# Chapter 14 ANOVA 1.

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Chapter 14 ANOVA 1

Referred to as treatments
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

One Way ANOVA Example 14.1 (Stock Ownership) (1) 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) The age categories are: Young (Under 35) Early middle-age (35 to 49) Late middle-age (50 to 65) Senior (Over 65)

Example 14.1 (Stock Ownership) (2)
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: H0:µ1 = µ2 = µ3 = µ4 i.e. there are no differences between population means H1: at least two means differ

Example 14.1 (Stock Ownership) (3)
Rule….F-stat macam chapter 13 (ada paham?) Rejection Region…. F > F ,k–1, n–k F-stat

Example 14.1 (Stock Ownership) (4)
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) (5)
Hence, the between-treatments variation SST, is Sample variances: Then SSE: 161,871

F >F ,(k–1),(n–k) = F5%, (4 –1), (366–4) = 2.61
Example 14.1 (Stock Ownership) (6) B-12 F >F ,(k–1),(n–k) = F5%, (4 –1), (366–4) = 2.61 H0:µ1 = µ2 = µ3 = µ4 H1: at least two means differ  2.79 > thus reject H0

Example 14.1 (Stock Ownership) (7)
Using Excel: Click Data, Data Analysis, Anova: Single Factor

Example 14.1 (Stock Ownership) (8)
p-value = < 0.05 Reject the null hypothesis (H0:µ1 = µ2 = µ3 = µ4) in favor of the alternative hypothesis (H1: 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.

Multiple Comparisons If the conclusion is “at least two treatment means differ” i.e. reject the H0: 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

Fisher’s Least Significant Difference
A better estimator of the pooled variances = MSE Substitute MSE in place of sp2 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

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.

Example 14.2 (Car bumper Quality) (2)
The cost of repairing the damage in each case was assessed. Xm14-02 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) (3)
F = 4.06, p-value = If 5% SL, then….Reject Ho 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) (4)
The sample means are…. We calculate the absolute value of the differences between means and compare them.

Example 14.2 (Car bumper Quality) (5)
Click Add-Ins > Data Analysis Plus > Multiple Comparisons

Example 14.2 (Car bumper Quality) (6)
Hence, µ1 and µ2, µ1 and µ3, µ2 and µ4, and µ3 and µ4 differ. The other two pairs µ1 and µ4, and µ2 and µ3 do not differ.

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 >

Example 14.2 k = 4 N1 = n2 = n3 = n4 = ng = 10 Ν = 40 – 4 = 36
MSE = 12,399 Thus,

Example 14.1 (Tukey’s Method)
Using Tukey’s method µ2 and µ4, and µ3 and µ4 differ.