Independent Samples:
Comparing
Proportions
Lecture 35
Section 11.5
Mon, Nov 20, 2006
Comparing Proportions



We now wish to compare proportions between
two populations.
Normally, we would be measuring proportions
for the same attribute.
For example, we could measure the proportion
of NC residents living below the poverty level
and the proportion of VA residents living below
the poverty level.
Examples


The “gender gap” – the proportion of men who
vote Republican vs. the proportion of women
who vote Republican.
The proportion of teenagers who smoked
marijuana in 1995 vs. the proportion of
teenagers who smoked marijuana in 2000.
Examples

The proportion of patients who recovered,
given treatment A vs. the proportion of patients
who recovered, given treatment B.

Treatment A could be a placebo.
Comparing proportions


To estimate the difference between population
proportions p1 and p2, we need the sample
proportions p1^ and p2^.
The difference p1^ – p2^ is an estimator of the
difference p1 – p2.
Hypothesis Testing

See Example 11.8, p. 721 – Perceptions of the
U.S.: Canadian versus French.
 p1
= proportion of Canadians who feel positive
about the U.S..
 p2 = proportion of French who feel positive about
the U.S..
Hypothesis Testing

The hypotheses.
H0: p1 – p2 = 0 (i.e., p1 = p2)
 H1: p1 – p2 > 0 (i.e., p1 > p2)




The significance level is  = 0.05.
What is the test statistic?
That depends on the sampling distribution of
p1^ – p2^.
The Sampling Distribution of
p1^ – p2^

If the sample sizes are large enough, then p1^ is
N(p1, 1), where
p 1  p1 
  1
1

n1
Similarly, p2^ is N(p2, 2), where
2 
p2 1  p2 
n2
The Sampling Distribution of
p1^ – p2^

The sample sizes will be large enough if
n1p1  5, and n1(1 – p1)  5, and
 n2p2  5, and n2(1 – p2)  5.

The Sampling Distribution of
p1^ – p2^

Therefore,
ˆp1  pˆ 2 is N  p1  p2 ,  12   2 2 


where
1   2
2
2
p1 (1  p1 ) p2 (1  p2 )


n1
n2
1   2 
2
2
p1 (1  p1 ) p2 (1  p2 )

n1
n2
The Test Statistic

Therefore, the test statistic would be
Z
 pˆ1  pˆ 2   0
p1 1  p1  p2 1  p2 

n1


n2
if we knew the values of p1 and p2.
We could estimate them with p1^ and p2^.
But there may be a better way…
Pooled Estimate of p


In hypothesis testing for the difference between
proportions, typically the null hypothesis is
H0: p1 = p2
Under that assumption, p1^ and p2^ are both
estimators of a common value (call it p).
Pooled Estimate of p


Rather than use either p1^ or p2^ alone to
estimate p, we will use a “pooled” estimate.
The pooled estimate is the proportion that we
would get if we pooled the two samples together.
Pooled Estimate of p
(The “Batting-Average” Formula)
x1
pˆ 1   x1  n1 pˆ 1
n1
x2
pˆ 2   x2  n2 pˆ 2
n2
x1  x2 n1 pˆ 1  n2 pˆ 2
pˆ 

n1  n2
n1  n2
The Standard Deviation of p1 –
^

^
p2
This leads to a better estimator of the standard
deviation of p1^ – p2^.
1   2
2
2
1 1
 pˆ (1  pˆ )  
 n1 n2 
1   2 
2
2
1 1
ˆp (1  pˆ )  
 n1 n2 
Caution

If the null hypothesis does not say
H0: p1 = p2
then we should not use the pooled estimate p^,
but should use the unpooled estimate
 pˆ  pˆ 
1
2
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n2
The Test Statistic

So the test statistic is
Z
 pˆ1  pˆ 2   0
1 1
pˆ 1  pˆ   
 n1 n2 
The Value of the Test Statistic

Compute p^:
pˆ 

366  611
977

 0.2741.
1017  2547 3564
Now compute z:
0.3599  0.2399
0.12
Z

 7.253.
1  0.01655
 1
(0.2741)(0.7259)


 1017 2547 
The p-value, etc.



Compute the p-value:
P(Z > 7.253) = 2.059  10-13.
Reject H0.
The data indicate that a greater proportion of
Canadians than French have a positive feeling
about the U.S.
Exercise 11.34



Sample #1: 361 “Wallace” cars reveal that 270
have the sticker.
Sample #2: 178 “Humphrey” cars reveal that
154 have the sticker.
Do these data indicate that p1  p2?
Example

State the hypotheses.
H0: p1 = p2
 H1: p1  p2


State the level of significance.

 = 0.05.
Example

Write the test statistic.
z
 pˆ1  pˆ 2   0
1 1
pˆ 1  pˆ   
 n1 n2 
Example

Compute p1^, p2^, and p^.
270
pˆ 1 
 0.7479
361
154
pˆ 2 
 0.8652
178
270  154 424
pˆ 

 0.7866
361  178 539
Example

Now we can compute z.
0.7479  0.8652
 0.1173
Z

 3.126
1  0.03752
 1
(0.7866)(1  0.7866)


 361 178 
Example

Compute the p-value.



p-value = 2  normalcdf(-E99, -3.126)
= 2(0.0008861)
= 0.001772.
The decision is to reject H0.
State the conclusion.

The data indicate that the proportion of Wallace cars
that have the sticker is different from the proportion
of Humphrey cars that have the sticker.
TI-83 – Testing Hypotheses
Concerning p1^ – p2^


Press STAT > TESTS > 2-PropZTest...
Enter
x1, n1
 x2, n2



Choose the correct alternative hypothesis.
Select Calculate and press ENTER.
TI-83 – Testing Hypotheses
Concerning p1^ – p2^

In the window the following appear.
The title.
 The alternative hypothesis.
 The value of the test statistic z.
 The p-value.
 p1^.
 p2^.
 The pooled estimate p^.
 n1.
 n2.

Example

Work Exercise 11.34 using the TI-83.
Confidence Intervals for p1 –
^

The formula for a confidence interval for
p1^ – p2^ is
 pˆ1  pˆ 2   z

p1 1  p1  p2 1  p2 

.
n1
n2
Caution: Note that we do not use the pooled
estimate for p^.
^
p2
TI-83 – Confidence Intervals for
p1^ – p2^


Press STAT > TESTS > 2-PropZInt…
Enter
x1, n1
 x2, n2
 The confidence level.


Select Calculate and press ENTER.
TI-83 – Confidence Intervals for
p1^ – p2^

In the window the following appear.
The title.
 The confidence interval.
 p1^.
 p2^.
 n1.
 n2.

Example

Find a 95% confidence interval using the data in
Exercise 11.34.