The t-test
Inferences about Population Means
Questions
• What is the main use of the t-test?
• How is the distribution of t related to the unit
normal?
• When would we use a t-test instead of a ztest? Why might we prefer one to the other?
• What are the chief varieties or forms of the ttest?
• What is the standard error of the difference
between means? What are the factors that
influence its size?
More Questions
• Identify the appropriate to version of t
to use for a given design.
• Compute and interpret t-tests
appropriately.
• Given that
H 0 :   75; H1 :   75; s y  14; N  49; t(.05, 48)  2.01
construct a rejection region. Draw a
picture to illustrate.
Background
• The t-test is used to test hypotheses
about means when the population
variance is unknown (the usual case).
Closely related to z, the unit normal.
• Developed by Gossett for the quality
control of beer.
• Comes in 3 varieties:
• Single sample, independent samples,
and dependent samples.
What kind of t is it?
• Single sample t – we have only 1 group; want
to test against a hypothetical mean.
• Independent samples t – we have 2 means, 2
groups; no relation between groups, e.g.,
people randomly assigned to a single group.
• Dependent t – we have two means. Either
same people in both groups, or people are
related, e.g., husband-wife, left hand-right
hand, hospital patient and visitor.
Single-sample z test
• For large samples (N>100) can use z to
test hypotheses about means.
( X  X )2
( X  )
zM 
est. M
• Suppose

est . M
sX


N
N 1
N
H 0 :   10; H1 :   10; s X  5; N  200
• Then
est. M
• If
sX
5
5



 .35
N
200 14.14
X  11  z 
(11  10)
 2.83; 2.83  1.96  p  .05
.35
The t Distribution
We use t when the population variance is unknown (the
usual case) and sample size is small (N<100, the usual
case). If you use a stat package for testing hypotheses
about means, you will use t.
The t distribution is a short, fat relative of the normal. The shape of t depends on
its df. As N becomes infinitely large, t becomes normal.
Degrees of Freedom
For the t distribution, degrees of freedom are always a
simple function of the sample size, e.g., (N-1).
One way of explaining df is that if we know the total or
mean, and all but one score, the last (N-1) score is not free to
vary. It is fixed by the other scores. 4+3+2+X = 10. X=1.
Single-sample t-test
With a small sample size, we compute the same numbers
as we did for z, but we compare them to the t distribution
instead of the z distribution.
H 0 :   10; H1 :   10; s X  5; N  25
est. M
sX
5


1
N
25
X  11  t 
t (.05,24)  2.064 (c.f. z=1.96)
Interval =
(11  10)
1
1
1<2.064, n.s.
X  tˆ M
11  2.064(1)  [8.936, 13.064]
Interval is about 9 to 13 and contains 10, so n.s.
Review
 How are the distributions of z and t related?
 Given that
H 0 :   75; H1 :   75; s y  14; N  49; t(.05, 48)  2.01
construct a rejection region. Draw a picture
to illustrate.
Difference Between Means (1)
• Most studies have at least 2 groups
(e.g., M vs. F, Exp vs. Control)
• If we want to know diff in population
means, best guess is diff in sample
means.
• Unbiased: E( y1  y2 )  E( y1)  E( y2 )  1  2
• Variance of the Difference: var( y  y )  
2
2





• Standard Error: diff
M1
M2
1
2
2
M1
  M2 2
Difference Between Means (2)
• We can estimate the standard error of
the difference between means.
est. diff  est. M2 1  est. M2 2
• For large samples, can use z
( X1  X 2 ) ( 1  2 ) H 0 : 1   2  0; H1 : 1   2  0
zdiff 
est diff
X  10; N  100; SD  2
1
1
1
X 2  12; N 2  100; SD2  3
est. diff
z diff 
4
9
13



 .36
100 100
100
2
(10  12)  0

 5.56; p  .05
.36
.36
Independent Samples t (1)
( y1  y2 ) ( 1  2 )
est diff
• Looks just like z: tdiff 
• df=N1-1+N2-1=N1+N2-2
• If SDs are equal, estimate is:
 diff 
 1
1 

   

N1 N 2
N
N
2
 1
2
2
2
Pooled variance estimate is weighted average:
 2  [( N1  1)s12  ( N 2  1)s22 ] /[1 /( N1  N 2  2)]
Pooled Standard Error of the Difference (computed):
est. diff
( N1  1) s12  ( N 2  1) s22  N1  N 2 



N1  N 2  2
N
N
 1 2 
Independent Samples t (2)
est. diff
tdiff 
( N1  1) s12  ( N 2  1) s22

N1  N 2  2
( y1  y2 ) ( 1  2 )
est diff
 N1  N 2 


N
N
 1 2 
H 0 : 1   2  0; H1 : 1   2  0
y1  18; s12  7; N1  5
y2  20; s22  5.83; N 2  7
est. diff
t diff
4(7)  6(5.83) 12 

 1.47


5  7  2  35 
(18  20)  0  2


 1.36; n.s.
1.47
1.47
tcrit = t(.05,10)=2.23
Review
What is the standard error of the
difference between means? What are
the factors that influence its size?
Describe a design (what IV? What
DV?) where it makes sense to use the
independent samples t test.
Dependent t (1)
Observations come in pairs. Brother, sister, repeated measure.
2
 diff
  M2 1   M2 2  2 cov( y1 , y2 )
Problem solved by finding diffs between pairs Di=yi1-yi2.
(D )

D
i
N
s 
2
D
D  E(D )
t
est. MD
2
(
D

D
)
 i
N 1
est. MD
df=N(pairs)-1
sD

N
Dependent t (2)
Brother
5
7
3
y 5
sD 
Sister
7
8
3
y6
2
(
D

D
)

N 1
1
Diff
2
1
0
(D  D )2
1
0
1
D 1
est. MD  1 / 3  .58
D  E(D ) 1
t

 1.72
est. MD
.58
Assumptions
• The t-test is based on assumptions of
normality and homogeneity of variance.
• You can test for both these (make sure
you learn the SAS methods).
• As long as the samples in each group
are large and nearly equal, the t-test is
robust, that is, still good, even tho
assumptions are not met.
Review
• Describe a design where it makes sense
to use a single-sample t.
• Describe a design where it makes sense
to use a dependent samples t.
Strength of Association (1)
• Scientific purpose is to predict or
explain variation.
• Our variable Y has some variance that
we would like to account for. There are
statistical indexes of how well our IV
accounts for variance in the DV. These
are measures of how strongly or closely
associated our Ivs and DVs are.
• Variance accounted for:
2
2
2



(



)
 2  Y 2 Y|X  1 2 2
Y
4 Y
Strength of Association (2)
• How much of variance in Y is
 
associated with the IV?   
2
2
Y
2
Y|X
2
Y
( 1   2 ) 2

4 Y2
Compare the 1st (left-most) curve with the curve in the
middle and the one on the right.
In each case, how
much of the variance
in Y is associated
with the IV, group
membership? More
in the second
comparison. As
mean diff gets big, so
does variance acct.
0.4
0.3
0.2
0.1
0.0
-4
-2
0
2
4
6
Association & Significance
• Power increases with association (effect
size) and sample size.
• Effect size: d  ( X  X ) / 
• Significance = effect size X sample
(X   )
(
X

X
)
t
size.
t
1
1
(independent
samples)
p
2
1
1 


N
N
2
 1
 p2 
2
(single
sample)
 2 
N
 
td N
Increasing sample size does not increase effect size
(strength of association). It decreases the standard
error so power is greater, |t| is larger.
Estimating Power (1)
• If the null is false, the statistic is no
longer distributed as t, but rather as
noncentral t. This makes power
computation difficult.
• Howell introduces the noncentrality
parameter delta to use for estimating
power. For the one-sample t,
 d n
Recall the relations between t
and d on the previous slide
Estimating Power (2)
• Suppose (Howell, p. 231) that we have
25 people, a sample mean of 105, and a
hypothesized mean and SD of 100 and
15, respectively. Then
d
105  100
 1 / 3  .33
15
Howell presents an appendix
where delta is related to
  d n  .33 25  1.65 power. For power = .8, alpha
= .05, delta must be 2.80. To
power  .38
solve for N, we compute:
    2.8 
2
  d n; n     
  8.48  71.91
 d   .33 
2
2
Estimating Power (3)
• Dependent t can be cast as a single
sample t using difference scores.
• Independent t. To use Howell’s
method, the result is n per group, so
double it. Suppose d = .5 (medium
effect) and n =25 per group.
 d
n
25
 .50
 .5 12.5  1.77
2
2
 
 2.8 
n  2   2
  62.72
d 
 .5 
2
2
Need 63 per group.
From Howell’s appendix, the
value of delta of 1.77 with
alpha = .05 results in power of
.43. For a power of .8, we
need delta = 2.80
SAS Proc Power – single
sample example
proc power;
onesamplemeans test=t
nullmean = 100
The POWER Procedure
mean = 105
One-sample t Test for Mean
stddev = 15
Fixed Scenario Elements
power = .8
Distribution
Normal
Method
Exact
ntotal = . ;
Null Mean
100
run;
Mean
105
Standard Deviation
15
Nominal Power
0.8
Number of Sides
2
Alpha
0.05
Computed N Total
Actual
N
Power Total
0.802
73
;
2 sample t Power
Calculate sample size
proc power;
twosamplemeans
meandiff= .5
stddev=1
power=0.8
ntotal=.;
run;
Two-sample t Test for Mean Difference
Fixed Scenario Elements
Distribution
Normal
Method
Exact
Mean Difference
0.5
Standard Deviation
1
Nominal Power
0.8
Number of Sides
2
Null Difference
0
Alpha
0.05
Group 1 Weight
1
Group 2 Weight
1
Computed N Total
Actual
N
Power Total
0.801 128
2 sample t Power
• proc power;
• twosamplemeans
• meandiff = 5 [assumed
difference]
• stddev =10 [assumed SD]
• sides = 1
[1 tail]
• ntotal = 50 [25 per
group]
• power = .; *[tell me!];
• run;
The POWER Procedure
Two-Sample t Test for Mean Difference
Fixed Scenario Elements
Distribution
Method
Number of Sides
Mean Difference
Standard Deviation
Total Sample Size
Null Difference
Alpha
Group 1 Weight
Group 2 Weight
Normal
Exact
1
5
10
50
0
0.05
1
1
Computed Power
Power
0.539
Typical Power in Psych
• Average effect size is about d=.40.
• Consider power for effect sizes between
.3 and .6. What kind of sample size do
we need for power of .8?
proc power;
twosamplemeans
meandiff= .3 to .6 by .1
stddev=1
power=.8
ntotal=.;
plot x= power min = .5 max=.95;
run;
Typical studies are underpowered.
Two-sample t Test for 1
Computed N Total
Mean Actual
N
Index Diff Power Total
1 0.3 0.801 352
2 0.4 0.804 200
3 0.5 0.801 128
4 0.6 0.804
90
Power Curves
Why a whopper of an IV is helpful.
600
Total Sample Size
500
400
300
200
100
0
0.5
0.6
0.7
0.8
Power
Mean Diff
0.3
0.4
0.5
0.6
0.9
1.0
Review
• About how many people total will you
need for power of .8, alpha is .05 (two
tails), and an effect size of .3?
• You can only afford 40 people per
group, and based on the literature, you
estimate the group means to be 50 and
60 with a standard deviation within
groups of 20. What is your power
estimate?