Chapter 10
Statistical Inference About Means and
Proportions With Two Populations
Slide 1
Learning objectives
1. Understand inferences About the Difference
Between Two Population Means: σ 1 and σ 2 Known
2. Understand Inferences About the Difference Between
Two Population Means: σ 1 and σ 2 Unknown
3. Understand Inferences About the Difference Between
Two Population Means: Matched Samples
4. Understand Inferences About the Difference Between
Two Population Proportions
Slide 2
1
Overview
Tests for two population
Two means
(Point estimator = x1 − x2 )
Two proportions
(Point estimator = p1 − p2)
σ1 and σ2 are known
Standard error
will be different
for each case.
σ1 and σ2 are unknown
Matched (or paired) sample
Slide 3
Comparison between one population case and
two population case.
nProcedures
of getting confidence interval and of
testing for two population case is very similar as those
of one population case
nConfidence
interval:
Point estimator ± Significance coefficient * Standard Error
nTest
Statistics
Point estimator - Hypothesized value of parameter
Standard error
nPoint
estimators and standard errors of two
population case are different from those of one
population case.
Slide 4
2
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Estimating the Difference Between
Two Population Means
Let µ1 equal the mean of population 1 and µ2 equal
the mean of population 2.
n
The difference between the two population means is
µ1 - µ2.
n
Let x1 equal the mean of sample 1 and x2 equal the
mean of sample 2.
n The point estimator of the difference between the
means of the populations 1 and 2 is x1 − x2.
Slide 5
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Sampling Distribution of x1 − x2 : σ known
Expected Value
E ( x1 − x 2 ) = µ 1 − µ 2
n
Standard Deviation (Standard Error)
σ x1 − x2 =
σ12 σ 22
+
n1 n2
where: σ1 = standard deviation of population 1
σ2 = standard deviation of population 2
n1 = sample size from population 1
n2 = sample size from population 2
Slide 6
3
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Interval Estimation of µ1 - µ2:
σ 1 and σ 2 Known
Interval Estimate
x1 − x2 ± zα / 2
σ12 σ 22
+
n1 n2
where:
1 - α is the confidence coefficient
Slide 7
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Interval Estimation of µ1 - µ2:
σ 1 and σ 2 Known
n Example: Par, Inc.
Par, Inc. is a manufacturer
of golf equipment and has
developed a new golf ball
that has been designed to
provide “extra distance.”
In a test of driving distance using a mechanical
driving device, a sample of Par golf balls was
compared with a sample of golf balls made by Rap,
Ltd., a competitor. The sample statistics appear on the
next slide.
Slide 8
4
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Interval Estimation of µ1 - µ2:
σ 1 and σ 2 Known
Example: Par, Inc.
Sample Size
Sample Mean
Sample #1
Par, Inc.
120 balls
275 yards
Sample #2
Rap, Ltd.
80 balls
258 yards
Based on data from previous driving distance
tests, the two population standard deviations are
known with σ 1 = 15 yards and σ 2 = 20 yards.
Slide 9
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Interval Estimation of µ1 - µ2:
σ 1 and σ 2 Known
Let us develop a 95% confidence interval estimate
of the difference between the mean driving distances of
the two brands of golf ball.
Point estimate of µ1 − µ2 = x1 − x2
= 275 − 258
= 17 yards
where:
µ1 = mean distance for the population
of Par, Inc. golf balls
µ2 = mean distance for the population
of Rap, Ltd. golf balls
Slide 10
5
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Interval Estimation of µ1 - µ2:
σ 1 and σ 2 Known
x1 − x2 ± zα / 2
σ12 σ 22
(15) 2 ( 20) 2
+
= 17 ± 1. 96
+
n1 n2
120
80
17 + 5.14 or 11.86 yards to 22.14 yards
We are 95% confident that the difference between
the mean driving distances of Par, Inc. balls and Rap,
Ltd. balls is 11.86 to 22.14 yards.
Slide 11
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Known
n Hypotheses
H0 : µ1 − µ2 ≥ D0 H0 : µ1 − µ2 ≤ D0 H0 : µ1 − µ2 = D0
Ha : µ1 − µ2 < D0 Ha : µ1 − µ2 > D0 Ha : µ1 − µ2 ≠ D0
Left-tailed
Right-tailed
Two-tailed
n Test Statistic
z=
( x1 − x2 ) − D0
σ 12 σ 22
+
n1 n2
Slide 12
6
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Known
Example: Par, Inc.
Can we conclude, using
α = .01, that the mean driving
distance of Par, Inc. golf balls
is greater than the mean driving
distance of Rap, Ltd. golf balls?
Slide 13
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Known
n p –Value and Critical Value Approaches
1. Develop the hypotheses.
H0 : µ 1 - µ 2 < 0
Ha : µ 1 - µ 2 > 0
where:
µ1 = mean distance for the population
of Par, Inc. golf balls
µ2 = mean distance for the population
of Rap, Ltd. golf balls
2. Specify the level of significance.
α = .01
Slide 14
7
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Known
n p –Value and Critical Value Approaches
3. Compute the value of the test statistic.
z=
z=
( x 1 − x 2 ) − D0
σ 12 σ 22
+
n1
n2
(235 − 218) − 0
2
(15)
(20 )
+
120
80
2
=
17
= 6.49
2.62
Slide 15
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Known
n p –Value Approach
4. Compute the p–value.
For z = 6.49, the p –value < .0001.
5. Determine whether to reject H0.
Because p–value < α = .01, we reject H0.
At the .01 level of significance, the sample evidence
indicates the mean driving distance of Par, Inc. golf
balls is greater than the mean driving distance of Rap,
Ltd. golf balls.
Slide 16
8
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Known
n Critical Value Approach
4. Determine the critical value and rejection rule.
For α = .01, z.01 = 2.33
Reject H0 if z > 2.33
5. Determine whether to reject H0.
Because z = 6.49 > 2.33, we reject H0.
The sample evidence indicates the mean driving
distance of Par, Inc. golf balls is greater than the mean
driving distance of Rap, Ltd. golf balls.
Slide 17
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
n
In-class exercise
#2 (p.400) (confidence interval)
#3 (p.400) (hypothesis testing)
Slide 18
9
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Interval Estimation of µ1 - µ2:
σ 1 and σ 2 Unknown
When σ 1 and σ 2 are unknown, we will:
• use the sample standard deviations s1 and s2
as estimates of σ 1 and σ 2 , and
• replace zα/2 with tα/2.
Slide 19
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Interval Estimation of µ1 - µ2:
σ 1 and σ 2 Unknown
Interval Estimate
x1 − x2 ± tα / 2
s12 s22
+
n1 n2
Where the degrees of freedom for tα/2 are:
2
 s12 s22 
 + 
 n1 n2 
df =
2
2
1  s12 
1  s22 
  +
 
n1 −1  n1  n2 −1  n2 
Slide 20
10
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Difference Between Two Population Means:
σ 1 and σ 2 Unknown
Example: Specific Motors
Specific Motors of Detroit
has developed a new automobile
known as the M car. 24 M cars
and 28 J cars (from Japan) were road
tested to compare miles-per-gallon (mpg) performance.
The sample statistics are shown on the next slide.
Slide 21
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Difference Between Two Population Means:
σ 1 and σ 2 Unknown
Example: Specific Motors
Sample #1
M Cars
24 cars
29.8 mpg
2.56 mpg
Sample #2
J Cars
28 cars
27.3 mpg
1.81 mpg
Sample Size
Sample Mean
Sample Std. Dev.
Slide 22
11
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Difference Between Two Population Means:
σ 1 and σ 2 Unknown
Example: Specific Motors
Let us develop a 90% confidence
interval estimate of the difference
between the mpg performances of
the two models of automobile.
Slide 23
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Point Estimate of µ 1 − µ 2
Point estimate of µ1 − µ2 = x1 − x2
= 29.8 - 27.3
= 2.5 mpg
where:
µ1 = mean miles-per-gallon for the
population of M cars
µ2 = mean miles-per-gallon for the
population of J cars
Slide 24
12
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Interval Estimation of µ 1 − µ 2:
σ 1 and σ 2 Unknown
The degrees of freedom for tα/2 are:
2
 (2.56)2 (1.81)2 
+


24
28 

df =
= 24.07 = 24
2 2
2 2
1  (2.56) 
1  (1.81) 

 +


24 −1 24  28 −1 28 
With α/2 = .05 and df = 24, tα/2 = 1.711
Slide 25
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Interval Estimation of µ 1 − µ 2:
σ 1 and σ 2 Unknown
x1 − x2 ± tα / 2
s12 s22
(2.56)2 (1.81)2
+ = 29.8 − 27.3 ± 1.711
+
n1 n2
24
28
2.5 + 1.069 or
1.431 to 3.569 mpg
We are 90% confident that the difference between
the miles-per-gallon performances of M cars and J cars
is 1.431 to 3.569 mpg.
Slide 26
13
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Unknown
Hypotheses
n
H0 : µ1 − µ2 ≥ D0 H0 : µ1 − µ2 ≤ D0 H0 : µ1 − µ2 = D0
Ha : µ1 − µ2 < D0 Ha : µ1 − µ2 > D0 Ha : µ1 − µ2 ≠ D0
Left-tailed
Right-tailed
Two-tailed
Test Statistic
n
t=
( x1 − x2 ) − D 0
s12 s 22
+
n1 n2
Slide 27
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Unknown
Example: Specific Motors
Can we conclude, using a
.05 level of significance, that the
miles-per-gallon (mpg) performance
of M cars is greater than the miles-pergallon performance of J cars?
Slide 28
14
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Unknown
n p –Value and Critical Value Approaches
1. Develop the hypotheses.
H0 : µ 1 - µ 2 < 0
Ha : µ 1 - µ 2 > 0
where:
µ1 = mean mpg for the population of M cars
µ2 = mean mpg for the population of J cars
Slide 29
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Unknown
n p –Value and Critical Value Approaches
2. Specify the level of significance.
α = .05
3. Compute the value of the test statistic.
t=
( x1 − x2 ) − D0
2
1
2
2
s
s
+
n1 n2
=
(29.8 − 27.3) − 0
(2.56) 2 (1.81) 2
+
24
28
= 4.003
Slide 30
15
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Unknown
n p –Value Approach
4. Compute the p –value.
The degrees of freedom for tα are:
2
df =
 (2.56)2 (1.81)2 
+


28 
 24
2
2
1  (2.56)2 
1  (1.81)2 
+




24 − 1  24  28 − 1  28 
= 24.07 = 24
Because t = 4.003 > t.005 = 2.797, the p–value < .005.
Slide 31
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Unknown
n p –Value Approach
5. Determine whether to reject H0.
Because p–value < α = .05, we reject H0.
We are at least 95% confident that the miles-pergallon (mpg) performance of M cars is greater than
the miles-per-gallon performance of J cars?.
Slide 32
16
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests About µ 1 − µ 2:
σ 1 and σ 2 Unknown
n Critical Value Approach
4. Determine the critical value and rejection rule.
For α = .05 and df = 24, t.05 = 1.711
Reject H0 if t > 1.711
5. Determine whether to reject H0.
Because 4.003 > 1.711, we reject H0.
We are at least 95% confident that the miles-pergallon (mpg) performance of M cars is greater than
the miles-per-gallon performance of J cars?.
Slide 33
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
n
In-class exercise
#9 (p.406) (confidence interval)
#10 (p.407) (hypothesis testing)
Slide 34
17
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Inferences About the Difference Between
Two Population Means: Matched Samples
n With a matched-sample design each sampled item
provides a pair of data values.
n This design often leads to a smaller sampling error
than the independent-sample design because
variation between sampled items is eliminated as a
source of sampling error.
Slide 35
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Inferences About the Difference Between
Two Population Means: Matched Samples
Example: Express Deliveries
A Chicago-based firm has
documents that must be quickly
distributed to district offices
throughout the U.S. The firm
must decide between two delivery
services, UPX (United Parcel Express) and INTEX
(International Express), to transport its documents.
Slide 36
18
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Inferences About the Difference Between
Two Population Means: Matched Samples
Example: Express Deliveries
In testing the delivery times
of the two services, the firm sent
two reports to a random sample
of its district offices with one
report carried by UPX and the
other report carried by INTEX. Do the data on the
next slide indicate a difference in mean delivery
times for the two services? Use a .05 level of
significance.
Slide 37
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Inferences About the Difference Between
Two Population Means: Matched Samples
Delivery Time (Hours)
District Office UPX INTEX Difference
Seattle
Los Angeles
Boston
Cleveland
New York
Houston
Atlanta
St. Louis
Milwaukee
Denver
32
30
19
16
15
18
14
10
7
16
25
24
15
15
13
15
15
8
9
11
7
6
4
1
2
3
-1
2
-2
5
Slide 38
19
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Inferences About the Difference Between
Two Population Means: Matched Samples
n p –Value and Critical Value Approaches
1. Develop the hypotheses.
H0 : µ d = 0
Ha : µ d ≠ 0
Let µd = the mean of the difference values for the
two delivery services for the population
of district offices
Slide 39
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Inferences About the Difference Between
Two Population Means: Matched Samples
n p –Value and Critical Value Approaches
2. Specify the level of significance.
α = .05
3. Compute the value of the test statistic.
d =
sd =
t=
∑ di ( 7 + 6 +... +5)
=
= 2. 7
n
10
2
76.1
∑ ( di − d )
=
= 2. 9
n −1
9
d − µd
2.7 − 0
=
= 2.94
sd n 2.9 10
Slide 40
20
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Inferences About the Difference Between
Two Population Means: Matched Samples
n p –Value Approach
4. Compute the p –value.
For t = 2.94 and df = 9, the p–value is between
.02 and .01. (This is a two-tailed test, so we double
the upper-tail areas of .01 and .005.)
5. Determine whether to reject H0.
Because p–value < α = .05, we reject H0.
We are at least 95% confident that there is a
difference in mean delivery times for the two
services?
Slide 41
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Inferences About the Difference Between
Two Population Means: Matched Samples
n Critical Value Approach
4. Determine the critical value and rejection rule.
For α = .05 and df = 9, t.025 = 2.262.
Reject H0 if t > 2.262
5. Determine whether to reject H0.
Because t = 2.94 > 2.262, we reject H0.
We are at least 95% confident that there is a
difference in mean delivery times for the two
services?
Slide 42
21
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
n
In-class exercise
#21 (p.414) (hypothesis testing)
#22 (p.414) (confidence interval)
Slide 43
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Sampling Distribution of p1 − p2
Expected Value
E ( p1 − p2 ) = p1 − p2
n
Standard Deviation (Standard Error)
σ p1 − p2 =
p1 (1 − p1 ) p2 (1 − p2 )
+
n1
n2
where: n1 = size of sample taken from population 1
n2 = size of sample taken from population 2
Slide 44
22
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Sampling Distribution of p1 − p2
If the sample sizes are large, the sampling distribution
of p1 − p2 can be approximated by a normal probability
distribution.
The sample sizes are sufficiently large if all of these
conditions are met:
n1p1 > 5
n1(1 - p1) > 5
n2p2 > 5
n2(1 - p2) > 5
Slide 45
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Sampling Distribution of p1 − p2
σ p1 − p2 =
p1 (1 − p1 ) p2 (1 − p2 )
+
n1
n2
p1 − p2
p1 – p2
Slide 46
23
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Interval Estimation of p1 - p2
Interval Estimate
p1 − p2 ± zα / 2
p1 (1 − p1 ) p2 (1 − p2 )
+
n1
n2
Slide 47
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Interval Estimation of p1 - p2
Example: Market Research Associates
Market Research Associates is
conducting research to evaluate the
effectiveness of a client’s new advertising campaign. Before the new
campaign began, a telephone survey
of 150 households in the test market
area showed 60 households “aware” of
the client’s product.
The new campaign has been initiated with TV and
newspaper advertisements running for three weeks.
Slide 48
24
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Interval Estimation of p1 - p2
Example: Market Research Associates
n
A survey conducted immediately
after the new campaign showed 120
of 250 households “aware” of the
client’s product.
Does the data support the position
that the advertising campaign has
provided an increased awareness of
the client’s product?
Slide 49
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Point Estimator of the Difference Between
Two Population Proportions
p1 = proportion of the population of households
“aware” of the product after the new campaign
p2 = proportion of the population of households
“aware” of the product before the new campaign
p1 = sample proportion of households “aware” of the
product after the new campaign
p2 = sample proportion of households “aware” of the
product before the new campaign
p1 − p2 =
120 60
−
= .48 − .40 = .08
250 150
Slide 50
25
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Interval Estimation of p1 - p2
For α = .05, z.025 = 1.96:
.48 − .40 ± 1.96
.48(.52) .40(.60)
+
250
150
.08 + 1.96(.0510)
.08 + .10
Hence, the 95% confidence interval for the difference
in before and after awareness of the product is
-.02 to +.18.
Slide 51
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
Hypothesis Tests about p1 - p2
Hypotheses
We focus on tests involving no difference between
the two population proportions (i.e. p1 = p2)
H0 : p1 − p2 ≥ 0
Ha : p1 − p2 < 0
Left-tailed
H
H00::
H
Ha::
a
pp1 −
- p2 < 0
1 p2 ≤ 0
pp1 −
- pp2 >> 00
1
2
Right-tailed
H0 : p1 − p2 = 0
Ha : p1 − p2 ≠ 0
Two-tailed
Slide 52
26
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests about p1 - p2
n Pooled Estimate of Standard Error of p1 − p2
1 1
σ p1− p2 = p(1− p)  + 
 n1 n2 
where:
p=
n1 p1 + n2 p2
n1 + n2
Slide 53
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests about p1 - p2
n Test Statistic
z=
( p1 − p 2 )
 1
1 
p (1 − p ) 
+

 n1 n2 
Slide 54
27
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests about p1 - p2
Example: Market Research Associates
Can we conclude, using a .05 level
of significance, that the proportion of
households aware of the client’s product
increased after the new advertising
campaign?
n
Slide 55
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests about p1 - p2
n p -Value and Critical Value Approaches
1. Develop the hypotheses.
H0: p1 - p2 < 0
Ha: p1 - p2 > 0
p1 = proportion of the population of households
“aware” of the product after the new campaign
p2 = proportion of the population of households
“aware” of the product before the new campaign
Slide 56
28
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests about p1 - p2
n p -Value and Critical Value Approaches
2. Specify the level of significance.
α = .05
3. Compute the value of the test statistic.
p=
250(. 48) + 150(. 40) 180
=
= . 45
250 + 150
400
s pp1 −−pp2 = . 45(. 55)( 1
+ 1 ) =. 0514
250 150
1
2
z=
(.48 − .40) − 0
.08
=
= 1.56
.0514
.0514
Slide 57
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests about p1 - p2
n p –Value Approach
4. Compute the p –value.
For z = 1.56, the p–value = .0594
5. Determine whether to reject H0.
Because p–value > α = .05, we cannot reject H0.
We cannot conclude that the proportion of households
aware of the client’s product increased after the new
campaign.
Slide 58
29
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
Hypothesis Tests about p1 - p2
n Critical Value Approach
4. Determine the critical value and rejection rule.
For α = .05, z.05 = 1.645
Reject H0 if z > 1.645
5. Determine whether to reject H0.
Because 1.56 < 1.645, we cannot reject H0.
We cannot conclude that the proportion of households
aware of the client’s product increased after the new
campaign.
Slide 59
•L.O.
•Test for means
−σ known
−σ unknown
−Matched sample
•Test for proportions
n
n
In-class exercise
#30 (p.420) (confidence interval)
#31 (p.421) (hypothesis testing)
Slide 60
30
End of Chapter 10
Slide 61
Formulas for two population case
Confidence
Interval
Means:
σ known
Means:
σ unknown
Means:
matched
Proportion
Test
Statistics
x1 − x2 ± zα /2
σ12 σ22
+
n1 n2
z=
x1 − x2 ± tα / 2
s12 s22
+
n1 n2
t=
Other
formula
( x1 − x2 ) − D0
σ 12 σ 22
+
n1 n2
( x1 − x 2 ) − D 0
2
1
2
2
s
s
+
n1 n 2
2
df =
 s12 s22 
 + 
 n1 n2 
2
2
1  s12 
1  s22 
  +
 
n1 −1 n1  n2 −1 n2 
Same as one population case
p1 − p2 ±zα/2
p1(1− p1) p2(1− p2)
+
n1
n2
z=
( p1 − p2 )
 1
1 
p (1 − p ) 
+

 n1 n 2 
p=
n1p1 +n2p2
n1 +n2
Slide 62
31