Transcript Chapter 10 Comparisons Involving Means

Chapter 10
Comparisons Involving Means
Estimation of the Difference between the Means of
Two Populations: Independent Samples
Hypothesis Tests about the Difference between the
Means of Two Populations: Independent Samples
Inferences about the Difference between the Means
of Two Populations: Matched Samples
Introduction to Analysis of Variance (ANOVA)
ANOVA: Testing for the Equality of k Population
Means
1
Estimation of the Difference Between the Means
of Two Populations: Independent Samples
Point Estimator of the Difference between the Means
of Two Populations
Sampling Distribution x1  x2
Interval Estimate of Large-Sample Case
Interval Estimate of Small-Sample Case
2
Point Estimator of the Difference Between
the Means of Two Populations
Let 1 equal the mean of population 1 and 2 equal
the mean of population 2.
The difference between the two population means is
1 - 2.
To estimate 1 - 2, we will select a simple random
sample of size n1 from population 1 and a simple
random sample of size n2 from population 2.
Let x1 equal the mean of sample 1 and x2 equal the
mean of sample 2.
The point estimator of the difference between the
means of the populations 1 and 2 is x1  x2 .
3
Sampling Distribution of x1  x2
n
Properties of the Sampling Distribution of x1  x2
• Expected Value
E ( x1  x2 )  1   2
4
Sampling Distribution of
x1  x2
Properties of the Sampling Distribution of x1  x2
– Standard Deviation
 x1  x2 
12
n1

 22
n2
where: 1 = standard deviation of population 1
2 = standard deviation of population 2
n1 = sample size from population 1
n2 = sample size from population 2
5
Interval Estimate of 1 - 2:
Large-Sample Case (n1 > 30 and n2 > 30)
Interval Estimate with 1 and 2 Known
x1  x2  z / 2  x1  x2
where:
1 -  is the confidence coefficient (level).
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Interval Estimate of 1 - 2:
Large-Sample Case (n1 > 30 and n2 > 30)
n
Interval Estimate with 1 and 2 Unknown
x1  x2  z / 2 sx1  x2
where:
sx1  x2
s12 s22


n1 n2
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Example: Par, Inc.
Interval Estimate of 1 - 2: Large-Sample Case
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.
8
Example: Par, Inc.
Interval Estimate of 1 - 2: Large-Sample Case
– Sample Statistics
Sample Size
Mean
Standard Dev.
Sample #1
Par, Inc.
n1 = 120 balls
x1 = 235 yards
s1 = ___yards
Sample #2
Rap, Ltd.
n2 = 80 balls
x2 = 218 yards
s2 =____ yards
9
Example: Par, Inc.
Point Estimate of the Difference Between Two
Population Means
1 = mean distance for the population of
Par, Inc. golf balls
2 = mean distance for the population of
Rap, Ltd. golf balls
Point estimate of 1 - 2 = x1  x2 = 235 - 218 = 17 yards.
10
Point Estimator of the Difference Between
the Means of Two Populations
Population 1
Par, Inc. Golf Balls
Population 2
Rap, Ltd. Golf Balls
1 = mean driving
2 = mean driving
distance of Par
golf balls
distance of Rap
golf balls
1 – 2 = difference between
the mean distances
Simple random sample
of n1 Par golf balls
Simple random sample
of n2 Rap golf balls
x1 = sample mean distance
for sample of Par golf ball
x2 = sample mean distance
for sample of Rap golf ball
x1 - x2 = Point Estimate of 1 – 2
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Example: Par, Inc.
95% Confidence Interval Estimate of the Difference
Between Two Population Means: Large-Sample Case,
1 and 2 Unknown
Substituting the sample standard deviations for
the population standard deviation:
x1  x2  z / 2
12
 22
(15) 2 ( 20) 2

 17  1. 96

n1 n2
120
80
= ___________ 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 lies in the interval of _______________ yards.
12
Interval Estimate of 1 - 2:
Small-Sample Case (n1 < 30 and/or n2 < 30)
Interval Estimate with  2 Known (and equal)
x1  x2  z / 2  x1  x2
where:
 x1  x2
1 1
  (  )
n1 n2
2
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Interval Estimate of 1 - 2:
Small-Sample Case (n1 < 30 and/or n2 < 30)
Interval Estimate with  2 Unknown (and assumed
equal)
x1  x2  t / 2 sx1  x2
where:
sx1  x2
1 1
 s2 (  )
n1 n2
2
2
(
n

1
)
s

(
n

1
)
s
1
2
2
s2  1
n1  n2  2
and the degrees of freedom for the t-distribution is
n1+n2-2.
14
Example: Specific Motors
Specific Motors of Detroit has developed a new
automobile known as the M car. 12 M cars and 8 J cars
(from Japan) were road tested to compare miles-pergallon (mpg) performance. The sample statistics are:
Sample Size
Mean
Standard Deviation
Sample #1
M Cars
n1 = 12 cars
x1 = 29.8 mpg
s1 = ____ mpg
Sample #2
J Cars
n2 = 8 cars
x2 = 27.3 mpg
s2 = ____ mpg
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Example: Specific Motors
Point Estimate of the Difference Between Two
Population Means
1 = mean miles-per-gallon for the population of
M cars
2 = mean miles-per-gallon for the population of
J cars
Point estimate of 1 - 2 = x1  x2 = ________ = ___ mpg.
16
Example: Specific Motors
95% Confidence Interval Estimate of the Difference
Between Two Population Means: Small-Sample Case
We will make the following assumptions:
– The miles per gallon rating must be normally
distributed for both the M car and the J car.
– The variance in the miles per gallon rating must
be the same for both the M car and the J car.
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Example: Specific Motors
n
95% Confidence Interval Estimate of the Difference
Between Two Population Means: Small-Sample Case
Using the t distribution with n1 + n2 - 2 = ___ degrees
of freedom, the appropriate t value is t.025 = ______.
We will use a weighted average of the two sample
variances as the pooled estimator of  2.
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Example: Specific Motors
95% Confidence Interval Estimate of the Difference
Between Two Population Means: Small-Sample Case
2
2
2
2
(
n

1
)
s

(
n

1
)
s
11
(
2
.
56
)

7
(
1
.
81
)
1
2
2
s2  1

 5. 28
n1  n2  2
12  8  2
x1  x2  t.025
1 1
1 1
s (  )  2. 5  2.101 5. 28(  )
n1 n2
12 8
2
= _____________, or .3 to 4.7 miles per gallon.
We are 95% confident that the difference between the
mean mpg ratings of the two car types is from .3 to
4.7 mpg (with the M car having the higher mpg).
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Hypothesis Tests About the Difference
between the Means of Two Populations:
Independent Samples
Hypotheses
H0: 1 - 2 < 0
Ha: 1 - 2 > 0
H0: 1 - 2 > 0
Ha: 1 - 2 < 0
Test Statistic
Large-Sample
z
( x1  x2 )  ( 1   2 )
12 n1   22 n2
H0: 1 - 2 = 0
Ha: 1 - 2  0
Small-Sample
t
( x1  x2 )  ( 1   2 )
s2 (1 n1  1 n2 )
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Example: Par, Inc.
Hypothesis Tests About the Difference between the
Means of Two Populations: Large-Sample Case
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.
21
Example: Par, Inc.
Hypothesis Tests About the Difference Between the
Means of Two Populations: Large-Sample Case
– Sample Statistics
Sample Size
Mean
Standard Dev.
Sample #1
Par, Inc.
n1 = 120 balls
x1 = 235 yards
s1 = ____ yards
Sample #2
Rap, Ltd.
n2 = 80 balls
x2 = 218 yards
s2 = ____ yards
22
Example: Par, Inc.
Hypothesis Tests About the Difference Between the
Means of Two Populations: Large-Sample Case
Can we conclude, using a .01 level of
significance, that the mean driving distance of Par,
Inc. golf balls is greater than the mean driving
distance of Rap, Ltd. golf balls?
23
Example: Par, Inc.
n
Hypothesis Tests About the Difference Between the
Means of Two Populations: Large-Sample Case
1 = mean distance for the population of Par, Inc.
golf balls
2 = mean distance for the population of Rap, Ltd.
golf balls
• Hypotheses H0: 1 - 2 < 0
Ha: 1 - 2 > 0
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Example: Par, Inc.
Hypothesis Tests About the Difference Between the
Means of Two Populations: Large-Sample Case
– Rejection Rule
Reject H0 if z > ________
z
( x1  x2 )  ( 1   2 )
12
n1

 22
n2
( 235  218)  0
17


 6. 49
2
2
2. 62
(15) ( 20)

120
80
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Example: Par, Inc.
n
Hypothesis Tests About the Difference Between the
Means of Two Populations: Large-Sample Case
• Conclusion
Reject H0. We are at least 99% confident that the
mean driving distance of Par, Inc. golf balls is
greater than the mean driving distance of Rap,
Ltd. golf balls.
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Example: Specific Motors
Hypothesis Tests About the Difference Between the
Means of Two Populations: Small-Sample Case
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?
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Example: Specific Motors
n
Hypothesis Tests About the Difference Between the
Means of Two Populations: Small-Sample Case
1 = mean mpg for the population of M cars
2 = mean mpg for the population of J cars
• Hypotheses H0: 1 - 2 < 0
Ha: 1 - 2 > 0
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Example: Specific Motors
Hypothesis Tests About the Difference Between the
Means of Two Populations: Small-Sample Case
– Rejection Rule
Reject H0 if t > _______
(a = .05, d.f. = 18)
– Test Statistic
where:
t
( x1  x2 )  ( 1   2 )
s2 (1 n1  1 n2 )
(n1  1)s12  (n2  1)s22
s 
n1  n2  2
2
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Inference About the Difference between the
Means of Two Populations: Matched Samples
With a matched-sample design each sampled item
provides a pair of data values.
The matched-sample design can be referred to as
blocking.
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.
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Example: Express Deliveries
Inference About the Difference between the Means of
Two Populations: Matched Samples
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.
In testing the delivery times of the two services, the
firm sent two reports to a random sample of ten
district offices with one report carried by UPX and
the other report carried by INTEX.
Do the data that follow indicate a difference in
mean delivery times for the two services?
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Example: Express Deliveries
District Office
Seattle
Los Angeles
Boston
Cleveland
New York
Houston
Atlanta
St. Louis
Milwaukee
Denver
Delivery Time (Hours)
UPX
INTEX
Difference
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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
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Example: Express Deliveries
Inference About the Difference between the Means of
Two Populations: Matched Samples
Let d = the mean of the difference values for the
two delivery services for the population of
district offices
– Hypotheses
H0: d = 0, Ha: d 
33
Example: Express Deliveries
n
Inference About the Difference between the Means of
Two Populations: Matched Samples
• Rejection Rule
Assuming the population of difference values is
approximately normally distributed, the t
distribution with n - 1 degrees of freedom applies.
With  = .05, t.025 = 2.262 (9 degrees of freedom).
Reject H0 if t < _________ or if t > __________
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Example: Express Deliveries
Inference About the Difference between the Means of
Two Populations: Matched Samples
 di ( 7  6... 5)
d 

 2. 7
n
10
2
76.1
 ( di  d )
sd 

 2. 9
n 1
9
d  d
2. 7  0
t

 2. 94
sd n 2. 9 10
35
Example: Express Deliveries
n
Inference About the Difference between the Means of
Two Populations: Matched Samples
• Conclusion
Reject H0.
There is a significant difference between the mean
delivery times for the two services.
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Introduction to Analysis of Variance
Analysis of Variance (ANOVA) can be used to test
for the equality of three or more population means
using data obtained from observational or
experimental studies.
We want to use the sample results to test the
following hypotheses.
H0: 1 = 2 = 3 = . . . = k
Ha: Not all population means are equal
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Introduction to Analysis of Variance
n
n
If H0 is rejected, we cannot conclude that all
population means are different.
Rejecting H0 means that at least two population
means have different values.
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Assumptions for Analysis of Variance
For each population, the response variable is normally
distributed.
The variance of the response variable, denoted  2, is
the same for all of the populations.
The observations must be independent.
39
Analysis of Variance:
Testing for the Equality of k Population Means
Between-Treatments Estimate of Population Variance
Within-Treatments Estimate of Population Variance
Comparing the Variance Estimates: The F Test
The ANOVA Table
40
Between-Treatments Estimate
of Population Variance
A between-treatment estimate of  2 is called the
mean square treatment and is denoted MSTR.
k
MSTR 
2
n
(
x

x
)
 j j
j 1
k 1
The numerator of MSTR is called the sum of squares
treatment and is denoted SSTR.
The denominator of MSTR represents the degrees of
freedom associated with SSTR.
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Within-Samples Estimate
of Population Variance
The estimate of  2 based on the variation of the
sample observations within each sample is called the
mean square error and is denoted by MSE.
k
MSE 
2
(
n

1
)
s
 j
j
j 1
nT  k
The numerator of MSE is called the sum of squares
error and is denoted by SSE.
The denominator of MSE represents the degrees of
freedom associated with SSE.
42
Comparing the Variance Estimates: The F Test
If the null hypothesis is true and the ANOVA
assumptions are valid, the sampling distribution of
MSTR/MSE is an F distribution with MSTR d.f. equal
to k - 1 and MSE d.f. equal to nT - k.
If the means of the k populations are not equal, the
value of MSTR/MSE will be inflated because MSTR
overestimates  2.
Hence, we will reject H0 if the resulting value of
MSTR/MSE appears to be too large to have been
selected at random from the appropriate F
distribution.
43
Test for the Equality of k Population Means
Hypotheses
H0: 1 = 2 = 3 = . . . = k
Ha: Not all population means are equal
Test Statistic
F = MSTR/MSE
Rejection Rule
Reject H0 if F > F
where the value of F is based on an F distribution
with k - 1 numerator degrees of freedom and nT - 1
denominator degrees of freedom.
44
Sampling Distribution of MSTR/MSE
The figure below shows the rejection region associated
with a level of significance equal to  where F
denotes the critical value.
Do Not Reject H0
Reject H0
F
Critical Value
MSTR/MSE
45
ANOVA Table
Source of
Sum of
Variation Squares
Treatment SSTR
Error
SSE
Total
SST
Degrees of
Freedom
k-1
nT - k
nT - 1
Mean
Squares
F
MSTR MSTR/MSE
MSE
SST divided by its degrees of freedom nT - 1 is simply
the overall sample variance that would be obtained if
we treated the entire nT observations as one data set.
k
nj
SST   ( xij  x) 2  SSTR  SSE
j 1 i 1
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Example: Reed Manufacturing
Analysis of Variance
J. R. Reed would like to know if the mean
number of hours worked per week is the same for the
department managers at her three manufacturing
plants (Buffalo, Pittsburgh, and Detroit).
A simple random sample of 5 managers from
each of the three plants was taken and the number of
hours worked by each manager for the previous
week is shown on the next slide.
47
Example: Reed Manufacturing
Analysis of Variance
Observation
1
2
3
4
5
Sample Mean
Sample Variance
Plant 1
Buffalo
48
54
57
54
62
Plant 2
Pittsburgh
73
63
66
64
74
55
____
68
_____
Plant 3
Detroit
51
63
61
54
56
57
______
48
Example: Reed Manufacturing
Analysis of Variance
– Hypotheses
H0: 1 = 2 = 3
Ha: Not all the means are equal
where:
1 = mean number of hours worked per
week by the managers at Plant 1
2 = mean number of hours worked per
week by the managers at Plant 2
3 = mean number of hours worked per
week by the managers at Plant 3
49
Example: Reed Manufacturing
Analysis of Variance
– Mean Square Treatment
Since the sample sizes are all equal
x= = (55 + 68 + 57)/3 = ____
SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = ____
MSTR = 490/(3 - 1) = 245
– Mean Square Error
SSE = 4(26.0) + 4(26.5) + 4(24.5) = _____
MSE = 308/(15 - 3) = 25.667
50
Example: Reed Manufacturing
Analysis of Variance
– F - Test
If H0 is true, the ratio MSTR/MSE should be near
1 since both MSTR and MSE are estimating  2. If
Ha is true, the ratio should be significantly larger
than 1 since MSTR tends to overestimate  2.
51
Example: Reed Manufacturing
n
Analysis of Variance
• Rejection Rule
Assuming  = .05, F.05 = 3.89 (2 d.f. numerator,
12 d.f. denominator). Reject H0 if F > _______
• Test Statistic
F = MSTR/MSE = 245/25.667 = _______
52
Example: Reed Manufacturing
Analysis of Variance
– ANOVA Table
Source of
Variation
Treatments
Error
Total
Sum of Degrees of
Squares Freedom
490
308
798
2
12
14
Mean
Square
245
25.667
F
9.55
53
Example: Reed Manufacturing
n
Analysis of Variance
• Conclusion
F = 9.55 > F.05 = _____, so we reject H0. The mean
number of hours worked per week by department
managers is not the same at each plant.
54
End of Chapter 10
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