Presentation is loading. Please wait.

Presentation is loading. Please wait.

Copyright © 2009 Cengage Learning Chapter 14 ANOVA.

Similar presentations


Presentation on theme: "Copyright © 2009 Cengage Learning Chapter 14 ANOVA."— Presentation transcript:

1 Copyright © 2009 Cengage Learning Chapter 14 ANOVA

2 Copyright © 2009 Cengage Learning 14.2 Analysis of Variance  Allows us to compare ≥2 populations of interval data.  ANOVA is:.  a procedure which determines whether differences exist between population means.  a procedure which analyzes sample variance. Referred to as treatments Not necessarily equal

3 Copyright © 2009 Cengage Learning 14.3 One Way ANOVA New Terminology:  Population classification criterion is called a factor  Each population is a factor level Example 14.1 (Stock Ownership) (1)  The analyst was interested in determining whether the ownership of stocks varied by age. (Xm14-01)Xm14-01  The age categories are:  Young (Under 35)  Early middle-age (35 to 49)  Late middle-age (50 to 65)  Senior (Over 65)

4 Copyright © 2009 Cengage Learning 14.4 Question: Does the stock ownership between the four age groups differs? n=366; % of financial assets invested in equity Factor: Age Category “One Way” ANOVA since only 1 factor 4 factor levels: (1)Young; (2)Early middle age; (3)Late middle age; & (4)Senior Hypothesis: H 0 :µ 1 = µ 2 = µ 3 = µ 4 i.e. there are no differences between population means H 1 : at least two means differ Example 14.1 (Stock Ownership) (2)

5 Copyright © 2009 Cengage Learning Example 14.1 (Stock Ownership) (3)  Rule….F-stat macam chapter 13 (ada paham?)  Rejection Region…. F > F,k–1, n–k  F-stat

6 Copyright © 2009 Cengage Learning 14.6  SST gave us the between-treatments variation  SSE (Sum of Squares for Error) measures the within-treatments variation  Sample Statistics & Grand Mean  Example 14.1 (Stock Ownership) (4)

7 Copyright © 2009 Cengage Learning 14.7  Hence, the between-treatments variation SST, is  Sample variances:  Then SSE: 161,871 Example 14.1 (Stock Ownership) (5)

8 Copyright © 2009 Cengage Learning 14.8 Example 14.1 (Stock Ownership) (6) F >F, (k–1),(n–k) = F 5%, (4 –1), (366–4) = 2.61 H 0 :µ 1 = µ 2 = µ 3 = µ 4 H 1 : at least two means differ  2.79 > 2.61 thus reject H 0 B-12

9 Copyright © 2009 Cengage Learning 14.9 Using Excel: ClickData, Data Analysis, Anova: Single Factor Example 14.1 (Stock Ownership) (7)

10 Copyright © 2009 Cengage Learning  p-value = < 0.05  Reject the null hypothesis (H 0 :µ 1 = µ 2 = µ 3 = µ 4 ) in favor of the alternative hypothesis (H 1 : at least two population means differ).  Conclusion: there is enough evidence to infer that the mean percentages of assets invested in the stock market differ between the four age categories. Example 14.1 (Stock Ownership) (8)

11 Copyright © 2009 Cengage Learning Multiple Comparisons If the conclusion is “at least two treatment means differ” i.e. reject the H 0 : We often need to know which treatment means are responsible for these differences 3 statistical inference procedures highlights it:  Fisher’s least significant difference (LSD) method  Tukey’s multiple comparison method

12 Copyright © 2009 Cengage Learning Fisher’s Least Significant Difference  A better estimator of the pooled variances = MSE  Substitute MSE in place of s p 2  Compares the difference between means to the Least Significant Difference LSD, given by:  LSD will be the same for all pairs of means if all k sample sizes are equal.  If some sample sizes differ, LSD must be calculated for each combination. Differ if absolute value of difference between means > LSD

13 Copyright © 2009 Cengage Learning Example 14.2 ( Car bumper Quality ) (1) North American automobile manufacturers have become more concerned with quality because of foreign competition. One aspect of quality is the cost of repairing damage caused by accidents. A manufacturer is considering several new types of bumpers. To test how well they react to low-speed collisions, 10 bumpers of each of 4 different types were installed on mid-size cars, which were then driven into a wall at 5 miles per hour.

14 Copyright © 2009 Cengage Learning  The cost of repairing the damage in each case was assessed. Xm14-02Xm14-02  Questions…  Is there sufficient evidence to infer that the bumpers differ in their reactions to low-speed collisions?  If differences exist, which bumpers differ?  Objective: to compare 4 populations  Data: interval  Samples: independent  Statistical method: One-way ANOVA. Example 14.2 ( Car bumper Quality ) (2)

15 Copyright © 2009 Cengage Learning  F = 4.06, p-value =  If 5% SL, then….Reject H o  There is enough evidence to infer that a difference exists between the 4 bumpers  The question now is…….which bumpers differ? Example 14.2 ( Car bumper Quality ) (3)

16 Copyright © 2009 Cengage Learning The sample means are…. We calculate the absolute value of the differences between means and compare them. Example 14.2 ( Car bumper Quality ) (4)

17 Copyright © 2009 Cengage Learning Click Add-Ins > Data Analysis Plus > Multiple Comparisons Example 14.2 ( Car bumper Quality ) (5)

18 Copyright © 2009 Cengage Learning Hence, µ 1 and µ 2, µ 1 and µ 3, µ 2 and µ 4, and µ 3 and µ 4 differ. The other two pairs µ 1 and µ 4, and µ 2 and µ 3 do not differ. Example 14.2 ( Car bumper Quality ) (6)

19 Copyright © 2009 Cengage Learning Tukey’s Multiple Comparison Method As before, we are looking for a critical number to compare the differences of the sample means against. In this case: Note: is a lower case Omega, not a “w” Critical value of the Studentized range with n–k degrees of freedom Table 7 - Appendix B harmonic mean of the sample sizes Different if the pair means diff >

20 Copyright © 2009 Cengage Learning Example 14.2 k = 4 N 1 = n 2 = n 3 = n 4 = n g = 10 Ν = 40 – 4 = 36 MSE = 12,399 Thus,

21 Copyright © 2009 Cengage Learning Example 14.1 (Tukey’s Method) Using Tukey’s method µ 2 and µ 4, and µ 3 and µ 4 differ.


Download ppt "Copyright © 2009 Cengage Learning Chapter 14 ANOVA."

Similar presentations


Ads by Google