Kxu stat-anderson-ch10-student 2

1
Chapter 10
Comparisons Involving Means
µµ11
== µµ22
??
ANOVAANOVA
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

2
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
Interval Estimate of µ1−µ2: Large-Sample Case
Interval Estimate of µ1−µ2: Small-Sample Case
x x1 2−

3
Point Estimator of the Difference Between
the Means of Two Populations
Let µ1 equal the mean of population 1 and µ2 equal the
mean of population 2.
The difference between the two population means is
µ1 - µ2.
To estimate µ1 - µ2, 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 equal the mean of sample 1 and equal the
mean of sample 2.
The point estimator of the difference between the
means of the populations 1 and 2 is .
x x1 2−
x1 x2

4
E x x( )1 2 1 2− = −µ µ
n Properties of the Sampling Distribution ofProperties of the Sampling Distribution of
• Expected ValueExpected Value
Sampling Distribution ofSampling Distribution of x x1 2−
x x1 2−

5
Properties of the Sampling Distribution of
– Standard Deviation
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
Sampling Distribution of x x1 2−
x x1 2−
σ
σ σ
x x
n n1 2
1
2
1
2
2
2
− = +

6
Interval Estimate with σ1 and σ2 Known
where:
1 - α is the confidence coefficient (level).
Interval Estimate of µ1 - µ2:
Large-Sample Case (n1 > 30 and n2 > 30)
x x z x x1 2 2 1 2
− ± −α σ/

7
n Interval Estimate withInterval Estimate with σσ11 andand σσ22 UnknownUnknown
where:where:
Large-Sample Case (n1 > 30 and n2 > 30)
x x z sx x1 2 2 1 2
− ± −α/
s
s
n
s
n
x x1 2
1
2
1
2
2
2
− = +

8
Example: Par, Inc.
Interval Estimate of µ1 - µ2: 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.

9
Example: Par, Inc.
Interval Estimate of µ1 - µ2: Large-Sample Case
– Sample Statistics
Sample #1 Sample #2
Par, Inc. Rap, Ltd.
Sample Size n1 = 120 balls n2 = 80 balls
Mean = 235 yards = 218 yards
Standard Dev. s1 = ___yards s2 =____ yards
x1 2x

10
Point Estimate of the Difference Between Two
Population Means
µ1 = mean distance for the population of
Par, Inc. golf balls
µ2 = mean distance for the population of
Rap, Ltd. golf balls
Point estimate of µ1 - µ2 = = 235 - 218 = 17 yards.x x1 2−
Example: Par, Inc.

11
Point Estimator of the Difference Between
the Means of Two Populations
Population 1Population 1
Par, Inc. Golf BallsPar, Inc. Golf Balls
µµ11 = mean driving= mean driving
distance of Pardistance of Par
golf ballsgolf balls
Par, Inc. Golf BallsPar, Inc. Golf Balls
distance of Pardistance of Par
Rap, Ltd. Golf BallsRap, Ltd. Golf Balls
distance of Rapdistance of Rap
Rap, Ltd. Golf BallsRap, Ltd. Golf Balls
distance of Rapdistance of Rap
µµ11 –– µµ22 = difference between= difference between
the mean distancesthe mean distances
Simple random sampleSimple random sample
ofof nn11 Par golf ballsPar golf balls
xx11 = sample mean distance= sample mean distance
for sample of Par golf ballfor sample of Par golf ball
ofof nn11 Par golf ballsPar golf balls
for sample of Par golf ballfor sample of Par golf ball
ofof nn22 Rap golf ballsRap golf balls
for sample of Rap golf ballfor sample of Rap golf ball
ofof nn22 Rap golf ballsRap golf balls
for sample of Rap golf ballfor sample of Rap golf ball
xx11 -- xx22 = Point Estimate of= Point Estimate of µµ11 –– µµ22

12
95% Confidence Interval Estimate of the Difference
Between Two Population Means: Large-Sample Case,
σ1 and σ2 Unknown
Substituting the sample standard deviations for
the population standard deviation:
= ___________ 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.
x x z
n n
1 2 2
1
2
1
2
2
2
2 2
17 1 96
15
120
20
80
− ± + = ± +α
σ σ
/ .
( ) ( )
Example: Par, Inc.

13
Small-Sample Case (n1 < 30 and/or n2 < 30)
Interval Estimate with σ 2
Known (and equal)
where:
x x z x x1 2 2 1 2
− ± −α σ/
σ σx x
n n1 2
2
1 2
1 1
− = +( )

14
Interval Estimate with σ 2
Unknown (and assumed
equal)
where:
and the degrees of freedom for the t-distribution is
n1+n2-2.
Small-Sample Case (n1 < 30 and/or n2 < 30)
x x t sx x1 2 2 1 2
− ± −α /
s
n s n s
n n
2 1 1
2
2 2
2
1 2
1 1
2
=
− + −
+ −
( ) ( )
s s
n n
x x1 2
2
1 2
1 1
− = +( )

15
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-per-
gallon (mpg) performance. The sample statistics are:
Sample #1 Sample #2
M Cars J Cars
Sample Size n1 = 12 cars n2 = 8 cars
Mean = 29.8 mpg = 27.3 mpg
Standard Deviation s1 = ____ mpg s2 = ____ mpg
x2x1

16
Point Estimate of the Difference Between Two
Population Means
µ1 = mean miles-per-gallon for the population of
M cars
µ2 = mean miles-per-gallon for the population of
J cars
Point estimate of µ1 - µ2 = = ________ = ___ mpg.x x1 2−

17
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.

18
n 95% Confidence Interval Estimate of the Difference
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 σ 2
.

19
= _____________, 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).
s
n s n s
n n
2 1 1
2
2 2
2
1 2
2 2
1 1
2
11 2 56 7 1 81
12 8 2
5 28=
− + −
+ −
=
+
+ −
=
( ) ( ) ( . ) ( . )
.
x x t s
n n
1 2 025
2
1 2
1 1
2 5 2 101 5 28
1
12
1
8
− ± + = ± +. ( ) . . . ( )

20
Hypotheses
H0: µ1- µ2 < 0 H0: µ1 - µ2 > 0 H0: µ1- µ2 = 0
Ha: µ1- µ2 > 0 Ha: µ1 - µ2 < 0 Ha: µ1- µ2 ≠ 0
Test Statistic
Large-Sample Small-Sample
Hypothesis Tests About the Difference
between the Means of Two Populations:
Independent Samples
z
x x
n n
=
− − −
+
( ) ( )1 2 1 2
1
2
1 2
2
2
µ µ
σ σ
t
x x
s n n
=
− − −
+
( ) ( )
( )
1 2 1 2
2
1 21 1
µ µ

21
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.
Example: Par, Inc.

22
Hypothesis Tests About the Difference Between the
– Sample Statistics
Sample #1 Sample #2
Par, Inc. Rap, Ltd.
Sample Size n1 = 120 balls n2 = 80 balls
Mean = 235 yards = 218 yards
Standard Dev. s1 = ____ yards s2 = ____ yards
Example: Par, Inc.
x1 x2

23
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?
Example: Par, Inc.

24
n Hypothesis Tests About the Difference Between the
µ1 = mean distance for the population of Par, Inc.
golf balls
µ2 = mean distance for the population of Rap, Ltd.
golf balls
•Hypotheses H0: µ1 - µ2 < 0
Ha: µ1 - µ2 > 0
Example: Par, Inc.Example: Par, Inc.

25
– Rejection Rule
Reject H0 if z > ________
z
x x
n n
=
− − −
+
=
− −
+
= =
( ) ( ) ( )
( ) ( ) .
.1 2 1 2
1
2
1
2
2
2
2 2
235 218 0
15
120
20
80
17
2 62
6 49
µ µ
σ σ
Example: Par, Inc.

26
• 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.
Example: Par, Inc.Example: Par, Inc.

27
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-per-
gallon performance of J cars?

28
µ1 = mean mpg for the population of M cars
µ2 = mean mpg for the population of J cars
•Hypotheses H0: µ1 - µ2 < 0
Ha: µ1 - µ2 > 0
Example: Specific MotorsExample: Specific Motors

29
– Rejection Rule
Reject H0 if t > _______
(a = .05, d.f. = 18)
– Test Statistic
where:
t
x x
s n n
=
− − −
+
( ) ( )
( )
1 2 1 2
2
1 21 1
µ µ
2 2
2 1 1 2 2
1 2
( 1) ( 1)
2
n s n s
s
n n
− + −
=
+ −

30
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.

31
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?

32
Delivery Time (Hours)
District Office UPX INTEX Difference
Seattle 32 25 7
Los Angeles 30 24 6
Boston 19 15 4
Cleveland 16 15 1
New York 15 13 2
Houston 18 15 3
Atlanta 14 15 -1
St. Louis 10 8 2
Milwaukee 7 9 -2
Denver 16 11 5

33
Let µd= the mean of the difference values for the
two delivery services for the population of
district offices
– Hypotheses
H0: µd= 0, Ha: µd ≠ 0

34
n Inference About the Difference between the Means of
• Rejection Rule
Assuming the population of difference values is
approximately normally distributed, the t
distribution with n - 1 degrees of freedom applies.
With α = .05, t.025 = 2.262 (9 degrees of freedom).
Reject H0 if t < _________ or if t > __________
Example: Express DeliveriesExample: Express Deliveries

35
d
d
n
i
=
∑
=
+ + +
=
( ... )
.
7 6 5
10
2 7
s
d d
n
d
i
=
−∑
−
= =
( ) .
.
2
1
76 1
9
2 9
t
d
s n
d
d
=
−
=
−
=
µ 2 7 0
2 9 10
2 94
.
.
.

36
n Inference About the Difference between the Means of
• Conclusion
Reject H0.
There is a significant difference between the mean
delivery times for the two services.
Example: Express DeliveriesExample: Express Deliveries

37
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: µ1 = µ2 = µ3 = . . .= µk
Ha: Not all population means are equal

38
Introduction to Analysis of Variance
n If H0 is rejected, we cannot conclude that all
population means are different.
n Rejecting H0 means that at least two population
means have different values.

39
Assumptions for Analysis of Variance
For each population, the response variable is normally
distributed.
The variance of the response variable, denoted σ 2
, is
the same for all of the populations.
The observations must be independent.

40
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

41
A between-treatment estimate of σ 2
is called the
mean square treatment and is denoted MSTR.
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.
Between-Treatments Estimate
of Population Variance
1
)(
MSTR
1
2
−
−
=
∑=
k
xxn
k
j
jj

42
The estimate of σ 2
based on the variation of the
sample observations within each sample is called the
mean square error and is denoted by MSE.
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.
Within-Samples Estimate
of Population Variance
kn
sn
T
k
j
jj
−
−
=
∑=1
2
)1(
MSE

43
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 σ 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.

44
Test for the Equality of k Population Means
Hypotheses
H0: µ1 = µ2 = µ3 = . . .= µk
Ha: Not all population means are equal
Test Statistic
F = MSTR/MSE
Rejection Rule
Reject H0 if F > Fα
where the value of Fα is based on an F distribution with
k - 1 numerator degrees of freedom and nT - 1
denominator degrees of freedom.

45
Sampling Distribution of MSTR/MSE
The figure below shows the rejection region associated
with a level of significance equal to α where Fα denotes
the critical value.
Do Not Reject H0 Reject H0
MSTR/MSE
Critical Value
Fα

46
ANOVA Table
Source of Sum of Degrees of Mean
Variation Squares Freedom Squares F
Treatment SSTR k - 1 MSTR MSTR/MSE
Error SSE nT- k MSE
Total SST nT - 1
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
j
n
i
ij
j
xx
1 1
2
SSESSTR)(SST

47
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.

48
Plant 1 Plant 2 Plant 3
Observation Buffalo Pittsburgh Detroit
1 48 73 51
2 54 63 63
3 57 66 61
4 54 64 54
5 62 74 56
Sample Mean 55 68 57
Sample Variance ____ _____ ______

49
– Hypotheses
H0: µ1 = µ2 = µ3
Ha: Not all the means are equal
where:
µ1 = mean number of hours worked per
week by the managers at Plant 1
µ2= mean number of hours worked per
µ3 = mean number of hours worked per

50
– 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
=

51
– F - Test
If H0 is true, the ratio MSTR/MSE should be near
1 since both MSTR and MSE are estimating σ 2
. If
Hais true, the ratio should be significantly larger
than 1 since MSTR tends to overestimate σ 2
.

52
n Analysis of Variance
•Rejection Rule
Assuming α = .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 = _______
Example: Reed ManufacturingExample: Reed Manufacturing

53
– ANOVA Table
Source of Sum of Degrees of Mean
Variation Squares Freedom Square F
Treatments 490 2 245 9.55
Error 308 12 25.667
Total 798 14

54
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.
Example: Reed ManufacturingExample: Reed Manufacturing

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