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Chapter 14 ANOVA 1

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**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

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**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)

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**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

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**Example 14.1 (Stock Ownership) (3)**

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

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**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

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**Example 14.1 (Stock Ownership) (5)**

Hence, the between-treatments variation SST, is Sample variances: Then SSE: 161,871

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**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

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**Example 14.1 (Stock Ownership) (7)**

Using Excel: Click Data, Data Analysis, Anova: Single Factor

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**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.

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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

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**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

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**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.

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**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.

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**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?

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**Example 14.2 (Car bumper Quality) (4)**

The sample means are…. We calculate the absolute value of the differences between means and compare them.

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**Example 14.2 (Car bumper Quality) (5)**

Click Add-Ins > Data Analysis Plus > Multiple Comparisons

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**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.

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**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 >

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**Example 14.2 k = 4 N1 = n2 = n3 = n4 = ng = 10 Ν = 40 – 4 = 36**

MSE = 12,399 Thus,

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**Example 14.1 (Tukey’s Method)**

Using Tukey’s method µ2 and µ4, and µ3 and µ4 differ.

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Lecture 11 One-way analysis of variance (Chapter 15.2)

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