# 1 Chapter 10 Comparisons Involving Means  1 =  2 ? ANOVA Estimation of the Difference between the Means of Two Populations: Independent Samples Hypothesis.

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1 Chapter 10 Comparisons Involving Means  1 =  2 ? ANOVA 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      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 n 1 from population 1 and a simple random sample of size n 2 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 n Properties of the Sampling Distribution of Expected Value Expected Value Sampling Distribution of

5 Properties of the Sampling Distribution of –Standard Deviation where:  1 = standard deviation of population 1  2 = standard deviation of population 2 n 1 = sample size from population 1 n 2 = sample size from population 2 Sampling Distribution of

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

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

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 #1 Sample #2 Par, Inc. Rap, Ltd. Sample Size n 1 = 120 balls n 2 = 80 balls Mean = 235 yards = 218 yards Standard Dev. s 1 = ___yards s 2 =____ yards

10 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 = = 235 - 218 = 17 yards. Example: Par, Inc.

11 Point Estimator of the Difference Between the Means of Two Populations Population 1 Par, Inc. Golf Balls  1 = mean driving distance of Par distance of Par golf balls Population 1 Par, Inc. Golf Balls  1 = mean driving distance of Par distance of Par golf balls Population 2 Rap, Ltd. Golf Balls  2 = mean driving distance of Rap distance of Rap golf balls Population 2 Rap, Ltd. Golf Balls  2 = mean driving distance of Rap distance of Rap golf balls  1 –  2 = difference between the mean distances the mean distances Simple random sample Simple random sample of n 1 Par golf balls of n 1 Par golf balls x 1 = sample mean distance for sample of Par golf ball Simple random sample Simple random sample of n 1 Par golf balls of n 1 Par golf balls x 1 = sample mean distance for sample of Par golf ball Simple random sample Simple random sample of n 2 Rap golf balls of n 2 Rap golf balls x 2 = sample mean distance for sample of Rap golf ball Simple random sample Simple random sample of n 2 Rap golf balls of n 2 Rap golf balls x 2 = sample mean distance for sample of Rap golf ball x 1 - x 2 = Point Estimate of  1 –  2

12 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 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. Example: Par, Inc.

13 Interval Estimate of  1 -  2 : Small-Sample Case ( n 1 < 30 and/or n 2 < 30) Interval Estimate with  2 Known (and equal) where:

14 Interval Estimate with  2 Unknown (and assumed equal) where: and the degrees of freedom for the t-distribution is n 1 +n 2 -2. Interval Estimate of  1 -  2 : Small-Sample Case ( n 1 < 30 and/or n 2 < 30)

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 n 1 = 12 cars n 2 = 8 cars Mean = 29.8 mpg = 27.3 mpg Standard Deviation s 1 = ____ mpg s 2 = ____ mpg

16 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 = = ________ = ___ mpg. Example: Specific Motors

17 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. Example: Specific Motors

18 n 95% Confidence Interval Estimate of the Difference Between Two Population Means: Small-Sample Case Using the t distribution with n 1 + n 2 - 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. Example: Specific Motors

19 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). Example: Specific Motors

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

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 Means of Two Populations: Large-Sample Case –Sample Statistics Sample #1 Sample #2 Par, Inc. Rap, Ltd. Sample Size n 1 = 120 balls n 2 = 80 balls Mean = 235 yards = 218 yards Standard Dev. s 1 = ____ yards s 2 = ____ yards Example: Par, Inc.

23 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? Example: Par, Inc.

24 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 H 0 :  1 -  2 < 0 H a :  1 -  2 > 0 Example: Par, Inc.

25 Hypothesis Tests About the Difference Between the Means of Two Populations: Large-Sample Case –Rejection Rule Reject H 0 if z > ________ Example: Par, Inc.

26 n Hypothesis Tests About the Difference Between the Means of Two Populations: Large-Sample Case Conclusion Reject H 0. 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.

27 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? Example: Specific Motors

28 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 H 0 :  1 -  2 < 0 H a :  1 -  2 > 0 Example: Specific Motors

29 Example: Specific Motors Hypothesis Tests About the Difference Between the Means of Two Populations: Small-Sample Case –Rejection Rule Reject H 0 if t > _______ ( 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 Delivery Time (Hours) District OfficeUPX 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 Example: Express Deliveries

33 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 H 0 :  d = 0, H a :  d  Example: Express Deliveries

34 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 H 0 if t __________ Example: Express Deliveries

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

36 n Inference About the Difference between the Means of Two Populations: Matched Samples Conclusion Reject H 0. There is a significant difference between the mean delivery times for the two services. Example: 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. H 0 :  1  =  2  =  3  = ... =  k  H a : Not all population means are equal

38 Introduction to Analysis of Variance n If H 0 is rejected, we cannot conclude that all population means are different. n Rejecting H 0 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 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. Between-Treatments Estimate of Population Variance

42 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. Within-Samples Estimate of Population Variance

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 n T - 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 H 0 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 H 0 :  1  =  2  =  3  = ... =  k  H a : Not all population means are equal Test Statistic F = MSTR/MSE Rejection Rule Reject H 0 if F > F  where the value of F  is based on an F distribution with k - 1 numerator degrees of freedom and n T - 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 H 0 Reject H 0 MSTR/MSE Critical Value FF FF

46 ANOVA Table Source of Sum of Degrees of Mean Variation Squares Freedom Squares F TreatmentSSTR k - 1 MSTR MSTR/MSE Error SSE n T - k MSE Total SST n T - 1 SST divided by its degrees of freedom n T - 1 is simply the overall sample variance that would be obtained if we treated the entire n T 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 Analysis of Variance Plant 1 Plant 2Plant 3 ObservationBuffalo 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 ____ _____ ______ Example: Reed Manufacturing

49 Analysis of Variance –Hypotheses H 0 :  1  =  2  =  3  H a : 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 Example: Reed Manufacturing

50 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 = = Example: Reed Manufacturing

51 Analysis of Variance – F - Test If H 0 is true, the ratio MSTR/MSE should be near 1 since both MSTR and MSE are estimating  2. If H a is true, the ratio should be significantly larger than 1 since MSTR tends to overestimate  2. Example: Reed Manufacturing

52 n Analysis of Variance Rejection Rule Assuming  =.05, F.05 = 3.89 (2 d.f. numerator, 12 d.f. denominator). Reject H 0 if F > _______ Test Statistic F = MSTR/MSE = 245/25.667 = _______ Example: Reed Manufacturing

53 Analysis of Variance –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 Example: Reed Manufacturing

54 n Analysis of Variance Conclusion F = 9.55 > F.05 = _____, so we reject H 0. The mean number of hours worked per week by department managers is not the same at each plant. Example: Reed Manufacturing

55 End of Chapter 10

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