Download File

T- and Z-Tests for Hypotheses
about the Difference between
Two Subsamples
Random samples, partitioned into two independent
subsamples (e.g., men and women).
Question: Are the means of some variable (such as
salary) significantly different between the two
subsample?
Key: The sampling distribution of all theoretically
possible differences between subsample means.
For large samples (i.e., when the Central Limit Theorem
holds), this sampling distribution of mean differences is
normally shaped; for smaller samples, the sampling
distribution takes the shape of one of the Student’s t
distributions, identified by degrees of freedom.
The key is: The difference between two means is a single
value. In the case of these so-called “means difference
tests,” the null hypothesis is that the means in general
(i.e., in the universe) do NOT differ. Symbolically,
H0: 2 - 1 = 0.00
There are two possible alternate hypotheses:
nondirectional;
H1: 2 - 1  0.0
and directional, either
H1: 2 - 1 > 0.0
or
H1: 2 - 1  0.0
In the 1984 General Social Survey, female respondents
were asked whether or not their mothers had attended
college. Then these female respondents were asked
about their own education levels. A reasonable
(alternate) hypothesis (H1) would be: Women whose
mothers attended college will themselves have more
formal education than women whose mothers did not
attend college. Thus, our two hypotheses are :
H1: 2 - 1 > 0.0
H0: 2 - 1 = 0.00
Dividing respondents into women whose mothers
attended college and women whose mothers did not
led to the calculation of the following statistics for the
two subsamples:
Mothers
ATTENDED
College
_
Y2 = 15.24
s22 = 6.57
N2 = 90
Mothers
DID NOT
Attend
_
Y1 = 12.57
s12 = 9.82
N1 = 359
Notice that the sample means DO in fact differ. This is
NOT the question.
Because these two subsamples combined exceed 100,
we know that the Central Limit Theorem applies. We
can convert the difference between the value of the
sample mean differences and the presumed value of
the mean difference in the universe under the null
hypothesis (i.e., 0.0) to z-values by using the estimated
standard error of the difference (the standard
deviation of the sampling distribution of sample mean
differences). Recall that in general standard errors are
estimated by dividing the standard deviation of the
sample by the square root of the sample size,
̂ 
sY
N
However, in the case of this means difference test, we
have TWO subsamples and thus TWO standard
deviations (actually variances in this example), one for
each subsample. What do we do? Simply combine the
two subsample variances, as follows:
ˆ y
2
 y1

2
2
2
1
s
s

N 2 N1
With the information from above, this means
 6.57   9.82 
ˆ y2  y1  


 90   359 
ˆ y
2

 y1
ˆ y
ˆ y
2
2
 y1
0.073  0.0274
 0.1004
 y1
 0.317
The algorithm for converting mean differences into zunits should look familiar:

Y
z
2
 Y1   2  1 
ˆ Y  Y
2
1
In this example,

15.24  12.57   0.0
z
0.317
2.67
z 
0.317
z  8.43
Selecting alpha = 0.05 for a one-tailed test and looking
for a critical value in Appendix 1 (pp. 540-542), we again
interpolate between 0.4495 and 0.4505, making
z = 1.645.
Since 8.43 is GREATER THAN the critical value of
1.645, we REJECT the null hypothesis at the 0.05 level
and conclude that IN GENERAL women whose mothers
attended college have higher education levels than
women whose mothers did not attend college.
Selecting alpha = 0.05 for a one-tailed test and looking
for a critical value in Appendix 1 (pp. 540-542), we again
interpolate between 0.4495 and 0.4505, making
z = 1.645.
Since 8.43 is GREATER THAN the critical value of
1.645, we REJECT the null hypothesis at the 0.05 level
and conclude that IN GENERAL women whose mothers
attended college have higher education levels than
women whose mothers did not attend college.
Z = 0.0
Z = 1.645
Z = 8.43
For random samples whose combined size is less than
120, we cannot assume that the sampling distribution of
mean differences will be normally shaped. This is
because the Central Limit Theorem doesn't hold with
samples this small. Student's t distributions must be
used instead.
The only new wrinkle here is that the standard error of
the difference cannot be estimated as above. The
sample standard deviations must be pooled in a way
that is sensitive to the impact of even slight differences
in small numbers.
Consider the following example:
The 63-city data set that we are using this semester has
been divided into two subsamples, one consisting of
SUNBELT cities and the other of FROSTBELT cities.
The question is: Did frostbelt cities lose population at a
higher rate than sunbelt cities in the fifteen years
between 1960-1974? Sample statistics are these:
Frostbelt Cities
_
Y2 = - 4.14
s22 = 9.98
N2 =
37
Sunbelt Cities
_
Y1 = 2.84
s12 = 57.61
N1 =
26
In the sample, frostbelt cities clearly lost population
(population change - 4.14 percent) at a greater rate
than sunbelt cities (which GAINED population, + 2.84
percent). The question is, is this sample difference
sufficient to infer a similar trend in the universe of
American cities?
Our alternate hypothesis is that in general frostbelt
cities lost population at a higher rate than sunbelt cities,
hence
H1: 2 - 1 < 0.0
In other words, we expect 2 to be a larger negative
number than 1. Notice again that the presence of the
“less than” sign (“<”) dictates a one-tailed test, this time
in the left-hand tail where negative mean differences
are located.
Our null hypothesis is that there is no difference in
the rates of population change, or
H0: 2 - 1 = 0.00
Since the subsample sizes are relatively small, we can't
simply slam the standard deviations together to
estimate the standard error of the difference. We must
weight the subsample standard deviations (actually the
subsample variances) by the size of the subsamples
(actually by the number of degrees of freedom in the
subsamples) before estimating the standard error. This
is called pooling.
s pooled
[( N 2  1) s  ( N1  1) s ]

( N 2  N1  2)
2
2
In this example,
s pooled
[(37  1)(9.98)  (26  1)(57.61)]

(37  26  2)
s pooled 
s pooled
[(36)(9.98)  (25)(57.61)]
61
359.280  1440.250

61
2
1
s pooled
1799.53

61
s pooled  29.500
s pooled  5.431
Now we can use this “pooled” constant to estimate the
standard error of the difference. The estimation is
ˆ y2  y1
where s = spooled
s2 s2
1 1

 s

N 2 N1
N 2 N1
ˆ y
ˆ y
2
2
 y1
ˆ y
2
ˆ y
 y1
1
1

37 26
 (5.431) 0.027  0.039
 y1
2
 (5.431)
 y1
ˆ y
2
 (5.431) 0.066
 (5.431)(0.257)
 y1
 1.396
Now we have the value of the standard error of the
difference, 1.396; this is our “currency exchange rate” that
will allow us to determine the value of the test statistic.
The algorithm should look familiar:

Y
t
2
 Y1   2  1 
ˆ Y  Y
2
1
In this example,
[( 4.14)  (2.84)]  0.0
t
1.396
 6.98
t 
1.396
t  5.00
We now know that location - 6.98 on the sampling
distribution converts to a t-location of - 5.00 on the
underlying X-axis. But what is the exact SHAPE of this
sampling distribution?
It is the Student’s t distribution with 61 degrees of
freedom. We have 63 cities selected randomly, but the
cities have been subdivided into two subsamples
(frostbelt and sunfelt cities). The 37 frostbelt-city
subsample has 36 degrees of freedom (37 - 1 = 36), and
the 26 sunbelt-city subsample has 25 degrees of
freedom (26 - 1 = 25). More generally, the number of
degrees of freedom in the t-test is
df = N2 + N1 - 2
Consulting Appendix 2 (p. 543) in search of the critical
value for df = 61, we once again find only critical values
for df = 60 and for df = 120.
Let's assume that we want to make our test with
alpha = 0.05. We could interpolate, by finding the value
that is 1 / 60th of the way from 1.671 to 1.645. This
value is extremely small and would round back to 1.671.
But we need the critical value for the left (negative) tail,
and Appendix 2 has only values for the right (positive)
tail. Because Student's t distributions are all
SYMMETRICAL, we simply add the negative sign to
1.671. Thus, our critical value is t0.05 = –1.671.
Since t = - 5.00 is GREATER THAN t = - 1.671, this
means that our sample difference lies INSIDE the region
of rejection in the left-hand tail. Thus, we REJECT the
null hypothesis at the 0.05 level and conclude that in
general frostbelt cities in the U.S. probably did lose
population at a greater rate than sunbelt cities in the
period 1960 and 1974.
t = - 5.00 t = - 1.671
t = 0.0
Using SAS to Produce T-Tests
libname old 'a:\';
libname library 'a:\';
options nonumber nodate ps=66;
proc ttest data=old.cities;
class agecity2;
var manufpct;
title1 'An Example of a T-Test';
title2 'SAS Version 8.1';
title3 'PPD 404';
run;
An Example of a T-Test
SAS Version 8.1
PPD 404
The TTEST Procedure
Statistics
Variable
Class
N
MANUFPCT
MANUFPCT
MANUFPCT
Newer
Older
Diff (1-2)
Lower CL
Mean
Mean
Upper CL
Mean
Lower CL
Std Dev
Std Dev
Upper CL
Std Dev
Std Err
22.928
20.1
-2.706
26.526
23.92
2.6063
30.125
27.74
7.9183
8.9262
7.2268
8.7659
10.949
9.2553
10.316
14.165
12.875
12.536
1.7761
1.8511
2.6565
38
25
T-Tests
Variable
Method
Variances
MANUFPCT
MANUFPCT
Pooled
Satterthwaite
Equal
Unequal
DF
t Value
Pr > |t|
61
57.1
0.98
1.02
0.3304
0.3139
Equality of Variances
Variable
Method
MANUFPCT
Folded F
Num DF
Den DF
F Value
Pr > F
37
24
1.40
0.3897
Reject the null hypothesis (H0)
when either:
1. the value of the statistical
test (2, z, t, F', or F) exceeds
the critical value at the
chosen -level; or,
2. the p-value for the statistical
test is smaller than the chosen
value of .
Do NOT reject the null
hypothesis (H0) when either:
1. the value of the statistical
test (2, z, t, F', or F) is less
than the critical value at the
chosen -level; or,
2. the p-value for the statistical
test is greater than the chosen
value of .
t = - 5.00 t = - 1.671 t = - 0.25 t = 0.0
Exercise 1
Means Difference Test
In the 1984 General Social Survey (GSS), 234 male respondents
with at least some college had a mean occupational prestige
score of 47.49 (with a variance of 213.28). In contrast, 351
male respondents with only a high school education or less had
an average occupational prestige score of 34.04 (with a
variance of 132.30). Test the null hypothesis (H 0) that there
is no statistically significant difference in occupational
prestige between these two groups. Assume that  = 0.05.
Perform a two-tailed test. Make your decision regarding the
null hypothesis using z-values in Appendix 1 (“Proportions of
Area under Standard Normal Curve"), pp. 540-542.
1.
What is the value of the standard error?
2.
What is the value of Z?
3.
What are the values of Z at the 2.5 percent and
97.5 percent areas under the normal curve?
4.
Do you reject or accept this null hypothesis?
Exercise 1 Answers
Means Difference Test
1.
What is the value of the standard error?
2.
What is the value of Z?
11.850
3.
What are the values of Z at the 2.5 percent and
97.5 percent areas under the normal curve?
 1.96
Do you reject or accept this null hypothesis?
Reject
4.
1.135
Exercise 2
Two Independent Samples t-test
In an experiment to determine the effects of hunger on handeye coordination, the following results, representing the
number of tasks completed success-fully, were obtained:
Experimental Group (#1)
(Hungry)
Mean
S.D.
N
14.0
2.449
10
Control Group (#2)
(Normal)
19.0
3.873
12
Calculate the estimated standard error of the difference,
obtain the value of t, and test the hypothesis that the
normally-fed (control) group performed better than did the
hungry (experimental) group. Use Student's t distribution
(Appendix 2, p. 543), and assume that  = 0.05. Perform a
one-tailed test.
Exercise 2 (continued)
Two Independent Samples t-test
1.
Expressed symbolically, what is the alternate
hypothesis?
2.
Expressed symbolically, what is the null
hypothesis?
3.
What is the value of the standard error?
4.
What is the value of t?
5.
How many degrees of freedom in this problem?
6.
What is the critical value of tdf?
7.
Do you reject or accept the null hypothesis?
Exercise 2 Answers
Two Independent Samples t-test
1.
Expressed symbolically, what is the alternate
hypothesis?
2 - 1 > 0.0
2.
Expressed symbolically, what is the null
hypothesis?
2 - 1 = 0.0
3.
What is the value of the standard error?
1.416
4.
What is the value of t?
3.530
5.
How many degrees of freedom in this problem?
6.
What is the critical value of tdf?
7.
Do you reject or accept the null hypothesis? Reject
20
+ 1.725