Transcript Chapter 10: The t Test For Two Independent Samples

Chapter 10: The t Test For
Two Independent Samples
Independent-Measures Designs
• Allows researchers to evaluate the mean
difference between two populations using data
from two separate samples.
– The identifying characteristic of the independentmeasures or between-subjects design is the
existence of two separate or independent
samples.
– Thus, an independent-measures design can be
used to test for mean differences between two
distinct populations (such as men versus women)
or between two different treatment conditions
(such as drug versus no-drug).
Independent-Measures Designs
(cont'd.)
• The independent-measures design is used in
situations where a researcher has no prior
knowledge about either of the two populations
(or treatments) being compared.
– In particular, the population means and standard
deviations are all unknown.
– Because the population variances are not known,
these values must be estimated from the sample
data.
The t Statistic for an Independent-Measures
Research Design
• As with all hypothesis tests, the general purpose
of the independent-measures t test is to
determine whether the sample mean difference
obtained in a research study indicates a real
mean difference between the two populations
(or treatments) or whether the obtained
difference is simply the result of sampling error.
• Remember, if two samples are taken from the
same population and are given exactly the same
treatment, there will still be some difference
between the sample means. (i.e. sampling error)
The t Statistic for an IndependentMeasures Research Design (cont'd.)
• This difference is called sampling error
• The hypothesis test provides a standardized,
formal procedure for determining whether the
mean difference obtained in a research study is
significantly greater than can be explained by
sampling error
n = n1 = n2 or n1 ≠ n2
• if n = n1 = n2 , s 
2
p
(n  1) s12  (n  1) s 22
1 2 2
 ( s1  s 2 )
2
2(n  1)

   s
2 1
2
2
1
1
 s p  n  n   n s  s 
s(2M1 M 2 )  s2M  s 2p  1n  1n   s 2p  n2   12 s12  s 22
• so
2
M
s

s12
n1

s22
n2
2
n
1
n
1
2
• i.e. (10.1) = (10.4) (from p. 321 & p. 324)
• if n1 ≠ n2 , s 2p 
(n1  1) s 12  (n2  1) s 22
2
( M1  M 2 )
s
(n1  n2  2)
s
2
M
s

2 1
p n1
 n12

2
1
 s 22

Independent-measures t statistic
• The formula for t statistic:
( M 1  M 2 )  ( 1  2 )
t

sM
M1  M 2
1 1
s   
 n1 n2 
2
p
• df = df1 + df2 = n1 + n2 – 2
• (p. 322) another formula for pooled variance:
SS1  SS 2
s 
(n1  n2  2)
2
p
Comparing Population Means: Unknown
Population Standard Deviations (The Pooled
t-test)
Finding the value of the test
statistic requires two steps:
1.Pool the sample standard
deviations.
2.Use the pooled standard
deviation to compute the tstatistic.
2
2
(
n

1
)
s

(
n

1
)
s
1
2
2
s 2p  1
n1  n2  2
t
M1  M 2
1 1
s   
 n1 n2 
2
p
s2(∆M)  Var( M1 – M2 )
• E(M1 – M2) = μ1 – μ2
• With independent samples / populations:
• Var (M1 – M2) = Var (M1) + Var (M2)
• Box 10.1 (p. 321)
pop 1: X’s range = 20
pop 2: X’s range = 10
max(X1 – X2) – Min(X1 – X2) = 50 – 20 = 30
p. 325
3. I: n=4, SS=100, II: n=8, SS=140
SS1  SS 2
100  140 240
a. s 


 24
(n1  n2  2) 4  8  2
10
2
p
2
s

s

s
b. ( M1  M 2 )
M
p

1
n1

 n12  24 14  18   3
4. I: n=9, s2 =35, II: n=3, s2 =40
2
2
(
n

1
)
s

(
n

1
)
s
8  35  2  40
2
1
2

 36
a. s 2p  1
(n1  n2  2)
93 2
b. s( M1  M 2 )  s 2p

1
n1

 n12  36 19  13   4
5. df = 8 + 12 – 2 = 18
Hypothesis Tests and Effect Size with
the Independent Measures t Statistic
• To prepare the data for analysis, the first step is
to compute the sample mean (M) and SS (or s,
or s2) for each of the two samples.
• The hypothesis test follows the same four-step
procedure outlined in Chapters 8 and 9.
Hypothesis Testing with the
Independent-Measures t Statistic
(cont'd.)
• Step 1.
– State the hypotheses and select the α level. For
the independent-measures test, H0 states that
there is no difference between the two population
means.
• Step 2.
– Locate the critical region. The critical values for
the t statistic are obtained using degrees of
freedom that are determined by adding together
the df value for the first sample and the df value
for the second sample.
Hypothesis Testing with the
Independent-Measures t Statistic
(cont'd.)
• Step 3.
– Compute the test statistic. The t statistic for the
independent-measures design has the same
structure as the single sample t introduced in
Chapter 9. However, in the independentmeasures situation, all components of the t
formula are doubled: there are two sample
means, two population means, and two sources
of error contributing to the standard error in the
denominator.
Hypothesis Testing with the
Independent-Measures t Statistic
(cont'd.)
• Step 4.
– Make a decision. If the t statistic ratio indicates
that the obtained difference between sample
means (numerator) is substantially greater than
the difference expected by chance (denominator),
we reject H0 and conclude that there is a real
mean difference between the two populations or
treatments.
Example 10.1 (p.326)
•
•
•
•
•
I: n=10, M=93, SS=200 (watched..SS..)
Il: n=10, M=85, SS=160 (did not watch..SS..)
H1 : μ1 ≠ μ2 (two-tailed test)
df = 10+10-2 = 18, α = 0.01
critical value to mark the “rejection region”:
t0.005 (18) =  2.878
SS1  SS 2
200  160
s 

 20
(n1  n2  2)
18
2
p
s( M1  M 2 )  s 2p

1
n1

 n12  20101  101   2
• t = (93-85)/2 = 4 > 2.878  reject H0
• This is non-experimental case  cannot obtain the
causal relationship
Measuring Effect Size for the
Independent-Measures t
• Effect size for the independent-measures t is
measured in the same way that we measured
effect size for the single-sample t in Chapter 9.
• Specifically, you can compute an estimate of
Cohen’s d or you can compute r2 to obtain a
measure of the percentage of variance
accounted for by the treatment effect.
estimated d and r2
M1  M 2
• estimated d: d 
sp
• so,
d = (93 – 85) /√20 = 1.79 (very large effect!)
2
t
and r 2  2
= 4/(4+18) = 0.47 (very large!)
t  df
check p. 264 for d’s scale evaluation
check p. 299 for r’s scale evaluation
Another formula for r2
• Example 10.2 (p. 329)
If H0 is true, then there’s no difference between 2
populations and samples  combine them into
one sample: n=20, M=89, SS=680
 89 is the estimator of overall μ
unknown population  use M=89 as μ
SS1 = Σ(X-μ)2 = 680,
Calculate SS2 and r2
• and for each sample, i = 1,2
SS2  {[ X i  (M i   )]  }2   ( X i  M i    ) 2   ( X i  M i ) 2
• so
SS2   ( X1  M1 ) 2   ( X 2  M 2 ) 2 =200+160=360
Another formula for r2
with treatment effect: SS1 = 680
treatment effect removed: SS2 = 360
SS1  SS 2
• r 
= 320/680 = 0.47
SS1
2
Confidence Interval for μ1 – μ2
• Same as chapter 9:
 t  ,df
2
( M 1  M 2 )  ( 1   2 )

sM
 1  2  M 1  M 2  t  ,df sM
2
Example 10.3 (p.331)
• 95% confidence interval for example 10.1
• 1st, find the boundary t value: check t table
(p.703)  t0.025, 18 =  2.101
• so,
1  2  M 1  M 2  t  ,df sM
2
= (93 – 85)  2.1012
= 8  4.202  CI = (3.798, 12.202)
Confidence Intervals and Hypothesis
Tests
• If CI includes 0  μ1 – μ2 = 0 is acceptable
• If 0 is not included in CI  μ1 – μ2 = 0 is not
acceptable  reject it with 95% confidence
Confidence Intervals and Hypothesis
Tests
Write a research report....
• Should include the following information in your
report:
- t statistic with df
- p value
- Cohen’s d
- CI
- M, SD, etc.
p. 333
1. df = 16+16-2=30, α=0.05, t0.025, 30 =  2.042
SS1  SS 2
1005  1155
s 2p 

 72
(n1  n2  2) 16  16  2
s( M1  M 2 )  s 2p

1
n1

 n12  72161  161   3
t = (86-82.5)/3 = 1.17 < 2.042  failed to reject Ho (p > 0.05)
d = 3.5/√72 = 0.41248
90%CI = (86-82.5)  t0.05, 30 3=3.5 1.6973(-1.591, 8.591)
2. df = 28  Σn = 28+2 = 30
if α=0.05, t0.025, 28 =  2.048
t statistic = 2.27 > 2.048  reject Ho (p < 0.05)
One-tailed test (example 10.4) (p. 334)
• H1 : μ1 > μ2 (right-tailed test)
• α = 0.01, t0.01, 18 = 2.552
• t statistic = 4 > 2.552  reject Ho
 significant effect (getting higher grade)
The role of s2 and n in the independentmeasures t test
• s2 ↑  s∆M ↑  t ↓  less likely to reject Ho
• n ↑  s∆M ↓  t ↑  More likely to reject Ho
• s2 ↑  d, r2 ↓
• n has no effect on d
• n has small effect on r2
Sample Variance and Sample Size
Example 10.5 (p. 335-336)
•
•
•
•
Fig 10.7
sp = 1.22, sM  s p n11  n12  1.22 92  0.57511
df = 9+9-2 = 16,
t = (13-8)/0.58 = 8.62  reject Ho
• if s↑, other factors remain unchanged
• Fig 10.8: sp = 6.65  sM  6.65 92  3.13484
• t = (13-8)/3.14 = 1.59  p > 0.05
 failed to reject Ho
Comparing Population Means: Equal,
Unknown Population Standard Deviations
(The Pooled t-test)
The t distribution is used as the test statistic if one or more
of the samples have less than 30 observations. The
required assumptions are:
Both populations must follow the normal distribution.
The samples are from independent populations.
The populations must have equal standard deviations.
(homogeneity of variance) i.e. σ1 = σ2 = σ
Assume: treatment added constant value to each score 
μ was changed, but σ was unchanged
The Homogeneity of Variance
Assumption
• Most hypothesis tests usually work reasonably
well even if the set of underlying assumptions
are violated.
• The one notable exception is the assumption of
homogeneity of variance for the independentmeasures t test.
– Requires that the two populations from which the
samples are obtained have equal variances
– Necessary in order to justify pooling the two
sample variances and using the pooled
variance in the calculation of the t statistic
The Homogeneity of Variance
Assumption (cont'd.)
• If the assumption is violated, then the t statistic
contains two questionable values: (1) the value
for the population mean difference which comes
from the null hypothesis, and (2) the value for
the pooled variance.
• The problem is that you cannot determine which
of these two values is responsible for a t statistic
that falls in the critical region.
• In particular, you cannot be certain that rejecting
the null hypothesis is correct when you obtain an
extreme value for t.
The Homogeneity of Variance
Assumption (cont'd.)
• If the two sample variances appear to be
substantially different, you should use Hartley’s
F-max test to determine whether or not the
homogeneity assumption is satisfied.
• If homogeneity of variance is violated, Box 10.2
presents an alternative procedure for computing
the t statistic that does not involve pooling the
two sample variances.
2
s (l arg est )
F max  2
s ( smallest )
Hartley’s F-max test
• Procedure for the test:
1. compute s2 for each sample
2. compute F-max
large F-max  unequal variances
F-max close to 1  equal variances
3. find critical value in Table B.3 (p.704)
sample F-max > table value  unequal variances
Note:
k: # of samples (here, k=2)
df = n-1 for each sample (Hartley test assumes all samples are
the same size)
Example 10.6 (p. 339)
• n=10, s2 = 12.34 and 9.15
F-max = 12.34/9.15 = 1.35
if α = 0.05, check table (p. 704): critical value of F-max = 4.03; if
α = 0.01, critical value of F-max = 6.54
 data do not provide evidence that homogeneity of variance
assumption has been violated
• If that homogeneity of variance assumption has been violated
 use equation (10.1) with different df
sM 
df 
s12
n1

(V1  V2 )
V12
n1 1

s22
n2
( M 1  M 2 )  ( 1   2 )
t
sM
2
V22
n2 1
V1 
s12
n1
, V2 
s22
n2
Comparing Population Means with Unknown
AND Unequal Population Standard Deviations
Use the formula for the t-statistic shown if it is not reasonable to
assume the population standard deviations are equal.
The degrees of freedom are adjusted downward by a rather
complex approximation formula. The effect is to reduce the
number of degrees of freedom in the test, which will require a
larger value of the test statistic to reject the null hypothesis.
11-41
p. 340
1. n=8, I: M=63, s2 = 18 II: M=58, s2 = 14
a. α = 0.05 one-tailed test, df = 14, t0.05,14 = 1.761
sM 
s12
n1

s22
n2

1
8
(18  14)  2
 t = (63-58)/2 = 2.5 > 1.761  reject Ho
s12
n1
s22
n2
  18 (68  60)  4
b. s2 = 68, and s2 = 60 sM 
 t = (63-58)/4 = 1.25 < 1.761  failed to reject Ho
s12
n1
s22
n2
  321 (18  14)  1
c. n = 32  sM 
 t = (63-58)/1 = 5 > 1.761  reject Ho