Transcript p 2

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Chapter 10: Comparing Two Populations or Groups
Section 10.1
Comparing Two Proportions
The Practice of Statistics, 4th edition – For AP*
STARNES, YATES, MOORE
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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 these two
situations.
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.
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 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 mpˆ = 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 s 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?
Comparing Two Proportions
Sampling Distribution of a Difference
Between Two Proportions
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 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
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 The
Sampling Distribution of a Difference
Between Two Proportions
pˆ1 - pˆ 2
pˆ1
pˆ 2
1
2
Center The mean of the sampling distribution is p1 - p2 . That is,
2
2
thesdifference
sample
proportions is an unbiased estimator of
= s 2 in+ s
pˆ1 - pˆ 2
pˆ1
pˆ 2
the difference in population
propotions. 2
2
æ p1 (1- p1)deviation
ö æ p2of
(1-the
p2 sampling
)ö
Spread The
distribution of pˆ1 - pˆ 2 is
= ç standard
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÷ ç
÷
n1
è
ø è p (1n-2 p ) ø p (1 - p )
1
1
2
+ 2
n2
p (1- p ) p (1- p ) n1
1
1
2
2
=
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as long as eachnsample is no
n 2 more than 10% of its population (10% condition).
1
s pˆ
ˆ2
1 -p
=
p1 (1- p1 ) p2 (1- p2 )
+
n1
n2
Comparing Two Proportions
The p
Proportions
ˆSampling
ˆ Distribution of the Difference Between
ˆ Sample
ˆ
Both
1 and p2 are random variables. The statistic p1 - p2 is the difference
of these
random
In Chapter1 6,
weproportion
learned that
for any two
Choose
antwo
SRS
of sizevariables.
n1 from Population
with
of successes
random variables
X and
Y, Population 2 with proportion of
p1independent
and an independent
SRS of size
n2 from
mX -Y = mX -p2m.Y and s X2 -Y = s X2 + s Y2
successes
Shape When n1 p1, n1 (1 - p1 ), n 2 p2 and n2 (1 - p2 ) are all at least 10, the
Therefore,
sampling distribution of pˆ1 - pˆ 2 is approximately Normal.
m
= m -m = p - p
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 The
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Sampling Distribution of the Difference between 2 proportions:
Shape Approximately Normal if n1p1 ≥ 10 and n1(1 – p1) ≥ 10, and
n2p2 ≥ 10 and n2(1 – p2) ≥ 10
Center m p̂ = p1 - p2
Spread s p̂ =
p1 (1- p1 ) p2 (1- p2 )
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if the sample is no more than 10% of the population
n1
n2
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, n 2 p2 = 200(0.5) =100
and n 2 (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)
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= 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
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 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.
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.
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 Example:
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, if the true difference in
proportions is 0.20. This does seem suspicious!
Intervals for p1 – p2
can use our familiar formula to calculate a confidence interval for p1 - p 2 :
statistic ± (critical value)× (standard deviation of statistic)
When the Independent condition is met, the standard deviation of the statistic
pˆ1 - pˆ 2 is :
p1 (1- p1 ) p2 (1- p2 )
+
ˆ2
1-p
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
s pˆ
=
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
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 Confidence
pˆ1 (1 - pˆ1 ) pˆ 2 (1 - pˆ 2 )
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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
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n1
n2
z Interval for p1 – p2
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 n 2 from Population 2 or by
two groups of size n1 and n 2 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).
Comparing Two Proportions
Two-Sample z Interval for a Difference Between Proportions
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 Two-Sample
Teens and Adults on Social Networks
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 Þ 1059
n1(1- pˆ1) = 800(1- 0.73) = 216
n2 (1- pˆ 2 ) = 2253(1- 0.47) =1194.09 Þ 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.
Comparing Two Proportions
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.
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 Example:
Teens and Adults on Social Networks
( pˆ1 - pˆ 2 ) ± z *
pˆ1 (1- pˆ1 ) pˆ 2 (1- pˆ 2 )
0.73(0.27) 0.47(0.53)
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= (0.73 - 0.47) ± 1.96
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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.
Take Pulse and Record!
Comparing Two Proportions
Do: Since the conditions are satisfied, we can construct a twosample z interval for the difference p1 – p2.
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 Example:
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
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 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 X1 + X 2
=
count of individuals in both samples combined n1 + n 2
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 :
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 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
twoconditions
proportions:
Suppose
Random,
and between
Independent
are met. To
test the hypothesis H 0 : p1 - p2 = 0, first find the pooled proportion pˆ C of
successes
bothare
samples
combined.
Thensample
compute
Random
Theindata
produced
by a random
of the
sizez nstatistic
1 from
Population 1 and a random sample of size n 2 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 )
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Normal The counts of "successes"
in each sample or
n1 and " failures"
n2
group - - n1 pˆ1, n1 (1 - pˆ1 ), n 2 pˆ 2 and n 2 (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
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 Two-Sample
State: Our hypotheses are

.
H0: p1 - p2 = 0
Ha: p1 - p2 ≠ 0
where p1 = the true proportion of U.S. adults that approve of Obama in 2009 and p2
= the true proportion of U.S. adults that approve of the job Obama is doing in 2010.
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 1010 adults
in August 2009 and 1024 in September 2010.
 Normal We check the counts of “successes” and “failures” and note the Normal
condition is met since they are all at least 10:
n2009 pˆ 2009 = 535, n2009 (1- pˆ 2009 ) = 475,n2010 pˆ 2010 = 512, n 2010 (1- pˆ 2010 ) = 512
 Independent We have two independent samples—one from each year. The
researchers are sampling without replacement, so we check the 10% condition: there
are at least 10(1024) = 10240 adults in 2010 and at least 10(1010) = 10100 adults in
2009.
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Comparing Two Proportions
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? A CNN poll of 1010 randomly selected U.S.
adults on August 28-30, 2009 showed that 53% approved of Obama’s job performance.
According to a CNN poll of 1024 randomly selected U.S. adults on September 1-2,
2010, 50% approved of Obama’s job performance. Determine if there has been a
change in President Obama’s approval rating. Use a significance level of 0.10.
Presidential Approval
pˆ C =
X1 + X 2
535 + 512
1047
=
=
= 0.5147
n1 + n2 1010 + 1024 2034
Test statistic :
( pˆ1 - pˆ 2 ) - 0
z=
=
ˆpC (1- pˆ C ) pˆ C (1- pˆ C )
+
n1
n2
(0.50 - 0.53) - 0
= -1.35
0.5147(1- 0.5147) 0.5147(1- 0.5147)
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1024
1010
P-value Using Table A or normalcdf, the
desired P-value is
2P(z ≤ -1.35) = 2(0.0885) = 0.1770.
Calculator will give slightly different answer: 0.1802
Conclude: Since our P-value, 0.1770, is
greater than the chosen significance level
of α = 0.10,we fail to reject H0. There is not
sufficient evidence to conclude that the
proportions of U.S. adults that approve of
Obama is different in 2010.
Comparing Two Proportions
Do: Since the conditions are satisfied, we can perform a two-sample z test for the
difference p1 – p2.
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 Example:
Example: Presidential Approval
Create a 90% confidence interval to confirm our findings from the significance test.
statistic ± (critical value)× (standard deviation of statistic)
pˆ (1- pˆ1 ) pˆ 2 (1- pˆ 2 )
( pˆ1 - pˆ 2 ) ± z * 1
+
n1
n2
(0.50 - 0.53) ± 1.645
0.53(0.47) 0.50(0.50)
+
1010
1024
-0.03 ± 0.036
(-0.066,0.006)
Since 0 falls in the 90% confidence interval, we fail to reject the hypothesis that there is no
difference in approval ratings.
Comparing Two Proportions
Do: Since the conditions are satisfied (previously), we can calculate a 90%
confidence interval for p1 – p2.
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

All drugs created Equal?
State: Our hypotheses are
H0: p1 - p2 = 0
Ha: p1 - p2 ≠ 0
where p1 = the true proportion of women that caught a cold and p2 = the true
proportion of men who caught a cold.
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 100 women
and 200 men.
 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 = 38, n1(1- pˆ1) = 62, n2 pˆ 2 =102, n2 (1- pˆ 2 ) = 98
 Independent We clearly have two independent samples—one from men and one
from women. Individual responses in the two samples also have to be independent. It is
safe to assume that one volunteers response to the drug is independent of another
volunteer.
Comparing Two Proportions
Suppose the Acme Drug Company develops a new drug, designed to prevent colds.
The company states that the drug is equally effective for men and women. To test this
claim, they choose a a simple random sample of 100 women and 200 men from a
population of 100,000 volunteers. At the end of the study, 38% of the women caught a
cold; and 51% of the men caught a cold. Based on these findings, can we reject the
company's claim that the drug is equally effective for men and women? Use a 0.05
level of significance.
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 Example:
All drugs created equal?
pˆ C =
X1 + X 2 38 + 102 140
=
=
= 0.467
n1 + n2 100 + 200 300
Test statistic :
( pˆ1 - pˆ 2 ) - 0
z=
=
ˆpC (1- pˆ C ) pˆ C (1- pˆ C )
+
n1
n2
(0.38 - 0.51) - 0
= -2.13
0.467(1- 0.467) 0.467(1- 0.467)
+
100
200
P-value Using Table A or normalcdf, the
desired P-value is
2P(z ≤-2.13) = 2(0.017) = 0.034
Conclude: Since our P-value, 0.034, is
less than the chosen significance level of α
= 0.05,we reject H0. There is evidence to
conclude there is a difference in
effectiveness of the drug between men and
women.
Comparing Two Proportions
Do: Since the conditions are satisfied, we can perform a two-sample z test for the
difference p1 – p2.
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 Example:

Significance Test in an Experiment
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.
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 Example:
What are the implications of a Type I error?
We reject the null hypothesis when we should not have. This implies Gemfibrozil reduces
heart attacks when it actually does not.
What are the implications of a Type II error?
We fail to reject the null hypothesis when we should have. This implies Gemfibrozil does not
reduce heart attacks when it actually does.
Considering these implications, what is an appropriate significance level?
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.
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), (84,1946). 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 n 2 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 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).

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 X1 + X 2
=
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.
+
Chapter 10
Comparing Two Populations or Groups
 10.1
Comparing Two Proportions
 10.2
Comparing Two Means
+ 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
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 mx = m
s
Spread sx =
if the sample is no more than 10% of the population
n
Comparing Two Means
Sampling Distribution of a Difference Between
Two Means
+
 The
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?
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,
mX -Y = mX - mY and s X2-Y = s X2 + s Y2
The Sampling Distribution of the Difference Between Sample Means
Therefore,
2
2
2
Choose an SRS of size n1 from Population
1
with
mean
µ1 and standard
s
=
s
+
s
x -x
x
x
mx1 -x2 = mx1 - mx2 = m1 - m2
1
2
1
2
Comparing Two Means
Sampling Distribution of a Difference
Between Two Means
+
 The
deviation σ1 and an independent SRS of size n2 from
Population
2 with mean
2
2
æ s1 ö æ s 2 ö
µ2 and standard deviation σ2.
= çNormal,
+the
÷
ç n ÷
Shape When the population distributions are
distribution
è n1 ø è sampling
2ø
of x1 - x 2 is approximately Normal. In other cases, the sampling distribution will
2
be approximately Normal if the sample sizes s
are
large
s 2 2enough ( n1 ³ 30,n2 ³ 30).
1
= is m+1 - m2 . That is, the difference
Center The mean of the sampling distribution
n1
n2
in sample means is an unbiased estimator of the
difference
in population means.
Spread The standard deviation of the sampling distribution of x1 - x 2 is
s12
s22
s 12
s 22
+ s
=
+
n1 n 2 x1 -x 2
n1
n2
as long as each sample is no more than 10% of its population (10% condition).
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, the sampling distribution
of x f - x m is Normal.
Its mean is m f - mm = 56.4 - 55.7 = 0.7 inches.
Its standard deviation is
2.7 2 3.8 2
+
= 1.55 inches.
12
8
Comparing Two Means

Who’s Taller at Ten, Boys or Girls?
+
 Example:
Standardize : When x f - x m = 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
 Example: Who’s Taller at Ten, Boys or Girls?
b) Find the probability of getting a difference in sample means
x f - x m that is less than 0.
(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 :
sx
1 -x 2
=
s12
n1
+
s2 2
n2
Since we don't know the values of the parameters s1 and s2 , 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 m1 -m2 .
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 ) - ( m1 - m2 )
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.
Intervals for µ1 – µ2
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 n 2 from Population 2 or by
two groups of size n1 and n 2 in a randomized experiment.
Normal Both population distributions are Normal OR both sample
group sizes are large (n1 ³ 30 and n2 ³ 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).
Comparing Two Means
Two-Sample t Interval for a Difference Between 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
2
southern trees and the actual mean
of the northern trees.
s1 sDBH
14.26 This
17.50 2
2
(x1 -the
x 2 )mean
± t * diameter
+
= (34.5
- 23.70)
± 1.699trees is+
interval suggests that
of
the
southern
n1 n 2
30
30
between 3.83 and 17.83 cm larger than= 10.83
the mean
of the
± 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
Problem:
(a)Construct and interpret a 99% confidence interval for the difference in mean
capacity of plastic grocery bags from Target and Walmart.
+
Comparing Two Means
Example: Plastic Grocery Bags
Do plastic bags from Target or plastic bags from Walmart hold more weight? A
group of AP Statistic students decided to investigate by filling a random sample
of 5 bags from each store with common grocery items until the bags ripped.
Then they weighed the contents of items in each bag to determine its capacity.
Here are their results, in grams:
Target: 12,572
13,999
11,215
15,447
10,896
Walmart: 9552
10,896
6983
8767
9972
State: We want to estimate at the 99% confidence level where mT = the mean
capacity of plastic bags from Target (in grams) and W = the mean capacity of plastic bags
from Walmart (in grams).
m
Plan: If the conditions are met, we will calculate a two-sample t interval
Random: The students selected a random sample of bags from each store.
Normal: Since the sample sizes are small, we must graph the data to see if it
is reasonable to assume that the population distributions are approximately
Normal.
Independent: The samples were selected independently and it is reasonable to
assume that there are more than 10(5) = 50 plastic grocery bags at each
store.
XT = 12825.8, sT = 1912.5, XB = 9234, sB = 1474.2
+
Do:
Conservative df = 5 – 1 = 4, the critical value for 99% confidence is t* = 4.604.
The confidence interval is:
1474.2 2 1912.5 2
(12826 - 9234) ± 4.604
+
=
5
5
3592 ± 4972 = (-1380,8564)
With technology and df =7.5, CI = (–101, 7285). Notice how much narrower this interval is.
Conclude: We are 99% confident that the interval from –1380 to 8564 grams captures
the true difference in the mean capacity for plastic grocery bags from Target and from
Walmart.
(b) Does your interval provide convincing evidence that there is a difference in the mean
capacity among the two stores?
Since the interval includes 0, it is plausible that there is no difference in the two means.
Thus, we do not have convincing evidence that there is a difference in mean capacity.
However, if we increased the sample size we would likely find a convincing difference since
it seems pretty clear that Target bags have a bigger capacity.
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
Tests for µ1 – µ2
statistic - parameter
test statistic =
standard deviation of statistic
t=
(x1 - x 2 ) - (m1 - m2 )
s12 s2 2
+
n1 n 2
To find the P-value, use the t distribution with degrees of freedom given by
technology or by the conservative approach (df = smaller of n1 - 1 and n2 - 1).
Comparing Two Means
To do a test, standardize x1 - x 2 to get a two - sample t statistic :
+
 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
t test for
between
two compute
means: the t statistic
testsample
the hypothesis
H the
: m difference
- m = hypothesized
value,

0
1
2
Random The data are produced by a random sample of size n1 from
(x - x 2 ) -size
( m1 n-2 mfrom
Population 1 and a randomt sample
Population 2 or by
2)
= 1 of
2
2
two groups of size n1 and n 2 in a randomized
s1
s2 experiment.
+
n1 n 2
Normal Both population distributions (or the true distributions
responses
two treatments)
are Normal
OR both
samplethis large
Findofthe
P - valuetobythe
calculating
the probabilty
of getting
a t statistic
groupin
sizes
are large specified
( n1 ³ 30 and
n 2 ³alternative
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 ) - ( m1 - m2 ) [5.000 - (-0.273)] - 0
t= 1 22
=
= 1.604
2
2
2
8.743 5.901
s1 s2
+
+
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:
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:
s12 s2 2
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.
Airplanes...which flies farther?
2. State the hypotheses you are interested in testing
3. Carry out your plan and collect the necessary data
4. Compare the flight distances for the two models (SOCS)
5. Perform a significance test (at the 0.05 level) to
determine if the difference in means is significant
6. Calculate a 95% confidence interval for the average
flight distance for YOUR PLANE
7. Construct a 95% confidence interval for the DIFFERENCE
in means.
Comparing Two Means
1. Design an experiment to determine which of two
airplane models flies farthest. Follow principals of
experimental design (Control, Random, Replication)
+
 Paper
Bounty:
+
Example: The stronger picker-upper?
In commercials for Bounty paper towels, the manufacturer claims that they are the
“quicker picker-upper.” But are they also the stronger picker upper? Two AP Statistics
students, Wesley and Maverick, decided to find out. They selected a random sample of
30 Bounty paper towels and a random sample of 30 generic paper towels and measured
their strength when wet. To do this, they uniformly soaked each paper towel with 4 ounces
of water, held two opposite edges of the paper towel, and counted how many quarters each
paper towel could hold until ripping, alternating brands. Here are their results:
106, 111, 106, 120, 103, 112, 115, 125, 116, 120, 126, 125, 116, 117, 114
118, 126, 120, 115, 116, 121, 113, 111, 128, 124, 125, 127, 123, 115, 114
Generic: 77, 103, 89, 79, 88, 86, 100, 90, 81, 84, 84, 96, 87, 79, 90
86, 88, 81, 91, 94, 90, 89, 85, 83, 89, 84, 90, 100, 94, 87
Problem:
The five-number summary for the Bounty paper towels is (103, 114, 116.5, 124, 128) and
(a)Display these distributions using parallel boxplots and briefly compare these distributions.
the five-number summary for the generic paper towels is (77, 84, 88, 90, 103).
Based only on the boxplots, discuss whether or not you think the mean for Bounty is significantly
higher than the mean for generic.
Both distributions are roughly symmetric, but the generic brand has two high outliers. The
center of the Bounty distribution is much higher than the center of the generic distribution.
Although the range of each distribution is roughly the same, the interquartile range of the
Bounty distribution is larger.
Since the centers are so far apart and there is almost no overlap in the two distributions, the
Bounty mean is almost certain to be significantly higher than the generic mean. If the means
were really the same, it would be virtually impossible to get so little overlap.
+
(b) Use a significance test to determine if there is convincing evidence that wet
Bounty paper towels can hold more weight, on average, than wet generic paper towels.
State: We want to perform a test of H0 : µB - µG = 0 versus Ha : µB - µG > 0 at the 5%
level of significance where µB = the mean number of quarters a wet Bounty paper towel
can hold and µG = the mean number of quarters a wet generic paper towel can hold.
Plan: If the conditions are met, we will conduct a two-sample t test for µB - µG .
Random: The students used a random sample of paper towels from each brand.
Normal: Even though there were two outliers in the generic distribution, both distributions
were reasonably symmetric and the sample sizes are both at least 30, so it is safe
to use t procedures.
Independent: The samples were selected independently and it is reasonable to assume
there are more than 10(30) = 300 paper towels of each brand.
- 88.1) - 0 evidence
Conclude:
Do:
the P-value
is lessXGthan
0.05, we
very convincing
XBSince
= 117.6
sB = 6.64
= 88.1
sG reject
= 6.30. Theret =is(117.6
= 17.64
2
that wet Bounty paper towels can hold more weight, on average, than wet
6.64generic
6.30 2paper towels.
+
30
30
P-value: Using either the conservative df = 30 – 1 = 29 or from technology (df = 57.8),
the P-value is approximately 0.
Since the P-value is approximately 0, it is almost impossible to get a difference in means of at
least 29.5 quarters by random chance, assuming that the two brands of paper towels can hold
the same amount of weight when wet.
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 ³ 30).
Center The mean m1 - m2 .
Spread As long as each sample is no more than 10% of its population
(10% condition), its standard deviation is
s 12
nn
+
s 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 ) - ( m1 - m2 )
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 n 2 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 ³ 30 and n 2 ³ 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
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 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 ) - ( m1 - m2 )
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.