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2 Sample Tests – Small Samples
1. Small sample, independent groups
a. Test of equality of population variances
b. If variances are equal, t-test
c. If variances are not equal, Wilcoxon Rank
Sum Test
2. Examples
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1b. Small samples, independent groups
We now turn to the case of comparing means for two
independent, small samples (ns < 30).
* There are 2 ways to do this – depending upon
whether the two population variances are equal
or different.
* In order to know which method we should use,
we have to test the hypothesis H0: 12 = 22
* So for small, independent samples, there are
always 2 steps– test the variances, then test the
means.
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1b. Small samples, independent groups
VERY IMPORTANT POINT:
* We can only use the independent groups t-test
when the two population variances are equal.
* We must not assume that 12 = 22.
* We must test H0: 12 = 22.
* The test of hypothesis about the two population
variances uses the ratio F= (12 / 22).
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1b. Small samples, independent groups
On an exam, you must test the hypothesis of equal
variances before doing the independent groups ttest!
If H0: 12 = 22 is rejected, we use the Wilcoxon
Rank Sum test instead of the t-test.
Note: before t-test only; not before Z test.
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Test of hypothesis of equal variances
Notes:
* Next slide shows a formal statement of test of
hypothesis about two population variances.
* Both one-tailed and two-tailed tests are shown.
* When you test equality of variances before
doing small sample, independent groups t-test,
always do a two-tailed test.
* One-tailed test of equality of variances has
other uses.
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Test of hypothesis of equal variances
H0: 12 = 22
HA: 12 < 22
or 12 > 22
Test statistic:
H0: 12 = 22
HA: 12 ≠ 22
F=
S12
S22
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Test of hypothesis of equal variances
Rejection region:
One-tailed
Two-tailed:
F > Fα
F > Fα/2
d.f. = (n1 – 1), (n2 – 1)
(See note on next slide.)
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α given in F table is
value for upper tail.
Since the F distribution
is not symmetric, we
have to compute critical
F for lower tail.
α
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Computing critical F values for lower tail
Critical F for upper tail of distribution is found in
Table VII, using α and d.f.
Critical F for lower tail of distribution:
1
Fα, n2-1, n1-1
Note that d.f. are inverted!
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1b. Small samples, independent groups
Now – back to our t-test.
* If you do NOT reject H0 in the test of equality of
variances, then you can pool the two sample
variances:
Sp2 = (n1-1)s12 + (n2-1)s22
n1 + n 2 - 2
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1b. Small samples, independent groups
H0: 1 – 2 = D0
HA: 1 – 2 > D0
or: 1 – 2 < D0
Test statistic:
H0: 1 – 2 = D0
HA: 1 – 2 ≠ D0
t = ( X1 X2) – D0
Sp2 1
n1
(
1
n2
)
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1b. Small samples, independent groups
Rejection region:
One-tailed:
t < -tα
or t > tα
Two-tailed:
│t│>tα/2
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1b. Small samples, independent groups
Wilcoxon Rank Sum Test
* first, combine the two samples and rank order
all the observations.
* smallest number has rank 1, largest number
has rank N (= sum of n1 and n2).
* separate samples and add up the ranks for the
smaller sample. (If n1 = n2, choose either one.)
* test statistic : rank sum T for smaller sample.
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1b. Small samples, independent groups
Wilcoxon – One-tailed Hypotheses:
H0: Prob. distributions for 2 sampled populations
are identical.
HA: Prob. distribution for Population A shifted to
right of distribution for Population B. (Note: could
be to the left, but must be one or the other, not
both.)
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1b. Small samples, independent groups
Wilcoxon – Two-tailed Hypotheses:
H0: Prob. distributions for 2 sampled populations
are identical.
HA: Prob. distribution for Population A shifted to
right or left of distribution for Population B.
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1b. Small samples, independent groups
Wilcoxon – Rejection region:
(With Sample taken from Population A being
smaller than sample for Population B) – reject H0
if
TA ≥ TU or TA ≤ TL
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1b. Small samples, independent groups
Wilcoxon for n1 ≥ 10 and n2 ≥ 10:
Test statistic:
Z=
TA – n1(n1 + n2 + 1)
2
n1n2(n1 + n2 + 1)
12
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Wilcoxon for n1≥ 10 and n2 ≥ 10
Rejection region:
One-tailed
Two-tailed
Z > Zα
│Z│ > Zα/2
Note: use this only when n1≥ 10 and n2 ≥ 10
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Example 1
A retail store sales consultant is asked to determine
whether customers at stores which require memberships
spend more on average than customers at ordinary stores
where you don’t have to buy a membership to shop. She
surveys as group of 7 randomly selected customers
leaving no-membership-required stores and 8 different
randomly selected customers leaving membersip-required
stores, recording the amount of their purchases shown on
their sales receipts. The data are shown below. Is there
evidence that customers at the membership-required
stores, on average, spend more money on a single trip to
the store than customers at no-membership-required
stores? (α =.05)
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Example 1
Amount spent
No membership required
41
28
38
32
24
36
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Membership required
44
36
35
43
39
36
50
37
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Example 1
This question involves a test of hypothesis about 2
population means, using the means of two small,
independent samples.
Thus, the first step is the test of hypothesis about
the population variances, σ12 and σ22.
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Example 1
No membership required:
X = 32.57
σ2 = 7646 – 7426.29 = 36.62
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Membership required:
X = 40.0
σ2 = 12992 – 12800
= 27.42
7
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Example 1
HO: σ12 = σ22
HA: σ12 ≠ σ22
F critical = F (6, 7, .025) = 5.12 (from table)
Fobt = 36.62
27.42
= 1.355 < 5.12
Therefore, we can do a t-test.
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Example 1
HO: μnon2 = μmem2
HA: μnon2 < μmem2
t critical = t (n – 2, α) = t (13, .05) = 1.771
s2p = 6 (36.62) + 7 (27.42) = 31.67
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Example 1
tobt =
40 – 32.57
31.67 + 31.67
7
8
=
7.43
4.524 + 3.959
=
7.43
2.913
= 2.55
 reject HO.
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Example 2
Many people are confused about the distinction
between Cajun and Creole cooking. One possible
distinction is that Cajun food contains more
cayenne and other peppers and is thus “hotter”
than Creole food. To test this hypothesis, the
“heat” (measured in Scoville Units) of random
samples of Cajun and Creole dishes is compared,
yielding the data on the next slide. Do these data
support the hypothesis that Cajun dishes are
hotter than Creole dishes (α = .05)?
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Example 2
Scoville Heat Units
Cajun Dishes
Creole Dishes
3500
3100
4200
4700
4100
2700
4700
3500
4200
2000
3705
3100
4100
1550
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Example 2
Cajun:
σ2 = 890592.8572
6
= 148432.14
Membership required:
σ2 = 66334999.98
= 1055833.33
7
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Example 2
HO: σ12 = σ22
HA: σ12 ≠ σ22
F critical = F (6, 6, .025) = 5.82 (from table)
Fobt = 1055833.33
= 7.11 > 5.82
148432.14
Therefore, we cannot do a t-test. We do a
Wilcoxon.
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Example 2
HO: Cajun distribution and Creole distribution are
identical.
HA: Cajun distribution is shifted to the right relative
to the Creole distribution.
Test statistic: T (rank sum total) for Cajun dishes
T critical: TU = 66 or TL = 39
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Example 2
Scoville Heat Units
Cajun Rank
Creole Rank
3500 6.5
3100 4.5
4200 11.5
4700 13.5
4100 9.5
2700 3
4700 13.5
3500 6.5
4200 11.5
2000 2
3705 8
3100 4.5
4100 9.5
1550 1
70
35
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Example 2
Check accuracy of rank sum total T:
TA + TB = 105 = n*(n+1) = 14*15 = 105
2
2
TA > TU = 70. Reject HO – Cajun dishes are hotter
than Creole dishes.
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