1. 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 = 2 ?
ANOVA

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 Large-Sample Case
Interval Estimate of Small-Sample Case

3. 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 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

4. Sampling Distribution of
Properties of the Sampling Distribution of
Expected Value

5. Sampling Distribution of
Properties of the Sampling Distribution of
Standard Deviation
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

6. Interval Estimate of 1 - 2: Large-Sample Case (n1 > 30 and n2 > 30)
Interval Estimate with 1 and 2 Known
where:
1 -  is the confidence coefficient

7. Interval Estimate of 1 - 2: Large-Sample Case (n1 > 30 and n2 > 30)
Interval Estimate with 1 and 2 Unknown
where:

8. 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.

9. Example: Par, Inc. Interval Estimate of 1 - 2: Large-Sample Case
Sample Statistics
Sample # Sample #2
Par, Inc Rap, Ltd.
Sample Size n1 = 120 balls n2 = 80 balls
Mean = 235 yards = 218 yards
Standard Dev. s1 = 15 yards s2 = 20 yards

10. 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 = = = 17 yards.

11. Point Estimator of the Difference Between the Means of Two Populations
Par, Inc. Golf Balls
m1 = mean driving
distance of Par
golf balls
Population 2
Rap, Ltd. Golf Balls
m2 = mean driving
distance of Rap
golf balls
m1 – m2 = difference between
the mean distances
Simple random sample
of n1 Par golf balls
x1 = sample mean distance
for sample of Par golf ball
Simple random sample
of n2 Rap golf balls
x2 = sample mean distance
for sample of Rap golf ball
x1 - x2 = Point Estimate of m1 – m2

12. 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:
= or yards to 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 to yards.

13. Interval Estimate of 1 - 2: Small-Sample Case (n1 < 30 and/or n2 < 30)
Interval Estimate with  2 Known
where:

14. Interval Estimate of 1 - 2: Small-Sample Case (n1 < 30 and/or n2 < 30)
Interval Estimate with  2 Unknown
where:

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 # Sample #2
M Cars J Cars
Sample Size n1 = 12 cars n2 = 8 cars
Mean = 29.8 mpg = 27.3 mpg
Standard Deviation s1 = 2.56 mpg s2 = 1.81 mpg

16. 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 = = = 2.5 mpg.

17. 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.

18. Example: Specific Motors
95% Confidence Interval Estimate of the Difference Between Two Population Means: Small-Sample Case
Using the t distribution with n1 + n2 - 2 = 18 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.

19. Example: Specific Motors
95% Confidence Interval Estimate of the Difference Between Two Population Means: Small-Sample Case
= 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).

20. Hypothesis Tests About the Difference between the Means of Two Populations: Independent Samples
Hypotheses
H0: 1 - 2 < H0: 1 - 2 > H0: 1 - 2 = 0
Ha: 1 - 2 > Ha: 1 - 2 < Ha: 1 - 2  0
Test Statistic
Large-Sample Small-Sample

21. 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.

22. Example: Par, Inc.
Hypothesis Tests About the Difference Between the Means of Two Populations: Large-Sample Case
Sample Statistics
Sample # Sample #2
Par, Inc Rap, Ltd.
Sample Size n1 = 120 balls n2 = 80 balls
Mean = 235 yards = 218 yards
Standard Dev. s1 = 15 yards s2 = 20 yards

23. 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?

24. Example: Par, Inc.
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.
Hypotheses H0: 1 - 2 < 0
Ha: 1 - 2 > 0

25. Example: Par, Inc.
Hypothesis Tests About the Difference Between the Means of Two Populations: Large-Sample Case
Rejection Rule
Reject H0 if z > 2.33

26. Example: Par, Inc.
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.

27. 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-per-gallon performance of J cars?

28. Example: Specific Motors
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

29. Example: Specific Motors
Hypothesis Tests About the Difference Between the Means of Two Populations: Small-Sample Case
Rejection Rule
Reject H0 if t > 1.734
(a = .05, d.f. = 18)
Test Statistic
where:

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. Example: Express Deliveries
Delivery Time (Hours)
District Office UPX INTEX Difference
Seattle
Los Angeles
Boston
Cleveland
New York
Houston
Atlanta
St. Louis
Milwaukee
Denver

33. 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 

34. Example: Express Deliveries
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 = (9 degrees of freedom).
Reject H0 if t < or if t > 2.262

35. Example: Express Deliveries
Inference About the Difference between the Means of Two Populations: Matched Samples

36. Example: Express Deliveries
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.

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: 1=2=3= = k
Ha: Not all population means are equal

38. Introduction to Analysis of Variance
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.

39. 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.

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. Between-Treatments Estimate of Population Variance
A between-treatment estimate of  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.

42. 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.
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.

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  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: 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.

45. 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
MSTR/MSE F
Critical Value

46. ANOVA Table Source of Sum of Degrees of Mean
Variation Squares Freedom Squares F
Treatment SSTR k 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.

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. Example: Reed Manufacturing
Analysis of Variance
Plant 1 Plant 2 Plant 3
Observation Buffalo Pittsburgh Detroit
Sample Mean
Sample Variance

49. 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

50. Example: Reed Manufacturing
Analysis of Variance
Mean Square Treatment
Since the sample sizes are all equal
x = ( )/3 = 60
SSTR = 5( )2 + 5( )2 + 5( )2 = 490
MSTR = 490/(3 - 1) = 245
Mean Square Error
SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308
MSE = 308/(15 - 3) = =

51. 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.

52. Example: Reed Manufacturing
Analysis of Variance
Rejection Rule
Assuming  = .05, F.05 = 3.89 (2 d.f. numerator,
12 d.f. denominator). Reject H0 if F > 3.89
Test Statistic
F = MSTR/MSE = 245/ = 9.55

53. Example: Reed Manufacturing
Analysis of Variance
ANOVA Table
Source of Sum of Degrees of Mean
Variation Squares Freedom Square F
Treatments
Error
Total

54. Example: Reed Manufacturing
Analysis of Variance
Conclusion
F = 9.55 > F.05 = 3.89, so we reject H0. The mean
number of hours worked per week by department
managers is not the same at each plant.