Overview
Inferences
On Two Samples
• We continue with confidence intervals and
hypothesis testing for more advanced
models
• Models comparing two means
– When the two means are dependent
– When the two means are independent
• Models comparing two proportions
Learning Objectives
Inference about Two Means:
Dependent/paired Samples
Two populations
• So far, we have covered a variety of
models dealing with one population
– The mean parameter for one population
– The proportion parameter for one population
• However, there are many real-world
applications that need techniques to
compare two populations
• Distinguish between independent and
dependent sampling
• Test hypotheses made regarding matchedpairs data
• Construct and interpret confidence intervals
about the population mean difference of
matched-pairs data
Examples
• Examples of situations with two
populations
– We want to test whether a certain treatment
helps or not … the measurements are the
“before” measurement and the “after”
measurement
– We want to test the effectiveness of Drug A
versus Drug B … we give 40 patients Drug A
and 40 patients Drug B … the measurements
are the Drug A and Drug B responses
1
Dependent Sample
• In certain cases, the two samples are very
closely tied to each other
• A dependent sample is one when each
individual in the first sample is directly matched
to one individual in the second
• Examples
– Before and after measurements (a specific person’s
before and the same person’s after)
– Experiments on identical twins (twins matched with
each other
Independent Sample
• On the other extreme, the two samples can be
completely independent of each other
• An independent sample is when individuals
selected for one sample have no relationship to
the individuals selected for the other
• Examples
– Fifty samples from one factory compared to fifty
samples from another
– Two hundred patients divided at random into two
groups of one hundred
Paired Samples
• The dependent samples are often called
matched-pairs
• Matched-pairs is an appropriate term because
each observation in sample 1 is matched to
exactly one in sample 2
– The person before the person after
– One twin the other twin
– An experiment done on a person’s left eye the
same experiment done on that person’s right eye
Analysis of Paired Samples
• The method to analyze matched-pairs is to
combine the pair into one measurement
– “Before” and “After” measurements – subtract the
before from the after to get a single “change”
measurement
– “Twin 1” and “Twin 2” measurements – subtract the 1
from the 2 to get a single “difference between twins”
measurement
– “Left eye” and “Right eye” measurements – subtract
the left from the right to get a single “difference
between eyes” measurement
Test hypotheses made regarding
matched-pairs sample
Compute Difference d
• Specifically, for the before and after
example,
– d1 = person 1’s after – person 1’s before
– d2 = person 2’s after – person 1’s before
– d3 = person 3’s after – person 1’s before
• This creates a new random variable d
• We would like to reformulate our problem
into a problem involving d (just one
variable)
2
Test for the True Difference µd
• How do our hypotheses translate?
– The two means are equal -> the mean difference is
zero -> µd = 0
– The two means are unequal -> the mean difference is
non-zero -> µd ≠ 0
• Thus our hypothesis test is
– H 0 : µd = 0
– H 1 : µd ≠ 0
– The standard deviation σd is unknown
• We know how to do this!
Test for the True Difference
• To solve
– H0: µd = 0
– H1: µd ≠ 0
– The standard deviation σd is unknown
• This is exactly the test of one population
mean with the standard deviation being
unknown
• This is exactly the subject covered in Unit
8
Assumptions
• In order for this test statistic to be used, the data
must meet certain conditions
– The sample is obtained using simple random
sampling
– The sample data are matched pairs
– The differences are normally distributed, or
the sample size (the number of pairs, n) is at
least 30
• These are the usual conditions we need to make
our Student’s t calculations
Example (continued)
• Hypotheses
– H0: µd = 0 … no difference
– H1: µd > 0 … helps
– (We’re only interested in if our treatment makes things
better or not)
– α = 0.01
• Calculations
– n = 5 (i.e. 5 pairs)
– d = .88 (mean of the paired-difference)
– sd = .83
Example
• An example … whether our treatment helps or not …
helps meaning a higher measurement
• The “Before” and “After” results
Before
After
Difference
7.2
6.6
6.5
5.5
5.9
8.6
7.7
6.2
5.9
7.7
1.4
1.1
– 0.3
0.4
1.8
Example (continued)
• Calculations
– n=5
– d = 0.88
– sd = 0.83
• The test statistic is
t0 =
d − µd
0.88 − 0
=
= 2.36
s/ n
0.83 / 5
• This has a Student’s t-distribution with 4 degrees
of freedom
3
Example (continued)
Example (continued)
• Use the Student’s t-distribution with 4 degrees of
freedom
• The right-tailed α = 0.01 critical value is 3.75 (i.e. t0.01;4 d.f.
= 3.75)
• 2.36 is less than 3.75 (the classical method)
• Thus we do not reject the null hypothesis
• There is insufficient evidence to conclude that our
method significantly improves the situation
• We could also have used the P-Value method. P value
is 0.039 (note: tcdf(2.36, E99, 4) = 0.039)
• Matched-pairs tests have the same various
versions of hypothesis tests
– Two-tailed tests
– Left-tailed tests (the alternatively hypothesis
that the first mean is less than the second)
– Right-tailed tests (the alternatively hypothesis
that the first mean is greater than the second)
• Each can be solved using the Student’s t
Classical and P-value Approaches
Summary of the Method
• Each of the types of tests can be solved using
either the classical or the P-value approach
• A summary of the method
– For each matched pair, subtract the first observation
from the second
– This results in one data item per subject with the data
items independent of each other
– Test that the mean of these differences is equal to 0
• Conclusions
– Do not reject that µd = 0
– Reject that µd = 0 ... Reject that the two populations
have the same mean
Confidence Interval for the Paired
Difference
Construct and interpret confidence
intervals about the population mean
difference of matched-pairs data
• We’ve turned the matched-pairs problem in one for a
single variable’s mean / unknown standard deviation
– We just did hypothesis tests
– We can use the techniques taught in Unit 7 (again,
single variable’s mean / unknown standard deviation)
to construct confidence intervals
• The idea – the processes (but maybe not the specific
calculations) are very similar for all the different models
4
Confidence Interval for the Paired
Difference
• Confidence intervals are of the form
Point estimate ± margin of error
• This is precisely an application of our results for a
population mean / unknown standard deviation
– The point estimate
d
• Thus a (1 – α) • 100% confidence interval for
the difference of two means, in the matchedpair case, is
s
d ± tα / 2 d
n
where tα/2 is the critical value of the Student’s
t-distribution with n – 1 degrees of freedom
and the margin of error
tα / 2
Confidence Interval for the Paired
Difference
sd
n
for a two-tailed test
Example
Example (continued)
Salt-free diets are often prescribed for people with high blood
pressure. The following data was obtained from an experiment
designed to estimate the reduction in diastolic blood pressure as a
result of following a salt-free diet for two weeks. Assume diastolic
readings to be normally distributed.
Before
93
106
87
92
102
95
88
110
After
92
102
89
92
101
96
88
105
1
4
-2
0
1
-1
0
5
Difference
Find a 99% confidence interval for the mean reduction
1. Population Parameter of Interest
The mean reduction (difference) in diastolic blood
pressure
2. The Confidence Interval Criteria
a. Assumptions: Both sample populations are assumed
normal
b.
Test statistic: t with df = 8 − 1 = 7
c.
Confidence level: 1 − α = 0.99
3. Sample evidence
Sample information:n = 8,
Example
Confidence coefficients:
Two-tailed situation, α/2 = 0.005
t(df, α/2) = t(7, 0.005) = 3.50
b.
Maximum error:
E = ( 3.50 )(
c.
2.39
) = 2.957
8
Confidence limits:
d −E
1 . 0 − 2 . 957
to
to
and
sd = 2.39
Summary
4. The Confidence Interval
a.
d = 1.0,
d+E
1 . 0 + 2 .957
− 1 .957 to 3 .957
5. The Results
−1.957 to 3.957 is the 99% confidence interval estimate for the
amount of reduction of diastolic blood pressure, µd..
• Two sets of data are dependent, or matchedpairs, when each observation in one is matched
directly with one observation in the other
• In this case, the differences of observation
values should be used
• The hypothesis test and confidence interval for
the difference is a “mean with unknown standard
deviation” problem, one which we already know
how to solve
5
Learning Objectives
Inference about Two Means:
Independent Samples
• Test hypotheses regarding the difference of
two independent means
• Construct and interpret confidence intervals
regarding the difference of two independent
means
Independent Samples
Independent Samples
• Two samples are independent if the values in
one have no relation to the values in the other
• Examples of not independent
– Data from male students versus data from
business majors (an overlap in populations)
– The mean amount of rain, per day, reported in
two weather stations in neighboring towns
(likely to rain in both places)
• A typical example of an independent samples
test is to test whether a new drug, Drug N,
lowers cholesterol levels more than the current
drug, Drug C
• A group of 100 patients could be chosen
– The group could be divided into two groups of
50 using a random method
– If we use a random method (such as a simple
random sample of 50 out of the 100 patients),
then the two groups would be independent
Test of Two Independent Samples
• The test of two independent samples is
very similar, in process, to the test of a
single population mean
• The only major difference is that a different
test statistic is used
• We will discuss the new test statistic
through an analogy with the hypothesis
test of one mean
Test hypotheses regarding the
difference of two independent
means
6
Test Statistic for a Single Mean
• For the test of one mean, we have the variables
– The hypothesized mean (µ)
– The sample size (n)
– The sample mean (x)
– The sample standard deviation (s)
• We expect that x would be close to µ
Test statistic for the Difference of Two
Means
• In the test of two means, we have two values for each
variable – one for each of the two samples
– The two hypothesized means µ1 and µ2
– The two sample sizes n1 and n2
– The two sample means x1 and x2
– The two sample standard deviations s1 and s2
• We expect that x1 – x2 would be close to µ1 – µ2
Standard Error of the Test Statistic for
a Single Mean
Standard Error of the Test Statistic for
the Difference of Two Means
• For the test of one mean, to measure the
deviation from the null hypothesis, it is logical to
take
• For the test of two means, to measure the
deviation from the null hypothesis, it is logical to
take
(x1 – x2) – (µ1 – µ2)
x–µ
which has a standard deviation/standard error of
approximately
2
which has a standard deviation/standard error of
approximately
s
n
s12
s2
+ 2
n1
n2
t -Test Statistic for a Single Mean
t - Test Statistic for the Difference
of Two Means
• For the test of one mean, under certain
appropriate conditions, the difference
x–µ
is Student’s t with mean 0, and the test statistic
t=
x−µ
s2
n
has Student’s t-distribution with n – 1 degrees of
freedom
• Thus for the test of two means, under certain
appropriate conditions, the difference
(x1 – x2) – (µ1 – µ2)
is approximately Student’s t with mean 0, and
the test statistic
t=
( x1 − x 2 ) − ( µ1 − µ 2 )
s12 s22
+
n1 n2
has an approximate Student’s t-distribution
7
Distribution of the t-statistic
• This is Welch’s approximation, that
t=
( x1 − x 2 ) − ( µ1 − µ 2 )
s12 s22
+
n1 n2
has approximately a Student’s t-distribution
• The degrees of freedom is the smaller of
n1 – 1 and n2 – 1
Note: Some computer or calculator calculates the degrees of freedom for this
t test statistic with a somewhat complicated formula. But, we’ll use the smaller
of n1 – 1 and n2 – 1 as the degrees of freedom.
General Test Procedure
• Now for the overall structure of the test
– Set up the hypotheses
– Select the level of significance α
– Compute the test statistic
– Compare the test statistic with the appropriate
critical values
– Reach a do not reject or reject the null
hypothesis conclusion
State Hypotheses & level of
significance
• State our two-tailed, left-tailed, or righttailed hypotheses
• State our level of significance α, often
0.10, 0.05, or 0.01
A Special Case
• For the particular case where be believe that the
two population means are equal, or µ1 = µ2, and
the two sample sizes are equal, or n1 = n2, then
the test statistic becomes
t=
( x1 − x 2 )
 s2 + s2 
2 
 1


n


with n – 1 degrees of freedom, where n = n1 = n2
Assumptions
• In order for this method to be used, the data
must meet certain conditions
– Both samples are obtained using simple
random sampling
– The samples are independent
– The populations are normally distributed, or
the sample sizes are large (both n1 and n2 are
at least 30)
• These are the usual conditions we need to make
our Student’s t calculations
Compute the Test Statistic
• Compute the test statistic
t=
( x1 − x 2) − ( µ1 − µ2 )
s12 s22
+
n1 n2
and the degrees of freedom, the smaller of
n1 – 1 and n2 – 1
• Compute the critical values (for the two-tailed,
left-tailed, or right-tailed test
8
Make a Statistical Decision
• Each of the types of tests can be solved using
either the classical or the P-value approach
• Based on either of these methods, do not reject
or reject the null hypothesis
Example
• We have two independent samples
– The first sample of n = 40 items has a sample mean of
7.8 and a sample standard deviation of 3.3
– The second sample of n = 50 items has a sample
mean of 11.6 and a sample standard deviation of 2.6
– We believe that the mean of the second population is
exactly 4.0 larger than the mean of the first population
– We use a level of significance α = .05
• We test H : µ − µ = −4 versus H : µ − µ ≠ −4
0
1
2
1
1
2
Example (continued)
• The test statistic is
t=
( x1 − x 2 ) − ( µ 1 − µ 2 )
s12 s 22
+
n1 n 2
=
( 7.8 − 12.9 ) − ( −4.0 )
3.3 2 2.6 2
+
40
50
= −1.72
• This has a Student’s t-distribution with 39 degrees of
freedom
• The two-tailed critical value is -2.02, so we do not reject
the null hypothesis (notice: invT(.025,39) = -2.02 or use
a t-table)
• Or, compute the p-value which is 0.093 greater than 0.05
level of significance. (Notice that: 2*tcdf(-E99,-1.72,39)
= 0.093)
• We do not have sufficient evidence to state that the
deviation from 4.0 is significant
Confidence Interval of µ1−µ2
• Confidence intervals are of the form
Point estimate ± margin of error
• We can compare our confidence interval with the
test statistic from our hypothesis test
– The point estimate is x1 – x2
– We use the denominator of the test statistic as
the standard error
– We use critical values from the Student’s t
Construct and interpret confidence
intervals regarding the difference of
two independent means
Confidence Interval of µ1−µ2
• Thus (1- α)100% confidence interval is
Point estimate ± margin of error
( x1 − x 2 ) ± tα / 2
s12 s22
+
n1 n2
Point estimate
Standard error
where tα/2 has the degrees of freedom that is the smaller of n1-1 and n2-1 .
9
Example
Example
A recent study reported the longest average workweeks for nonsupervisory employees in private industry to be chef and
construction
1. Parameter of interest
The difference between the mean hours/week for chefs and the
mean hours/week for construction workers, µ1 - µ2
2. The Confidence Interval Criteria
Industry
n
Chef
Construction
18
12
Average Hours/Week Standard Deviation
48.2
44.1
6.7
2.3
Find a 95% confidence interval for the difference in mean length of
workweek between chef and construction. Assume normality for the
sampled populations and that the samples were selected randomly.
Example
4.
The Confidence Interval
a. Confidence coefficients:
t0.025, 11d.f.= 2.20
b. Margin of error:
E = ( 2.20 )(
6.7 2 2.3 2
+
) = 3.77
18
12
c. Confidence limits:
5.
4.1 – 3.77 = 0.33 to 4.1 + 3.77 = 7.87
The Results
0.33 to 7.87 is a 95% confidence interval for the difference in mean
hours/week for chefs and construction workers. ( It also means that there
is a significant difference between the mean hours/week for chefs and the
mean hours/week for construction workers at 0.05 level of significance,
since the interval does not contain zero.)
a.
Assumptions: Both populations are assumed normal and
the samples were random and independently selected
b.
Test statistic: t with df = 11;
the smaller of n1 − 1 = 18 − 1 = 17 or n2 − 1 = 12 − 1 = 11
c. Confidence level: 1 − α = 0.95
3. The Sample Evidence
Sample information given in the table
Point estimate for µ1 - µ2:
x1 − x 2 = 48.2 − 44.1 = 4.1
Summary
• Two sets of data are independent when
observations in one have no affect on
observations in the other
• In this case, the differences of the two means
should be used in a Student’s t-test
• The overall process, other than the formula for
the standard error, are the general hypothesis
test and confidence intervals process
Learning Objectives
Inference about
Two Population Proportions
• Test hypotheses regarding two population
proportions
• Construct and interpret confidence intervals
for the difference between two population
proportions
10
Inference about Two Proportions
• This progression should not be a surprise
• One mean and one proportion
Test hypotheses regarding two
population proportions
– Unit 7 – confidence intervals
– Unit 8 – hypothesis tests
• Two means
– Unit 9 - hypothesis tests and confidence
intervals
• Now for two proportions …
Examples
• We now compare two proportions, testing
whether they are the same or not
• Examples
– The proportion of women (population one) who have
a certain trait versus the proportion of men
(population two) who have that same trait
– The proportion of white sheep (population one) who
have a certain characteristic versus the proportion of
black sheep (population two) who have that same
characteristic
Test of One Proportion
• For the test of one proportion, we had the
variables of
–
–
–
–
The hypothesized population proportion (p0)
The sample size (n)
The number with the certain characteristic (x)
The sample proportion (p̂ = x / n)
• We expect that p̂ should be close to p0
Two Population Proportions
• The test of two populations proportions is
very similar, in process, to the test of one
population proportion and the test of two
population means
• The only major difference is that a different
test statistic is used
• We will discuss the new test statistic
through an analogy with the hypothesis
test of one proportion
Test of Two Proportions
• In the test of two proportions, we have two
values for each variable – one for each of the
two samples
– The two hypothesized proportions (p1 and p2)
– The two sample sizes (n1 and n2)
– The two numbers with the certain characteristic (x1
and x2)
– The two sample proportions ( p̂1 = x1 / n1 and
p̂2 = x2 / n2 )
• We expect that p̂1 − p̂2 should be close to p1 – p2
11
Test Statistic of One Proportion
• For the test of one proportion, to measure the
deviation from the null hypothesis, we took
p̂ − p0
Test Statistic of Two Proportions
• For the test of two proportions, to measure the
deviation from the null hypothesis, it is logical to
take
( p̂1 − p̂2 ) − ( p1 − p2 )
which has a standard deviation of
which has a standard deviation of
p0 ( 1− p0 )
n
p1( 1 − p1 )
p ( 1 − p2 )
+ 2
n1
n2
Test Statistic for One Proportion
• For the test of one proportion, under certain
appropriate conditions, the difference
Test Statistic for Two Proportions
• Thus for the test of two proportions, under
certain appropriate conditions, the difference
( p̂1 − p̂2 ) − ( p1 − p2 )
p̂ − p0
is approximately normal with mean 0, and the
test statistic
z=
p̂ − p0
p0 ( 1 − p0 )
n
z=
has an approximate standard normal distribution
Test Statistic for Equal Proportions
• For the particular case where we believe that the two
population proportions are equal, or p1 = p2 (i.e.
p1 – p2 = 0). Thus
( p̂1 − p̂2 ) − ( p1 − p2 ) = p̂1 − p̂2
and
z=
( p̂1 − p̂ 2 ) − ( p1 − p 2 )
p̂ c ( 1 − p̂ c )
p̂ ( 1 − p̂ c )
+ c
n1
n2
is approximately normal with mean 0, and the
test statistic
p̂1 − p̂ 2
=
p̂ c ( 1 − p̂ c )
1
1
+
n1
n2
Here, since two population proportions are the same under the null
hypothesis, we use p̂ c , an estimated common proportion for both p1 and p2,
which is computed by combining two samples together to calculate an
x1 + x 2
estimated common sample proportion. That is,
p̂ c =
( p̂1 − p̂2 ) − ( p1 − p2 )
p1( 1 − p1 )
p ( 1 − p2 )
+ 2
n1
n2
has an approximate standard normal distribution
General Test Procedure
• Now for the overall structure of the test
– Set up the hypotheses
– Select the level of significance α
– Compute the test statistic
– Compare the test statistic with the appropriate
critical values
– Reach a do not reject or reject the null
hypothesis conclusion
n1 + n 2
12
Hypotheses and Level of
Significance
Assumptions
• In order for this method to be used, the
data must meet certain conditions
– Both samples are obtained independently
using simple random sampling
– Each sample size is large
• These are the usual conditions we need to
make our test of proportions calculations
• State our two-tailed, left-tailed, or right-tailed
hypotheses
• State our level of significance α, often 0.10,
0.05, or 0.01
Test Statistic and Critical Values
• Compute the test statistic
z=
( p̂1 − p̂ 2 ) − ( p1 − p 2 )
P̂c ( 1 − P̂c )
Make Statistical Decision
• Each of the types of tests can be solved using
either the classical or the P-value approach
1
1
+
n1 n 2
which has an approximate standard normal distribution
• Compute the critical values (for the two-tailed, lefttailed, or right-tailed test)
• Based on either of these two methods, do not
reject the null hypothesis
Example
• We have two independent samples
– 55 out of a random sample of 100 students at one
university are commuters
– 80 out of a random sample of 200 students at another
university are commuters
– We wish to know of these two proportions are equal
– We use a level of significance α = .05
• Both samples sizes are large so our method can
be used
Example (continued)
•
The test statistic is
z=
( p̂1 − p̂ 2 ) − ( p1 − p 2 )
p̂c ( 1 − p̂ c )
Notice that
p̂ c =
•
•
•
1
1
+
n1 n2
0.55 − 0.40
=
0.45( 1 − 0.45 )
1
1
+
100 200
= 2.46
55 + 80
= 0.45
100 + 200
The critical values for a two-tailed test using the normal distribution
are ± 1.96, thus we reject the null hypothesis
Or, we calculate P-value which is 0.014 less than the 0.05 level of
significance. ( Notice: 2*normalcdf(2.46,E99) = 0.014)
We conclude that the two proportions are significantly different
13
Confidence Interval of p1 – p2
• Thus confidence intervals are
Point estimate ± margin of error
( p̂1 − p̂2 ) ± zα / 2
p̂1( 1 − p̂1 )
p̂ ( 1 − p̂2 )
+ 2
n1
n2
Point estimate
Standard error
Here, for calculating the standard error, we use separate estimates of
the population proportions, p̂1 , p̂ 2 instead of the common estimate p̂ c
Example (continued)
1.
2.
3.
Population Parameter of Interest : The difference
between the proportion of microcomputers needing
service for manufacturer 1 and the proportion of
microcomputers needing service for manufacturer 2,
that is, p1- p2
Point estimate: p̂1 − p̂ 2 = 0.15 − 0.09 = 0.06
Confidence coefficients:
z(α/2) = z(0.01) = 2.33
0.01
A consumer group compared the reliability of two similar
microcomputers from two different manufacturers. The proportion
requiring service within the first year after purchase was determined
for samples from each of two manufacturers.
Find a 98% confidence interval for p1 − p2, the difference in
proportions needing service
Manufacturer Sample Size Proportion Needing Service
1
200
0.15
2
250
0.09
Example (continued)
• Margin of error:
E = 2.33
( 0.15 )( 0.85 ) ( 0.09 )( 0.91 )
+
= 0.0724
200
250
• Confidence limits:
0.06 – 0.0724 = -0.0124
to
0.06 + 0.0724 = 0.1324
Results
−0.0124 to 0.1324 is a 98% confidence interval for the
difference in proportions
0.01
0 .98
0
Example
z(0.01)
2 .33
z
Summary
• We can compare proportions from two
independent samples
• We use a formula with the combined
sample sizes and proportions for the
standard error
• The overall process, other than the
formula for the standard error, are the
general hypothesis test and confidence
intervals process
Inferences
on Two Samples
Summary
14
Summary
• The process of hypothesis testing is very similar
across the testing of different parameters
• The major steps in hypothesis testing are
– Formulate the appropriate null and alternative
hypotheses
– Calculate the test statistic
– Determine the appropriate critical value or
values
– Reach the reject / do not reject conclusions
Tests for Means and Proportions
• Similarities in hypothesis test processes
Parameter
Mean (one
population)
Two Means
(Independent)
Two Means
(Dependent)
Two
Proportions
H0:
µ = µ0
µ1 = µ2
µ1 = µ2
p1 = p2
(2-tailed) H1:
µ ≠ µ0
µ1 ≠ µ2
µ1 ≠ µ2
p1 ≠ p2
(L-tailed) H1:
µ < µ0
µ1 < µ2
µ1 < µ2
p1 < p2
(R-tailed) H1:
µ > µ0
µ1 > µ2
µ1 > µ2
p1 > p2
Test statistic
Difference
Difference
Difference
Difference
Critical value
Normal
Normal
Student t
Normal
Summary
• We can test whether sample data from two
different samples supports a hypothesis claim
about a population mean or proportion
• For two population means, there are two cases
– Dependent (or matched-pair) samples
– Independent samples
• All of these tests follow very similar processes,
differing only in their test statistics and the
distributions for their critical values
15