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Stat 250 Gunderson Lecture Notes
Learning about the Difference in Population Means
Part 2: Confidence Interval for a Difference in Population Means
Chapter 11: Section 4, CI Module 5
11.4 CI Module 5: Confidence Interval for the
Difference in Two Population Means
Lesson 1: The General (Unpooled) Case
 We have two populations or groups from which independent samples are available,
(or one population for which two groups can be formed using a categorical variable).
 The response variable is quantitative and we are interested in comparing the means
for the two populations.
A Typical Summary of the Responses for a Two Independent Samples Problem:
Population
Sample Size
Sample Mean
Sample Standard Deviation
1
n1
x1
s1
2
n2
x2
s2
Let 1 be the mean for the first population and  2 be the mean for the second population.
Parameter of interest: the difference in the population means 1   2 .
Sample estimate: the difference in the sample means x1  x2 .
Standard error: s.e. x1  x 2  
s12 s 22
where s1 and s2 are the sample standard deviations.

n1 n 2
So we have our estimate of the difference in the two population means, namely x1  x2 , and
we have its standard error. To make our confidence interval, we need to know the multiplier.
The multiplier t* is a t-value such that the area between –t* and t* equals the desired
confidence level. The degrees of freedom for the t-distribution will depend on whether we use
an ugly formula (used by software packages) or we use a conservative “by-hand” approach.
General Two Independent-Samples t Confidence Interval for 1 - 2
x1  x 2   t * s.e.( x1  x 2 )
where s.e. x1  x 2  
s12 s 22
and t * is the appropriate value for a t-distribution, and

n1 n 2
the df can be found using an approximation or conservatively as df = smaller of (n1 – 1 or n2 – 1)
This interval requires we have independent random samples from normal populations.
151
If the sample sizes are large (both > 30), the assumption of normality is not so crucial
and the result is approximate.
From Utts, Jessica M. and Robert F. Heckard. Mind on Statistics, Fourth Edition. 2012.
Used with permission.
Lesson
2:
The Pooled
Case
If we can further assume the population variances are (unknown but) equal, then there is
a procedure for which the t* multiplier is easier to find using an exact (not approximate)
t-distribution. It involves pooling the sample variances for an overall estimate and updating the
standard error accordingly.
It sometimes may be reasonable to assume that the measurements in the two populations have
the same variances …
so that ____________________ where _____ denotes the common population variance.
Since both sample variances would be estimating the common population variance, it would
make sense to combine or pool the two sample variances together to form an overall estimate.
Pooled standard deviation: s p 
(n1  1)s12  (n2  1)s22
n1  n2  2
Notes:
(1)
Each sample variance is weighted by the corresponding degrees of freedom.
So a larger sample size will result in a larger weight for that sample variance.
(2)
The denominator gives the total degrees of freedom:
df =
Replacing the individual standard deviations s1 and s2 with the pooled version sp in the formula
for the standard error leads to the pooled standard error of x1  x 2 is given by:
Pooled s.e.( x1  x 2 ) =
Pooled Two Independent-Samples t Confidence Interval for 1 - 2
x1  x 2   t * pooled s.e.( x1  x 2 )
where
pooled s.e.x1  x2   s p
1
1
(n1  1)s12  (n2  1)s22

and s p 
;
n1 n2
n1  n2  2
and t * is the appropriate value for a t(n1 + n2 – 2) distribution.
152
This interval requires we have independent random samples from normal populations with
equal population variances. If the sample sizes are large (both > 30), the assumption of
normality is not so crucial and the result is approximate.
Computer Package Notes:
(1) Most computer software provide results for both the unpooled and the pooled two
independent samples t procedures.
(2) Some computer software also provide the results of a Levene’s test for assessing if the
population variances can be assumed equal.
The null hypothesis for this test is that the population variances are equal. So a small pvalue for Levene’s test would lead to rejecting that null hypothesis and concluding that the
pooled procedure should not be used. Often a significance level of 10% is used for this
condition checking. Your lab module 8 provides more details about Levene’s test.
(3) Another graphical tool for assessing equal population variance is side-by-side boxplots and
comparing the IQRs. If the lengths of the boxes and overall ranges between the two groups
are very different, the pooled method may not be reasonable.
Here are some guidelines to help you assess which version to use (pages 432 - 433).

If the sample standard deviations are similar, the assumption of common population
variance is reasonable and the pooled procedure can be used. If the sample sizes
happen to be the same, the pooled and unpooled standard errors are equal anyway.
The advantage for the pooled version is that finding the df is simpler.
But the sample standard deviations are almost never exactly equal. So how similar is
similar enough?

If the larger standard deviation is from the group with the larger sample size, the
pooled procedure is acceptable because it will be conservative (produce a wider
interval).

If the smaller standard deviation is from the group with the larger sample size, the
pooled procedure can produce a misleading narrower interval.
Bottom-line: Pool if reasonable; but if the sample standard deviations are not similar,
we have the unpooled procedure that can be used.
153
Try It! Comparing Reading Scores
A study compared data on reading scores for students taught by certified teachers versus those
taught by uncertified teachers. Summaries of reading scores for both samples are provided.
Group Statistics
READING
SCORE
GROUP
1
2
N
Mean
34.21
28.20
10
10
Std. Deviation
9.118
7.036
Std. Error
Mean
2.883
2.225
Independent Sam ple s Te st
Levene's Test for
Equality of Varianc es
F
READING
SCORE
Equal variances
as sumed
Equal variances
not as sumed
.104
Sig.
.751
t-t est for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
St d. Error
Difference
1.651
18
.116
6.01
3.642
1.651
16.913
.117
6.01
3.642
Let 1 denote the average reading score for the population of students taught by a certified
teacher and 2 denote the average reading score for the population of students taught by an
uncertified teacher.
a. One of the assumptions for the pooled two independent samples confidence interval to be
valid is that the 2 populations (from which we took our samples) have the same standard
deviation. Look at the two sample standard deviations and the Levene's test result. Does
the assumption seem to hold (at the 10% level)? Explain.
b. Give a 95% confidence interval for the difference in the population mean reading scores,
that is, for 1 - 2.
c. Based on the interval, does there appear to be a difference in the mean reading scores for
the two populations? Explain.
154
Try It! Stroop’s Word Color Test
In Stroop’s Word Color Test, 100 words that are color names are shown in colors different from
the word. For example, the word red might be displayed in blue. The task is to correctly identify
the display color of each word; in the example just given the correct response would be blue.
Gustafson and Kallmen (1990) recorded the time needed to complete this test for n = 16
individuals after they had consumed alcohol and for n = 16 other individuals after they had
consumed a placebo drink flavored to taste as if it contained alcohol. Each group was balanced
with 8 men and 8 women.
In the alcohol group, the mean completion time was 113.75 seconds and the standard deviation
was 22.64 seconds. In the placebo group, the mean completion time was 99.87 seconds and the
standard deviation was 12.04 seconds.
Group
1 = alcohol
2 = placebo
Sample size
16
16
Sample mean
113.75
99.87
Sample standard deviation
22.64
12.04
We can assume that the two samples are independent random samples, that the model for
completion time is normal for each population.
a. What graph(s) would you make to check the normality condition? Be specific.
b. How did the two groups compare descriptively?
c. Which procedure? Pooled or unpooled? Why?
d. Calculate a 95% confidence interval for the difference in population means.
e. Based on the confidence interval, can we conclude that the population means for the two
groups are different? Why or why not?
155
What if?
Suppose the researchers Gustafson and Kallmen were convinced (based on past results) that
the underlying population variances were equal, so they prefer that a pooled confidence
interval be constructed.
The estimate of the common population standard deviation would be:
sp 
(n1  1) s12  (n2  1) s 22

n1  n2  2
(16  1)22.64  (16  1)12.04
 328.77  18.13
16  16  2
2
2
The pooled standard error for the difference in the two sample means would be:
pooled s.e.x1  x2   s p
1
1
1 1

 18.13

 6.41
n1 n2
16 16
which is the same as the unpooled standard error since the sample sizes were equal.
The t* multiplier would be based on df = 16 + 16 – 2 = 30, so t* = 2.04 (from Table A.2).
The 95% Pooled Confidence Interval would be:
x1  x 2   t * pooled s.e.( x1  x 2 )
 (13.88)  (2.04)(6.41)  13.88  13.08  (0.80, 26.96)
This interval still does not include 0, so the same decision would be made; however, the interval
is a bit narrower.
In this example, the unpooled interval may be a bit conservative (wider) but the evidence is still
strong to state the two population means appear to differ.
Note: Computer packages will compute both intervals for us, so we would not have to do all of
the computations above by hand!
156
Stat 250 Formula Card
Here is the summary of the confidence intervals for the big 5 parameters covered in
Chapters 10 and 11. We have now covered all 5 scenarios.
Population Proportion
Two Population Proportions
Population Mean
Parameter
p1  p 2
Parameter
pˆ 1  pˆ 2
Statistic
Standard Error
Parameter
p
p̂
Statistic
Standard Error
s.e.( pˆ ) 
pˆ (1  pˆ )
n
s.e.( pˆ 1  pˆ 2 ) 
Confidence Interval
x
Statistic
Standard Error
pˆ 1 (1  pˆ 1 ) pˆ 2 (1  pˆ 2 )

n1
n2
s.e.( x ) 
Confidence Interval
x  t *s.e.( x )
df = n – 1
*
 pˆ 1  pˆ 2   z *s.e. pˆ 1  pˆ 2 
Conservative Conf. Interval
pˆ 
s
n
Confidence Interval
pˆ  z s.e.( pˆ )

z*
Paired Confidence Interval
d  t *s.e.(d )
df = n – 1
2 n
 z* 

Sample Size n  

 2m 
2
Two Population Means
General
Parameter
Statistic
Standard Error
s.e.x1  x2  
1   2
x1  x 2
Pooled
Confidence Interval
x1  x2   t s.e.(x1  x2 )
x1  x 2
pooled s.e.x1  x2   s p
s12 s22

n1 n2
where s p 
*
1   2
Parameter
Statistic
Standard Error
1 1

n1 n2
(n1  1)s12  (n 2  1) s 22
n1  n 2  2
Confidence Interval
df = min( n1  1, n2  1)
x1  x2   t * pooled s.e.(x1  x2 )
157
df = n1  n2  2
Additional Notes
A place to … jot down questions you may have and ask during office hours, take a few extra notes, write
out an extra practice problem or summary completed in lecture, create your own short summary about
this chapter.
158