+
2, 4, 6, 8, 12, 16,
18, 24, 36, 38, 46,
54, 58, 60 ,66
Chapter 10: Comparing Two Populations or Groups
Section 10.1
Comparing Two Proportions
The Practice of Statistics, 4th edition – For AP*
STARNES, YATES, MOORE
+
Chapter 10
Comparing Two Populations or Groups
 10.1
Comparing Two Proportions
 10.2
Comparing Two Means
+ Section 10.1
Comparing Two Proportions
Learning Objectives
After this section, you should be able to…

DETERMINE whether the conditions for performing inference are
met.

CONSTRUCT and INTERPRET a confidence interval to compare
two proportions.

PERFORM a significance test to compare two proportions.

INTERPRET the results of inference procedures in a randomized
experiment.
+
What if we want to compare the effectiveness of Treatment 1 and
Treatment 2 in a completely randomized experiment? This time,
the parameters p1 and p2 that we want to compare are the true
proportions of successful outcomes for each treatment. We use
the proportions of successes in the two treatment groups to make
the comparison. Here’s a table that summarizes this situation,
whether it is a population or treatment.
Comparing Two Proportions
Suppose we want to compare the proportions of individuals with a
certain characteristic in Population 1 and Population 2. Let’s call
these parameters of interest p1 and p2. The ideal strategy is to
take a separate random sample from each population and to
compare the sample proportions with that characteristic.
+
 Introduction

In Chapter 7, we saw that the sampling distribution of a sample
proportion has the following properties:
Shape Approximately Normal if np ≥ 10 and n(1 - p) ≥ 10
Center  pˆ  p
p(1  p)
if the sample is no more than 10% of the population
n
To explore the sampling distribution of the difference between two
proportions, let’s start with two populations having a known proportion of
successes.
Spread  pˆ 
 At School 1, 70% of students did their homework last night
 At School 2, 50% of students did their homework last night.
Suppose the counselor at School 1 takes an SRS of 100 students and
records the sample proportion that did their homework.
School 2’s counselor takes an SRS of 200 students and records the
sample proportion that did their homework.
What can we say about the difference pˆ1  pˆ 2 in the sample proportions?
+
Sampling Distribution of a Difference
Between Two Proportions
Comparing Two Proportions

 The
Using Fathom software, we generated an SRS of 100 students from
School 1 and a separate SRS of 200 students from School 2. The
difference in sample proportions was then calculated and plotted. We
repeated this process 1000 times. The results are below:
What do you notice about the shape, center, and spread
of the sampling distribution of pˆ1  pˆ 2 ?
Comparing Two Proportions
Sampling Distribution of a Difference
Between Two Proportions
+
 The
Both pˆ1 and pˆ 2 are random variables. The statistic pˆ1  pˆ 2 is the difference
of these two random variables. In Chapter 6, we learned that for any two
independent random variables X and Y,
X Y  X  Y and  X2 Y   X2   Y2
The
Sampling Distribution of the Difference Between Sample Proportions
Therefore,
 pˆ1  pˆ 2 
size
p1 np2 from Population 1 with proportion of successes
ˆ1   pˆof
Choose
an pSRS
2
1
p1 and an independent SRS of size n2 from Population 2 with proportion of
 2pˆ1  pˆ 2  p2pˆ21 .  2pˆ 2
successes
2
2
 pn1(1
Shape When
p , n (1 p1
), np22p(1
p2 )n2 (1  p2 ) are all at least 10, the
2 and
1 1 p11 )
 
 ˆ  ˆ

sampling distribution
of
p

p
is
approximately
Normal.
n
n

 1 2 2

1
Center Thepmean
(1 p1of
) thep2sampling
(1 p2 ) distribution is p1  p2 . That is,
 1

the difference in
is an unbiased estimator of
n1 sample proportions
n2
the difference in population propotions.
p (1 p1 ) deviation
p (1 p2of) the sampling distribution of pˆ1  pˆ 2 is
Spread The standard
 pˆ1  pˆ 2  1
 2
n1
n 2 p (1  p ) p (1  p )
1
1
2
 2
n1
n2
as long as each sample is no more than 10% of its population (10% condition).
+
Sampling Distribution of a Difference
Between Two Proportions
Comparing Two Proportions

 The
Who Does More Homework?
a) Describe the shape, center, and spread of the sampling distribution of pˆ1  pˆ 2.
Because n1 p1 =100(0.7) =70, n1 (1 p1 )  100(0.30)  30, n2 p2 = 200(0.5) =100
and n2 (1 p2 )  200(0.5)  100 are all at least 10, the sampling distribution
of pˆ1  pˆ 2 is approximately Normal.
Its mean is p1  p2  0.70  0.50  0.20.
Its standard deviation is
0.7(0.3) 0.5(0.5)

 0.058.
100
200
Comparing Two Proportions
Suppose that there are two large high schools, each with more than 2000 students, in a certain
town. At School 1, 70% of students did their homework last night. Only 50% of the students at
School 2 did their homework last night. The counselor at School 1 takes an SRS of 100
students and records the proportion that did homework. School 2’s counselor takes an SRS of
200 students and records the proportion that did homework. School 1’s counselor and School
2’s counselor meet to discuss the results of their homework surveys. After the meeting, they
both report to their principals that p
ˆ1  pˆ 2 = 0.10.
+
 Example:
Who Does More Homework?
Standardize : When pˆ1  pˆ 2  0.10,
z
0.10  0.20
 1.72
0.058
Use Table A : The area to the left of z  1.72
under the standard Normal curve is 0.0427.
c) Does the result in part (b) give us reason to doubt
the counselors' reported value?
There is only about a 4% chance of getting a difference in sample proportions
as small as or smaller than the value of 0.10 reported by the counselors.
This does seem suspicious!
Comparing Two Proportions
b) Find the probability of getting a difference in sample proportions
pˆ1  pˆ 2 of 0.10 or less from the two surveys.
+
 Example:
Intervals for p1 – p2
can use our familiar formula to calculate a confidence interval for
p1  p2 :
statistic  (critical value)  (standard deviation of statistic)
When the Independent condition is met, the standard deviation of the statistic
pˆ1  pˆ 2 is :

p (1  p1 ) p2 (1  p2 )
 pˆ1  pˆ 2  1

n1
n2
Because we don' t know the values of the parameters p1 and p2 , we replace them
in the standard deviation formula with the sample proportions. The result is the
standard error of the statistic pˆ1  pˆ 2 :
Comparing Two Proportions
When data come from two random samples or two groups in a randomized
experiment, the statistic pˆ1  pˆ 2 is our best guess for the value of p1 p2 . We
+
 Confidence
pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )

n1
n2
If the Normal condition is met, we find the critical value z* for the given confidence
level from the standard Normal curve. Our confidence interval for p1 – p2 is:
statistic  (critical value)  (standard deviation of statistic)
pˆ (1 pˆ1) pˆ 2 (1 pˆ 2 )
( pˆ1  pˆ 2 )  z * 1

n1
n2


Two-Sample z Interval for a Difference Between Proportions
When the Random, Normal, and Independent conditions are met, an
approximate level C confidence interval for ( pˆ1  pˆ 2 ) is
( pˆ1  pˆ 2 )  z *
pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )

n1
n2
where z * is the critical value for the standard Normal curve with area C
between  z * and z * .
Random The data are produced by a random sample of size n1 from
Population 1 and a random sample of size n2 from Population 2 or by
two groups of size n1 and n2 in a randomized experiment.
Normal The counts of " successes" and " failures" in each sample or
group - - n1 pˆ1, n1 (1  pˆ1 ), n2 pˆ 2 and n2 (1  pˆ 2 ) - - are all at least 10.
Independent Both the samples or groups themselves and the individual
observations in each sample or group are independent. When sampling
without replacement, check that the two populations are at least 10 times
as large as the corresponding samples (the 10% condition).
+
z Interval for p1 – p2
Comparing Two Proportions

 Two-Sample
As part of the Pew Internet and American Life Project, researchers conducted two surveys in late
2009. The first survey asked a random sample of 800 U.S. teens about their use of social
media and the Internet. A second survey posed similar questions to a random sample of 2253
U.S. adults. In these two studies, 73% of teens and 47% of adults said that they use socialnetworking sites. Use these results to construct and interpret a 95% confidence interval for the
difference between the proportion of all U.S. teens and adults who use social-networking sites.
State: Our parameters of interest are p1 = the proportion of all U.S. teens who use
social networking sites and p2 = the proportion of all U.S. adults who use socialnetworking sites. We want to estimate the difference p1 – p2 at a 95% confidence
level.
Plan: We should use a two-sample z interval for p1 – p2 if the conditions are satisfied.
 Random The data come from a random sample of 800 U.S. teens and a separate
random sample of 2253 U.S. adults.
 Normal We check the counts of “successes” and “failures” and note the Normal
condition is met since they are all at least 10:
n1 pˆ1 = 800(0.73) = 584
n2 pˆ 2 = 2253(0.47) =1058.91  1059
n1(1 pˆ1)  800(1 0.73)  216
n2 (1 pˆ 2 )  2253(1 0.47)  1194.09  1194
 Independent We clearly have two independent samples—one of teens and one of
adults. Individual responses in the two samples also have to be independent. The
researchers are sampling without replacement, so we check the 10% condition: there
are at least 10(800) = 8000 U.S. teens and at least 10(2253) = 22,530 U.S. adults.
+
Teens and Adults on Social Networks
Comparing Two Proportions

 Example:
Do: Since the conditions are satisfied, we can construct a twosample z interval for the difference p1 – p2.
( pˆ1  pˆ 2 )  z *
pˆ1(1 pˆ1) pˆ 2 (1 pˆ 2 )
0.73(0.27) 0.47(0.53)

 (0.73  0.47)  1.96

n1
n2
800
2253
 0.26  0.037
 (0.223, 0.297)
Conclude: We are 95% confident that the interval from 0.223 to 0.297
captures the true difference in the proportion of all U.S. teens and
adults who use social-networking sites. This interval suggests that
more teens than adults in the United States engage in social
networking by between 22.3 and 29.7 percentage points.
+
Teens and Adults on Social Networks
Comparing Two Proportions

 Example:
A study by the National Athletic Trainers Association surveyed random
samples of 1679 high school freshmen and 1366 high school seniors
in Illinois. Results showed that 34 of the freshmen and 24 of the
seniors had used anabolic steroids. Steroids, which are dangerous,
are sometimes used to improve athletic performance. Define the
parameters, then construct and interpret a 95% confidence interval for
the difference between the population proportions. Explain why we
could say there is no difference between these two sample
proportions?
+

Interval (-0.007 to 0.0124)
We are 95% confident that the interval from -0.007 to 0.0124
captures the true difference in proportion of freshmen and
seniors in Illinois who use anabolic steroids.
Since 0 is a plausible value, we have evidence that there is
no difference between the two proportions.

Problem:

(a) Use the results of these polls to construct and interpret a 90%
confidence interval for the change in President Obama’s approval rating
among all US adults.

(b) Based on your interval, is there convincing evidence that President
Obama’s job approval rating has changed?
+
Many news organizations conduct polls asking adults in the United States
if they approve of the job the president is doing. How did President
Obama’s approval rating change from August 2009 to September 2010?
According to a CNN poll of 1024 randomly selected U.S. adults on
September 1-2, 2010, 50% approved of President Obama’s job
performance. A CNN poll of 1010 randomly selected U.S. adults on
August 28-30, 2009 showed that 53% approved of President Obama’s job
performance.
+
State: We want to estimate at the 90% confidence level where = the true
proportion of all U.S. adults who approved of President Obama’s job
performance in September 2010 and = the true proportion of all U.S. adults who
approved of President Obama’s job performance in August 2009.
Plan: We should use a two-sample z interval for if the conditions are satisfied.
Random: The data came from separate random samples.
Normal: = 512, = 512, = 535, = 475 are all at least 10.
Independent: The samples were taken independently and there are more than
10(1024) = 10,240 U.S. adults in 2010 and 10(1010) = 10,100 U.S. adults in
2009.
= –0.03 0.036 = (–0.066, 0.006)
We are 95% confident that the interval from –0.066 to 0.006 captures the true
change in the proportion of U.S. adults who approve of President Obama’s job
performance from August 2009 to September 2010. That is, it is plausible that
his job approval has fallen by up to 6.6 percentage points or increased by up to
0.6 percentage points.
(b) Since 0 is included in the interval, it is plausible that there has been no
change in President Obama’s approval rating. Thus, we do not have convincing
evidence that his approval rating has changed.
+
Tests for p1 – p2
We’ll restrict ourselves to situations in which the hypothesized difference is
0. Then the null hypothesis says that there is no difference between the two
parameters:
H0: p1 - p2 = 0 or, alternatively, H0: p1 = p2
The alternative hypothesis says what kind of difference we expect.
Ha: p1 - p2 > 0, Ha: p1 - p2 < 0, or Ha: p1 - p2 ≠ 0
If the Random, Normal, and Independent conditions are met, we can proceed
with calculations.
Comparing Two Proportions
An observed difference between two sample proportions can reflect an
actual difference in the parameters, or it may just be due to chance variation
in random sampling or random assignment. Significance tests help us
decide which explanation makes more sense. The null hypothesis has the
general form
H0: p1 - p2 = hypothesized value
+
 Significance
Tests for p1 – p2
test statistic 
statistic  parameter
standard deviation of statistic
z
( pˆ1  pˆ 2 )  0
standard deviation of statistic
If H0: p1 = p2 is true, the two parameters are the same. We call their
common value p. But now we need a way to estimate p, so it makes sense
to combine the data from the two samples. This pooled (or combined)
sample proportion is:
pˆ C 
count of successes in both samples combined
X  X2
 1
count of individuals in both samples combined
n1  n2
Use pˆ C in place of both p1 and p2 in the expression for the denominator of the test
statistic :
( pˆ1  pˆ 2 )  0

z
pˆ C (1  pˆ C ) pˆ C (1  pˆ C )

n1
n2
Comparing Two Proportions
To do a test, standardize pˆ1  pˆ 2 to get a z statistic :
+
 Significance
Two-Sample z Test for the Difference Between Proportions
If the following
conditions are met, we can proceed with a twosample the
z test
for theNormal,
difference
between two
proportions:
Suppose
Random,
and Independent
conditions
are met. To
test the hypothesis H 0 : p1  p2  0, first find the pooled proportion pˆ C of
successes
bothare
samples
combined.
Then sample
computeofthe
Random
Theindata
produced
by a random
size z nstatistic
1 from
Population 1 and a random sample of size n2 from Population 2 or by
( pˆ1  pˆ 2 )  0experiment.
two groups of size n1 andzn2 in a randomized
pˆ C (1  pˆ C ) pˆ C (1  pˆ C )

Normal The counts of " successes"
in each sample or
n1 and " failures"
n2
group - - n1 pˆ1, n1 (1  pˆ1 ), n2 pˆ 2 and n2 (1  pˆ 2 ) - - are all at least 10.
Find the P - value by calculating the probabilty of getting a z statistic this
large or larger in the direction specified by the alternative hypothesis H a :
Independent Both the samples or groups themselves and the individual
observations in each sample or group are independent. When sampling
without replacement, check that the two populations are at least 10 times
as large as the corresponding samples (the 10% condition).


Comparing Two Proportions
z Test for The Difference Between
Two Proportions
+
 Two-Sample
Researchers designed a survey to compare the proportions of children who come to school
without eating breakfast in two low-income elementary schools. An SRS of 80 students from
School 1 found that 19 had not eaten breakfast. At School 2, an SRS of 150 students included
26 who had not had breakfast. More than 1500 students attend each school. Do these data
give convincing evidence of a difference in the population proportions? Carry out a
significance test at the α = 0.05 level to support your answer.
State: Our hypotheses are
H0: p1 - p2 = 0
Ha: p1 - p2 ≠ 0
where p1 = the true proportion of students at School 1 who did not eat breakfast,
and p2 = the true proportion of students at School 2 who did not eat breakfast.
Plan: We should perform a two-sample z test for p1 – p2 if the conditions are satisfied.
 Random The data were produced using two simple random samples—of 80 students
from School 1 and 150 students from School 2.
 Normal We check the counts of “successes” and “failures” and note the Normal
condition is met since they are all at least 10:
n1 pˆ1 =19, n1(1 pˆ1)  61, n2 pˆ 2 = 26, n2 (1 pˆ 2 ) 124
 Independent We clearly have two independent samples—one from each school.
Individual responses in the two samples also have to be independent. The researchers
are sampling without replacement, so we check the 10% condition: there are at least

10(80) = 800 students at School 1 and at least 10(150) = 1500 students at School 2.
Comparing Two Proportions

Hungry Children
+
 Example:
Do: Since the conditions are satisfied, we can perform a two-sample z test for the
difference p1 – p2.
pˆ C 
X1  X 2 19  26
45


 0.1957
n1  n2 80 150 230
Test statistic :
( pˆ1  pˆ 2 )  0
z=

pˆ C (1pˆ C ) pˆ C (1 pˆ C )

n1
n2
(0.2375  0.1733)  0
 1.17
0.1957(1 0.1957) 0.1957(1 0.1957)

80
150
P-value Using Table A or normalcdf, the
desired P-value is
2P(z ≥ 1.17) = 2(1 - 0.8790) = 0.2420.
Conclude: Since our P-value, 0.2420, is
greater than the chosen significance level
of α = 0.05,we fail to reject H0. There is not
sufficient evidence to conclude that the
proportions of students at the two schools
who didn’t eat breakfast are different.
+
Hungry Children
Comparing Two Proportions

 Example:
High levels of cholesterol in the blood are associated with higher risk of heart attacks. Will
using a drug to lower blood cholesterol reduce heart attacks? The Helsinki Heart Study
recruited middle-aged men with high cholesterol but no history of other serious medical
problems to investigate this question. The volunteer subjects were assigned at random to one
of two treatments: 2051 men took the drug gemfibrozil to reduce their cholesterol levels, and a
control group of 2030 men took a placebo. During the next five years, 56 men in the
gemfibrozil group and 84 men in the placebo group had heart attacks. Is the apparent benefit
of gemfibrozil statistically significant? Perform an appropriate test to find out.
State: Our hypotheses are
H0: p1 - p2 = 0
Ha: p1 - p2 < 0
OR
H0: p1 = p2
Ha: p1 < p2
where p1 is the actual heart attack rate for middle-aged men like the ones in this
study who take gemfibrozil, and p2 is the actual heart attack rate for middle-aged
men like the ones in this study who take only a placebo. No significance level was
specified, so we’ll use α = 0.01 to reduce the risk of making a Type I error
(concluding that gemfibrozil reduces heart attack risk when it actually doesn’t).
Comparing Two Proportions

Significance Test in an Experiment
+
 Example:
Cholesterol and Heart Attacks
 Random The data come from two groups in a randomized experiment
 Normal The number of successes (heart attacks!) and failures in the two groups are
56, 1995, 84, and 1946. These are all at least 10, so the Normal condition is met.
 Independent Due to the random assignment, these two groups of men can be viewed
as independent. Individual observations in each group should also be independent:
knowing whether one subject has a heart attack gives no information about whether
another subject does.
Do: Since the conditions are satisfied, we can perform a two-sample z test for the
difference p1 – p2.
Test statistic :
X1  X 2
56  84

z=
n1  n 2 2051 2030
140

 0.0343
4081
pˆ C 
P-value Using Table A or
normalcdf, the desired
 Pvalue is 0.0068
( pˆ1  pˆ 2 )  0

pˆ C (1 pˆ C ) pˆ C (1 pˆ C )

n1
n2
(0.0273  0.0414)  0
 2.47
0.0343(1 0.0343) 0.0343(1 0.0343)

2051
2030
Comparing Two Proportions
Plan: We should perform a two-sample z test for p1 – p2 if the conditions are satisfied.
+
 Example:
Conclude: Since the P-value, 0.0068, is less
than 0.01, the results are statistically significant
at the α = 0.01 level. We can reject H0 and
conclude that there is convincing evidence of a
lower heart attack rate for middle-aged men like
these who take gemfibrozil than for those who
take only a placebo.
+ Section 10.1
Comparing Two Proportions
Summary
In this section, we learned that…

Choose an SRS of size n1 from Population 1 with proportion of successes p1 and
an independent SRS of size n2 from Population 2 with proportion of successes p2.
Shape When n1 p1, n1(1  p1), n2 p2 and n2 (1  p2 ) are all at least 10, the
sampling distribution of pˆ1  pˆ 2 is approximately Normal.
Center The mean of the sampling distribution is p1  p2 . That is,
the difference in sample proportions is an unbiased estimator of
the difference in population proportions.

Spread The standard deviation of the sampling distribution of pˆ1  pˆ 2 is
p1(1 p1 ) p2 (1 p2 )

n1
n2
as long as each sample is no more than 10% of its population (10% condition).


Confidence intervals and tests to compare the proportions p1 and p2 of successes
for two populations or treatments are based on the difference between the sample
proportions.

When the Random, Normal, and Independent conditions are met, we can use twosample z procedures to estimate and test claims about p1 - p2.

+ Section 10.1
Comparing Two Proportions
Summary
In this section, we learned that…

The conditions for two-sample z procedures are:
Random The data are produced by a random sample of size n1 from
Population 1 and a random sample of size n2 from Population 2 or by two
groups of size n1 and n2 in a randomized experiment.
Normal The counts of " successes" and " failures" in each sample or
group - - n1 pˆ1, n1 (1 pˆ1 ), n 2 pˆ 2 and n 2 (1 pˆ 2 ) - - are all at least 10.

Independent Both the samples or groups themselves and the individual
observations in each sample or group are independent. When sampling
without replacement, check that the two populations are at least 10 times
as large as the corresponding samples (the 10% condition).


An approximate level C confidence interval for p1 - p2 is

( pˆ1  pˆ 2 )  z *
pˆ1(1 pˆ1) pˆ 2 (1 pˆ 2 )

n1
n2
where z* is the standard Normal critical value. This is called a two-sample z
interval for p1 - p2.

+ Section 10.1
Comparing Two Proportions
Summary
In this section, we learned that…

Significance tests of H0: p1 - p2 = 0 use the pooled (combined) sample
proportion
pˆ C 

count of successes in both samples combined
X  X2
 1
count of individuals in both samples combined
n1  n2
The two-sample z test for p1 - p2 uses the test statistic
( pˆ1  pˆ 2 )  0
pˆ C (1 pˆ C ) pˆ C (1 pˆ C )

n1
n2
with P-values calculated from the standard Normal distribution.



z
Inference about the difference p1 - p2 in the effectiveness of two treatments in a
completely randomized experiment is based on the randomization
distribution of the difference of sample proportions. When the Random,
Normal, and Independent conditions are met, our usual inference procedures
based on the sampling distribution will be approximately correct.
+
Looking Ahead…
In the next Section…
We’ll learn how to compare two population means.
We’ll learn about
 The sampling distribution for the difference of
means
 The two-sample t procedures
 Comparing two means from raw data and
randomized experiments
 Interpreting computer output for two-sample t
procedures
+ Section 10.2
Comparing Two Means
Learning Objectives
After this section, you should be able to…

DESCRIBE the characteristics of the sampling distribution of the difference between
two sample means

CALCULATE probabilities using the sampling distribution of the difference between
two sample means

DETERMINE whether the conditions for performing inference are met

USE two-sample t procedures to compare two means based on summary statistics or
raw data

INTERPRET computer output for two-sample t procedures

PERFORM a significance test to compare two means

INTERPRET the results of inference procedures
+
σ known, use z*
(x1-x2) + z* σ12 + σ22
n1 n2
Intervals
σ unknown, use t*
(x1-x2) + t* Sx12 + Sx22
n1
n2
1. Decide on whether to use z* or t*
2. State parameters and Define the order you are subtracting at [C]
% confidence level.
3. State the type of interval. “We are performing a 2 sample z/t interval
for the difference of two means”
4. Random/Normal/Independent. Check both samples, use CLT for
Normality.
5. State values for x1, x2, Sx1, Sx2, and df if t interval (use more
conservative df value)
6. Find the z or t value
7. State correct formula and plug all values into formula and simplify to
find interval
8. Conclude on interval. “We are [C] % confident that the interval from
[here] to [there] captures the true difference in means of [context of the
question].
Our parameters of interest are the population means µ1 and µ2. Once
again, the best approach is to take separate random samples from each
population and to compare the sample means.
Suppose we want to compare the average effectiveness of two treatments
in a completely randomized experiment. In this case, the parameters µ1
and µ2 are the true mean responses for Treatment 1 and Treatment 2,
respectively. We use the mean response in the two groups to make the
comparison.
Here’s a table that summarizes these two situations:
Comparing Two Means
In the previous section, we developed methods for comparing two
proportions. What if we want to compare the mean of some quantitative
variable for the individuals in Population 1 and Population 2?
+
 Introduction

In Chapter 7, we saw that the sampling distribution of a sample mean has the
following properties:
Shape Approximately Normal if the population distribution is Normal or
n ≥ 30 (by the central limit theorem).
Center x  

Spread x 
if the sample is no more than 10% of the population
n
To explore the sampling distribution of the difference between two means, let’s start
with two Normally distributed populations having known means and standard
deviations.
Based on information from the U.S. National Health and Nutrition Examination
Survey (NHANES), the heights (in inches) of ten-year-old girls follow a Normal
distribution N(56.4, 2.7). The heights (in inches) of ten-year-old boys follow a Normal
distribution N(55.7, 3.8).
Suppose we take independent SRSs of 12 girls and 8 boys of this age and measure
their heights.
What can we say about the difference x f  x m in the average heights of the
sample of girls and the sample of boys?
+
Sampling Distribution of a Difference Between
Two Means
Comparing Two Means

 The
Using Fathom software, we generated an SRS of 12 girls and a separate
SRS of 8 boys and calculated the sample mean heights. The difference
in sample means was then calculated and plotted. We repeated this
process 1000 times. The results are below:
What do you notice about the shape, center, and spread
of the sampling distribution of x f  x m ?
Comparing Two Means
Sampling Distribution of a Difference
Between Two Means
+
 The
Both x1 and x 2 are random variables. The statistic x1 - x 2 is the difference
of these two random variables. In Chapter 6, we learned that for any two
independent random variables X and Y,
X Y  X  Y and  X2Y   X2   Y2
The Sampling Distribution of the Difference Between Sample Means
Choose an SRS of size n1 from Population 1 with mean µ1 and standard
Therefore,
deviation
σ1 and an independent SRS of size n2 from Population 2 with mean
2
2
2








µ2 
and
deviation
σ
.





x1 xstandard
x
x
1
2
2
2
1
2
x1 x 2
x1
x2
2
2
Shape When the population distributions are Normal, the
   sampling
   distribution
of x1  x 2 is approximately Normal. In other cases,
the
1 sampling2distribution will
 

 n enough
  
 ( n n
1  30,n 2  30).
be approximately Normal if the sample sizes are large

 

1
2
Center The mean of the sampling distribution is 12 2 . That
 1  12 is, the difference
in sample means is an unbiased estimator of the difference
in population means.

n
n
1
2
Spread The standard deviation of the sampling distribution
of
12

22
 12
 12
x1  x 2 is
n1 n 2 

x1 x 2 
n1
n 2 (10% condition).
as long as each sample is no more than 10% of its population
Comparing Two Means
Sampling Distribution of a Difference
Between Two Means
+
 The
Based on information from the U.S. National Health and Nutrition Examination Survey
(NHANES), the heights (in inches) of ten-year-old girls follow a Normal distribution N(56.4,
2.7). The heights (in inches) of ten-year-old boys follow a Normal distribution N(55.7, 3.8). A
researcher takes independent SRSs of 12 girls and 8 boys of this age and measures their
heights. After analyzing the data, the researcher reports that the sample mean height of the
boys is larger than the sample mean height of the girls.
a) Describe the shape, center, and spread of the sampling distribution of x f  x m .
Because both population distributions are Normal,
of x f  x m is Normal.
Its mean is  f  m  56.4  55.7  0.7 inches.
Its standard deviation is
2.7 2 3.8 2

 1.55 inches.
12
8
the sampling distribution
Comparing Two Means

Who’s Taller at Ten, Boys or Girls?
+
 Example:
Who’s Taller at Ten, Boys or Girls?
Standardize : When x1  x 2  0,
z
0  0.70
 0.45
1.55
Use Table A : The area to the left of z  0.45
under the standard Normal curve is 0.3264.
Comparing Two Means
b) Find the probability of getting a difference in sample means
x1  x 2 that is less than 0.
+
 Example:
(c) Does the result in part (b) give us reason to doubt the researchers’
stated results?
If the mean height of the boys is greater than the mean height of the girls,
x m  x f , That is
x f  x m  0. Part (b) shows that there' s about a 33% chance of getting a difference in
sample means that' s negative just due to sampling variability. This gives us little reason
to doubt the researcher' s claim.
Two-Sample t Statistic
+
 The
When the Independent condition is met, the standard deviation of the statistic
x1  x 2 is :
x
1 x 2

12
n1

2 2
n2
Since we don' t know the values of the parameters 1 and 2, we replace them
in the standard deviation formula with the sample standard deviations. The result
is the standard error of the statistic x1  x 2 :
Comparing Two Means
When data come from two random samples or two groups in a randomized
experiment, the statistic x1  x 2 is our best guess for the value of 1 2 .
s12 s2 2

n1 n 2
If the Normal condition is met, we standardize the observed difference to obtain
a t statistic that tells us how far the observed difference is from its mean in standard
deviation units:
t
(x1  x 2 )  ( 1  2 )
s12 s2 2

n1 n 2
The two-sample t statistic has approximately a t distribution.
We can use technology to determine degrees of freedom
OR we can use a conservative approach, using the smaller
of n1 – 1 and n2 – 1 for the degrees of freedom.

Two-Sample t Interval for a Difference Between Means
When the Random, Normal, and Independent conditions are met, an
approximate level C confidence interval for (x1  x 2 ) is
s12 s2 2
(x1  x 2 )  t *

n1 n 2
where t * is the critical value for confidence level C for the t distribution with
degrees of freedom from either technology or the smaller of n1 1 and n 2 1.
Random The data are produced by a random sample of size n1 from
Population 1 and a random sample of size n2 from Population 2 or by
two groups of size n1 and n2 in a randomized experiment.
Normal Both population distributions are Normal OR both sample
group sizes are large ( n1  30 and n2  30).
Independent Both the samples or groups themselves and the individual
observations in each sample or group are independent. When sampling
without replacement, check that the two populations are at least 10 times
as large as the corresponding samples (the 10% condition).
+
Intervals for µ1 – µ2
Comparing Two Means

 Confidence
Trees, Small Trees, Short Trees, Tall Trees
State: Our parameters of interest are µ1 = the true mean DBH of all trees in the
southern half of the forest and µ2 = the true mean DBH of all trees in the northern half
of the forest. We want to estimate the difference µ1 - µ2 at a 90% confidence level.
Comparing Two Means
The Wade Tract Preserve in Georgia is an old-growth forest of longleaf pines that has survived in
a relatively undisturbed state for hundreds of years. One question of interest to foresters who
study the area is “How do the sizes of longleaf pine trees in the northern and southern halves
of the forest compare?” To find out, researchers took random samples of 30 trees from each
half and measured the diameter at breast height (DBH) in centimeters. Comparative boxplots
of the data and summary statistics from Minitab are shown below. Construct and interpret a
90% confidence interval for the difference in the mean DBH for longleaf pines in the northern
and southern halves of the Wade Tract Preserve.
+
 Big
Trees, Small Trees, Short Trees, Tall Trees
 Random The data come from a random samples of 30 trees each from the
northern and southern halves of the forest.
 Normal The boxplots give us reason to believe that the population distributions of
DBH measurements may not be Normal. However, since both sample sizes are at
least 30, we are safe using t procedures.
 Independent Researchers took independent samples from the northern and
southern halves of the forest. Because sampling without replacement was used, there
have to be at least 10(30) = 300 trees in each half of the forest. This is pretty safe to
assume.
Do: Since the conditions are satisfied, we can construct a two-sample t interval for
the difference µ1 – µ2. We’ll use the conservative df = 30-1 = 29.
Conclude: We are 90% confident that the interval from 3.83 to 17.83
centimeters captures the difference in the actual mean DBH of the
2
2
southern trees and the actual mean
s1 s2DBH of the northern trees.
14.262This
17.502
(x1  x 2 )  t *

 (34.5  23.70)  1.699

interval suggests that
the meanndiameter
of
the
southern
trees
is
n2
30
30
1
between 3.83 and 17.83 cm larger than
the mean
of the
 10.83
 7.00 diameter
(3.83, 17.83)
northern trees.
Comparing Two Means
Plan: We should use a two-sample t interval for µ1 – µ2 if the conditions are satisfied.
+
 Big
+
+
Tests for µ1 – µ2
The alternative hypothesis says what kind of difference we expect.
Ha: µ1 - µ2 > 0, Ha: µ1 - µ2 < 0, or Ha: µ1 - µ2 ≠ 0
If the Random, Normal, and Independent conditions are met, we can proceed
with calculations.
Comparing Two Means
An observed difference between two sample means can reflect an actual
difference in the parameters, or it may just be due to chance variation in
random sampling or random assignment. Significance tests help us decide
which explanation makes more sense. The null hypothesis has the general
form
H0: µ1 - µ2 = hypothesized value
We’re often interested in situations in which the hypothesized difference is
0. Then the null hypothesis says that there is no difference between the two
parameters:
H0: µ1 - µ2 = 0 or, alternatively, H0: µ1 = µ2
+
 Significance
t Test for The Difference Between
Two-Sample t Test for the Difference Between Two Means
If the following conditions are met, we can proceed with a twoSuppose the Random, Normal, and Independent conditions are met. To
sample t test for the difference between two means:
test the hypothesis H :     hypothesized value , compute the t statistic

0
1
2
Random The data are produced by a random sample of size n1 from
(x  x )  (   2 ) Population 2 or by
Population 1 and a random tsample
 1 of2 size 1 n2 from
2
2
two groups of size n1 and n2 in a randomized
s1
s2 experiment.

n1 n 2
Normal Both population distributions (or the true distributions
responses
two treatments)
are Normal
OR both
Findofthe
P - valuetobythe
calculating
the probabilty
of getting
a tsample
statistic this large
groupin
sizes
are largespecified
( n1  30 and
n 2 alternative
 30).
or larger
the direction
by the
hypothesis H a . Use the
t distribution with degrees of freedom approximated by technology or the
smaller
of n1 1 andBoth
n 2 1.
Independent
the samples or groups themselves and the individual
observations in each sample or group are independent. When sampling
without replacement, check that the two populations are at least 10 times
as large as the corresponding samples (the 10% condition).


Comparing Two Means
Two Means
+
 Two-Sample
Calcium and Blood Pressure
Comparing Two Means
Does increasing the amount of calcium in our diet reduce blood pressure? Examination of a large
sample of people revealed a relationship between calcium intake and blood pressure. The
relationship was strongest for black men. Such observational studies do not establish
causation. Researchers therefore designed a randomized comparative experiment. The
subjects were 21 healthy black men who volunteered to take part in the experiment. They
were randomly assigned to two groups: 10 of the men received a calcium supplement for 12
weeks, while the control group of 11 men received a placebo pill that looked identical. The
experiment was double-blind. The response variable is the decrease in systolic (top number)
blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as
a negative response Here are the data:
+
 Example:
State: We want to perform a test of
H0: µ1 - µ2 = 0
Ha: µ1 - µ2 > 0
where µ1 = the true mean decrease in systolic blood
pressure for healthy black men like the ones in this study
who take a calcium supplement, and µ2 = the true mean
decrease in systolic blood pressure for healthy black men
like the ones in this study who take a placebo.
We will use α = 0.05.
Calcium and Blood Pressure
• Random The 21 subjects were randomly assigned to the two treatments.
• Normal With such small sample sizes, we need to examine the data to see if it’s
reasonable to believe that the actual distributions of differences in blood pressure when
taking calcium or placebo are Normal. Hand sketches of calculator boxplots and
Normal probability plots for these data are below:
The boxplots show no clear evidence of skewness and no outliers. The Normal
probability plot of the placebo group’s responses looks very linear, while the Normal
probability plot of the calcium group’s responses shows some slight curvature. With no
outliers or clear skewness, the t procedures should be pretty accurate.
• Independent Due to the random assignment, these two groups of men can be
viewed as independent. Individual observations in each group should also be
independent: knowing one subject’s change in blood pressure gives no information
about another subject’s response.
Comparing Two Means
Plan: If conditions are met, we will carry out a two-sample t test for µ1- µ2.
+
 Example:
Calcium and Blood Pressure
Test statistic :
(x  x 2 )  ( 1  2 ) [5.000  (0.273)]  0
t 1

 1.604
2
2
2
2
8.743 5.901
s1
s

 2
10
11
n1 n 2
P-value Using the conservative df = 10 – 1 = 9,
we can use Table B to show that the P-value is
between 0.05 and 0.10.
Comparing Two Means
Do: Since the conditions are satisfied, we can perform a two-sample t test for the
difference µ1 – µ2.
+
 Example:
Conclude: Because the P-value is greater than α = 0.05, we fail to reject H0. The
experiment provides some evidence that calcium reduces blood pressure, but the
evidence is not convincing enough to conclude that calcium reduces blood pressure
more than a placebo.
Calcium and Blood Pressure
With df = 9, the critical value for a 90% confidence interval is t* = 1.833.
The interval is:
2
Comparing Two Means
We can estimate the difference in the true mean decrease in blood pressure for
the calcium and placebo treatments using a two-sample t interval for µ1 - µ2. To
get results that are consistent with the one-tailed test at α = 0.05 from the
example, we’ll use a 90% confidence level. The conditions for constructing a
confidence interval are the same as the ones that we checked in the example
before performing the two-sample t test.
+
 Example:
2
s1
s2
8.7432 5.9012
(x1  x 2 )  t *

 [5.000  (0.273)]  1.833

n1 n 2
10
11
 5.273  6.027
 (0.754,11.300)
We are 90% confident that the interval from -0.754 to 11.300 captures the difference in true
mean blood pressure reduction on calcium over a placebo. Because the 90% confidence
interval includes 0 as a plausible value for the difference, we cannot reject H0: µ1 - µ2 = 0
against the two-sided alternative at the α = 0.10 significance level or against the one-sided
alternative at the α = 0.05 significance level.
Two-Sample t Procedures Wisely
Using the Two-Sample t Procedures: The Normal Condition
•Sample size less than 15: Use two-sample t procedures if the data in both
samples/groups appear close to Normal (roughly symmetric, single peak,
no outliers). If the data are clearly skewed or if outliers are present, do not
use t.
• Sample size at least 15: Two-sample t procedures can be used except in
the presence of outliers or strong skewness.
• Large samples: The two-sample t procedures can be used even for clearly
skewed distributions when both samples/groups are large, roughly n ≥ 30.
Comparing Two Means
The two-sample t procedures are more robust against non-Normality than
the one-sample t methods. When the sizes of the two samples are equal
and the two populations being compared have distributions with similar
shapes, probability values from the t table are quite accurate for a broad
range of distributions when the sample sizes are as small as n1 = n2 = 5.
+
 Using
Two-Sample t Procedures Wisely
 In planning a two-sample study, choose equal
sample sizes if you can.
 Do not use “pooled” two-sample t procedures!
 We are safe using two-sample t procedures for
comparing two means in a randomized experiment.
 Do not use two-sample t procedures on paired data!
 Beware of making inferences in the absence of
randomization. The results may not be generalized to
the larger population of interest.
Comparing Two Means
Here are several cautions and considerations to make when using twosample t procedures.
+
 Using
+ Section 10.2
Comparing Two Means
Summary
In this section, we learned that…

Choose an SRS of size n1 from Population 1 and an independent SRS of size n2
from Population 2. The sampling distribution of the difference of sample means
has:
Shape Normal if both population distributions are Normal; approximately
Normal otherwise if both samples are large enough ( n  30).
Center The mean 1  2 .
Spread As long as each sample is no more than 10% of its population

(10% condition), its standard deviation is

 12
nn

 22
n2
.

Confidence intervals and tests for the difference between the means of two
populations or the mean responses to two treatments µ1 – µ2 are based on the
difference between the sample means.

If we somehow know the population standard deviations σ1 and σ2, we can use a z
statistic and the standard Normal distribution to perform probability calculations.

+ Section 10.2
Comparing Two Means
Summary

Since we almost never know the population standard deviations in practice, we
use the two-sample t statistic
(x1  x 2 )  ( 1  2 )
t
s12 s2 2

n1 n 2

where t has approximately a t distribution with degrees of freedom found by
technology or by the conservative approach of using the smaller of n1 – 1 and

n2 – 1 .

The conditions for two-sample t procedures are:
Random The data are produced by a random sample of size n1 from
Population 1 and a random sample of size n2 from Population 2 or by two
groups of size n1 and n2 in a randomized experiment.


Normal Both population distributions (or the true distributions of responses
to the two treatments) are Normal OR both sample/group sizes are large
(n1  30 and n 2  30).
Independent Both the samples or groups themselves and the individual
observations in each sample or group are independent. When sampling
without replacement, check that the two populations are at least 10 times
as large as the corresponding samples (the 10% condition).
+ Section 10.2
Comparing Two Means
Summary

The level C two-sample t interval for µ1 – µ2 is
2
2
s1 s2
(x1  x 2 )  t *

n1 n2
where t* is the critical value for confidence level C for the t distribution with
degrees of freedom from either technology or the conservative approach.

To test H0: µ1 
- µ2 = hypothesized value, use a two-sample t test for µ1 - µ2.
The test statistic is
t
(x1  x 2 )  ( 1  2 )
s12 s2 2

n1 n 2
P-values are calculated using the t distribution with degrees of freedom from
either technology or the conservative approach.
+ Section 10.2
Comparing Two Means
Summary

The two-sample t procedures are quite robust against departures from
Normality, especially when both sample/group sizes are large.

Inference about the difference µ1 - µ2 in the effectiveness of two treatments in
a completely randomized experiment is based on the randomization
distribution of the difference between sample means. When the Random,
Normal, and Independent conditions are met, our usual inference procedures
based on the sampling distribution of the difference between sample means
will be approximately correct.

Don’t use two-sample t procedures to compare means for paired data.