'
I. Confidence intervals and tests for the
difference of two means (Section 9.3)
Often we’re interested in comparing two
groups. For example:
$
• Are the rates of teen smoking the same
among white and minority groups? If
not, how large is the difference?
• Is a new medication more effective at
reducing blood pressure than the
currently used medication? If so, how
much more effective?
We’ll learn how to answer questions about
the difference in means between the two
groups.
&
1
%
$
'
Basic setting
• Population 1 has mean µ1 and standard
deviation σ1 .
• Population 2 has mean µ2 and standard
deviation σ2 .
• We’re interested in µ1 − µ2 .
• We have independent samples of size n1
and n2 from the two populations.
• We typically will not know the
population standard deviations σ1 and
σ2 , so we’ll estimate them by the sample
standard deviations S1 and S2 .
• We’ll assume that the populations are
normally distributed, or that the sample
sizes n1 and n2 are large. (See the text
for details.)
&
2
%
$
'
Example:
• A new medication has been developed to
reduce blood pressure.
• Let µ1 represent the mean reduction in
blood pressure using the new medication
for people with high blood pressure.
• Let µ2 represent the mean reduction in
blood pressure using the old medication
for people with high blood pressure.
• We’re interested in testing the hypotheses
H0 : µ 1 − µ 2 ≤ 0
Ha : µ 1 − µ 2 > 0
and in forming a confidence interval for
µ1 − µ 2 .
• We randomly assign n1 = 20 people with
high blood pressure to use the new
medication, and n2 = 15 people with high
blood pressure to use the old medication.
&
3
%
'
The t distribution again
$
• The difference in sample means X 1 − X 2
is a good estimator of µ1 − µ2 .
• As usual, we’ll standardize the estimator.
• Fact: The standardized estimator
(X 1 − X 2 ) − (µ1 − µ2 )
p
S12 /n1 + S22 /n2
has a t distribution.
&
4
%
'
• We need to know how many degrees of
freedom!
$
• Answer 1: There is a complicated (but
accurate) formula that gives the degrees
of freedom in the text. Most software
(e.g. Minitab) uses this formula.
• Answer 2 (Easier but less accurate):
Use the minimum of n1 − 1 and n2 − 1 as
the degrees of freedom.
• We’ll always use Answer 2.
&
5
%
$
'
Formulas
• Confidence interval for µ1 − µ2 :
q
(X 1 − X 2 ) ± c S12 /n1 + S22 /n2 .
• We get the multiplier c from Table B.3,
the t table.
• Hypothesis test for whether µ1 − µ2 is 0:
– Test statistic is
X1 − X2
p
S12 /n1 + S22 /n2
– Compute p-values the usual way using
Table B.3.
&
6
%
'
Back to example
Recall that µ1 is the mean BP reduction
using the new medication and µ2 is the
mean BP reduction using the old
medication, and that n1 = 20 and n2 = 15.
$
• Collect data and find
X 1 = 19, S1 = 5
X 2 = 16, S2 = 3
• Want to test
H0 : µ 1 − µ 2 ≤ 0
Ha : µ 1 − µ 2 > 0
&
7
%
$
'
• Test statistic:
T =p
=p
19 − 16
52 /20 + 32 /15
3
25/20 + 9/15
≈ 2.206.
• Degrees of freedom: 14.
• The p-value is the area to the right of
2.206. From Table B.3 we can bound the
p-value to be between 0.01 and 0.025.
&
8
%
'
• A 99% confidence interval for the mean
difference µ1 − µ2 is
p
(19 − 16) ± 2.977 52 /20 + 32 /15
$
• This is about (−1.05, 7.05).
&
9
%
$
'
Example:
• Chapin Social Insight test (measures how
accurately a person appraises other
people) was given to a group of college
students.
• Question of interest: Do males and
females differ in average social insight?
• Formally we want to test
H0 : µ 1 − µ 2 = 0
Ha : µ1 − µ2 6= 0
where µ1 and µ2 are the mean scores for
males and females.
&
10
%
$
'
• Data:
Gender
Male
Female
Sample size
133
162
Sample mean
25.34
24.94
Sample s.d.
5.05
5.44
• Test statistic:
25.34 − 24.94
p
5.052 /133
+
5.442 /162
≈ 0.654.
• Compute a p-value from the t
distribution with 132 degrees of freedom.
• Since our table doesn’t have this, we’ll
use 120 degrees of freedom.
• We want twice the area to the right of
0.654.
• The area to the right of 0.845 is 0.2, so
the area to the right of 0.654 is more
than 0.2.
• So the p-value is more than 0.4.
&
11
%