0, nor does it give you the critical t.
To make your decision about the H0 you must compare the
p-value with your &alpha-level. If the p-value
is equal to or smaller than the your &alpha-level,
then you should reject the H0, otherwise you should fail to
reject H0.
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You may also
notice that there are two rows of numbers in the t-test output. One row
"assumes equal variance" the other doesn't. This is related to the
assumption of homogeneity of variance discussed above. If the
Levene's test is not significant (look at the Sig. value), then we can
assume equal variances and use the values in that row. If Levene's test
is significant, we must use the values in the second row.
In this case, the Levene's test is not significant (p = .981) so we
read the upper row (Levene's test is not significant most of the time
which means that the homogeneity of variance assumption has been met.).
The p-value of the t-test is .937 so the null hypothesis is retained
(i.e., the means are not significantly different).
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Download the worksheet
here to answer the questions.
Email it to your GA when you are finished.
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