Business Statistics CONTENTS Comparing two s Comparing more than - PowerPoint PPT Presentation

SEVERAL 𝜈 S: COMPARISON Business Statistics

CONTENTS Comparing two 𝜈 s Comparing more than two 𝜈 s Analysis of variance Testing significance of ANOVA Performing ANOVA The statistical model Equal variances Old exam question Further study

COMPARING TWO 𝜈 S Recall the comparison of the means of two independent samples ( 𝑍 1 and 𝑍 2 ) with the 𝑢 -test: We earlier wrote 𝑌 1 and 𝑌 2 or ▪ 𝐼 0 : 𝜈 1 = 𝜈 2 𝑌 and 𝑍 , so why not 𝑍 1 and 𝑍 2 ? 𝑍 1 −𝑍 ▪ under 𝐼 0 : ~𝑢 df , where df is the number of degrees 2 𝑡 𝑍 1 −𝑍 2 of freedom, depending on the assumption of the variances 𝑧 1 −𝑧 2 ▪ reject when 𝑢 calc = > 𝑢 crit 𝑡 𝑍1−𝑍2 Can we do this for three samples as well? ▪ 𝐼 0 : 𝜈 1 = 𝜈 2 = 𝜈 3

COMPARING MORE THAN TWO 𝜈 S A first attempt: 𝑍 1 −𝑍 ▪ think about how 𝐼 0 : 𝜈 1 = 𝜈 2 leads to 2 𝜏 diff ~𝑢 df 𝑍 1 −𝑍 2 −𝑍 ▪ try if 𝐼 0 : 𝜈 1 = 𝜈 2 = 𝜈 3 leads to 3 ~𝑢 df 𝜏 diff Very wrong! ▪ every year a few students try to do this at their exam, but this is a dead end road to take!

COMPARING MORE THAN TWO 𝜈 S A second attempt: ▪ pairwise comparisons: 𝑍 1 vs. 𝑍 2 and 𝑍 2 vs 𝑍 3 1 vs. 𝑍 3 not needed ▪ 𝑍 ▪ that implies doing two tests ▪ not one For each test the probability of incorrectly rejecting 𝐼 0 is 𝛽 ▪ so with 2 tests, it becomes 1 − 1 − 𝛽 2 ▪ example: 𝛽 = 0.05 gives 0.0975 , so almost double So that will not work ▪ think about testing 10 means (45 comparisons), the probability of a wrong decision becomes 90%

COMPARING MORE THAN TWO 𝜈 S A third attempt: ▪ We can partition the sum of squares into two parts: ▪ between the three groups ▪ within each group ▪ Try to identify sources of variation in a numerical dependent variable 𝑍 (the response variable) ▪ Variation in 𝑍 about its mean is partly explained by a categorical independent variable (the factor, with different levels) ▪ and partly unexplained (random error)

COMPARING MORE THAN TWO 𝜈 S Example ▪ Chips defect rates are different in every batch, but are there systematic differences between manufacturers? ▪ numerical dependent variable: chip defect rate ( 𝑍 ) ▪ categorical independent variable (one factor with four levels): manufacturer (1-4)

COMPARING MORE THAN TWO 𝜈 S Statistical model (formulation 1): ▪ manufaturer 1: 𝑍 𝑗1 = 𝜈 1 + 𝜁 𝑗1 ▪ ... ▪ manufaturer 4: 𝑍 𝑗4 = 𝜈 4 + 𝜁 𝑗4 Where: ▪ 𝑍 is the defect rate ▪ 𝜈 1 is the mean for manufacturer 1 ▪ ... ▪ 𝜈 4 is the mean for manufacturer 4 ▪ 𝜁 is the random, unexplained, part Null hypothesis: 𝜈 1 = 𝜈 2 = 𝜈 3 = 𝜈 4 ▪ 𝜈 1 = 𝜈 2 = 𝜈 3 = 𝜈 4

COMPARING MORE THAN TWO 𝜈 S Statistical model (formulation 2): This “group effect” 𝛽 has nothing to do with the ▪ manufaturer 1: 𝑍 𝑗1 = 𝜈 + 𝛽 1 + 𝜁 𝑗1 significance level 𝛽 ! ▪ ... ▪ manufaturer 4: 𝑍 𝑗4 = 𝜈 + 𝛽 4 + 𝜁 𝑗4 Where ▪ 𝑍 is the defect rate ▪ 𝜈 is the overall mean defect rate ▪ 𝛽 1 is manufacturer 1’s mean deviation from 𝜈 ▪ ... ▪ 𝛽 4 is manufacturer 4’s mean deviation from 𝜈 ▪ 𝜁 is the random, unexplained, part 𝜈 + 𝛽 1 = ⋯ = 𝜈 + 𝛽 4 Null hypothesis: ▪ 𝛽 1 = 𝛽 2 = 𝛽 3 = 𝛽 4 = 0

COMPARING MORE THAN TWO 𝜈 S ▪ What is the alternative hypothesis 𝐼 1 ? ▪ when 𝐼 0 : 𝜈 1 = 𝜈 2 = 𝜈 3 = 𝜈 4 ▪ Formulation 1: ▪ wrong: 𝐼 1 : 𝜈 1 ≠ 𝜈 2 ≠ 𝜈 3 ≠ 𝜈 4 ▪ correct: 𝐼 1 : 𝑜𝑝𝑢 𝜈 1 = 𝜈 2 = 𝜈 3 = 𝜈 4 ▪ or: at least one of the 𝜈 s differs from the other 𝜈 s 𝜈 1 = 𝜈 2 = 𝜈 3 ≠ 𝜈 4 𝜈 1 ≠ 𝜈 2 ≠ 𝜈 3 ≠ 𝜈 4

COMPARING MORE THAN TWO 𝜈 S ▪ What is the alternative hypothesis 𝐼 1 ? ▪ when 𝐼 0 : 𝛽 1 = 𝛽 2 = 𝛽 3 = 𝛽 4 = 0 ▪ Formulation 2: ▪ wrong: 𝐼 1 : 𝛽 1 ≠ 𝛽 2 ≠ 𝛽 3 ≠ 𝛽 4 ≠ 0 ▪ correct: 𝐼 1 : 𝑜𝑝𝑢 𝛽 1 = 𝛽 2 = 𝛽 3 = 𝛽 4 = 0 ▪ or: at least one of the 𝛽 s differs from 0 𝛽 1 = 𝛽 2 = 𝛽 3 ≠ 𝛽 4 𝛽 1 ≠ 𝛽 2 ≠ 𝛽 3 ≠ 𝛽 4

EXERCISE 1 We want to investigate possible differences in mean income in Atlanta, Boston, Chicago and Detroit. a. What is the null hypothesis? b. Suppose the null hypothesis is rejected. What can you conclude?

ANALYSIS OF VARIANCE ▪ Define notation: ▪ 𝑧 is the numerical value (e.g., chip defect rate) ▪ 𝑧 𝑗𝑘 is the value for observation # 𝑗 within treatment # 𝑘 (e.g., machine # 𝑘 ) 𝑧 ∙𝑘 is the average over all observations ( 𝑗 = 1, … , 𝑜 𝑘 ) within ▪ ത treatment # 𝑘 ▪ 𝑧 ∙∙ is the average over all observations within all treatments ( 𝑘 = ത ത 1, … , 𝑑 ) Observe the position of the ▪ Analysis of variance (ANOVA) model dots. A dot tells that index has been averaged over or 𝑍 𝑍 𝑗𝑘 = 𝜈 𝑘 + 𝜁 𝑗𝑘 𝑗𝑘 = 𝜈 + 𝛽 𝑘 + 𝜁 𝑗𝑘

ANALYSIS OF VARIANCE Compare variation within groups to variation between groups ▪ variation within group # 𝑘 : 𝑜 𝑘 2 ▪ 𝑇𝑇𝑋 𝑘 = σ 𝑗=1 𝑧 𝑗𝑘 − ത 𝑧 ∙𝑘 ▪ variation within all groups 𝑘 = 1, … , 𝑑 : 𝑑 ▪ 𝑇𝑇𝑋 = σ 𝑘=1 𝑇𝑇𝑋 𝑘 = 𝑜 𝑘 2 𝑑 ▪ σ 𝑘=1 σ 𝑗=1 𝑧 𝑗𝑘 − ത 𝑧 ∙𝑘 ▪ variation between the 𝑑 groups ▪ so due to the 𝛽 s: 2 𝑑 𝑧 ∙𝑘 − ത ▪ 𝑇𝑇𝐵 = σ 𝑘=1 𝑜 𝑘 ത 𝑧 ∙∙ ത So 𝑇𝑇𝐵 is the variation around the mean ത 𝑧 ∙∙ that is explained by the model, by factor “A”

ANALYSIS OF VARIANCE Together, 𝑇𝑇𝐵 and 𝑇𝑇𝑋 make up the total variation ▪ variation in entire data set: So 𝑇𝑇𝑈 is the total variation around the grand mean ഥ 𝑧 ∙∙ 𝑜 𝑘 2 𝑑 𝑧 𝑗𝑘 − ത ▪ 𝑇𝑇𝑈 = σ 𝑘=1 σ 𝑗=1 𝑧 ∙∙ ത ▪ so ▪ 𝑇𝑇𝑈 = 𝑇𝑇𝐵 + 𝑇𝑇𝑋 ▪ Think about the logic: we are comparing several means by comparing two variances ▪ analysis of variance is used to compare 𝜈 1 , 𝜈 2 , … , 𝜈 𝑑 𝑇𝑇𝑈 𝑇𝑇𝐵 𝑇𝑇𝐹

ANALYSIS OF VARIANCE

ANALYSIS OF VARIANCE Source of variation 𝑻𝑻 𝐞𝐠 𝑵𝑻 𝑮−ratio 𝑁𝑇𝐵 = 𝑇𝑇𝐵 𝐺 = 𝑁𝑇𝐵 𝑇𝑇𝐵 𝑑 − 1 between groups (due to factor “A”) 𝑑 − 1 𝑁𝑇𝑋 𝑁𝑇𝑋 = 𝑇𝑇𝑋 𝑇𝑇𝑋 𝑜 − 𝑑 within groups 𝑜 − 𝑑 𝑇𝑇𝑈 𝑜 − 1 total

EXERCISE 2 We sample from the four cities incomes from 100 persons ( 30 from Atlanta, 20 from Boston, 25 from Chicago and Detroit). a. What is 𝑜 and 𝑑 in the previous scheme? 𝑗𝑘 = 𝜈 𝑘 + 𝜁 𝑗𝑘 for the case of the 8 th respondent b. Specify 𝑍 from Chicago.

TESTING SIGNIFICANCE OF ANOVA What do we test? ▪ 𝐼 0 : 𝜈 1 = 𝜈 2 = ⋯ = 𝜈 𝑑 ▪ or equivalently 𝐼 0 : 𝛽 1 = 𝛽 2 = ⋯ = 𝛽 𝑑 = 0 How do we test? ▪ by comparing 𝑇𝑇𝐵 and 𝑇𝑇𝑋 ▪ or equivalently 𝑁𝑇𝐵 and 𝑁𝑇𝑋 (which are variances!) ▪ if 𝐼 0 is true, 𝑁𝑇𝐵 and 𝑁𝑇𝑋 are expected to be equal ▪ their ratio is the test statistic: 𝐺 = 𝑁𝑇𝐵 if this ratio is large, the group 𝑁𝑇𝑋 averages are likely to differ

TESTING SIGNIFICANCE OF ANOVA The test statistic 𝐺 ▪ is likely to be around 1 if 𝐼 0 is true ▪ is likely to be much larger than 1 if 𝐼 1 is true ▪ has a sampling distribution 𝐺 𝑑−1,𝑜−𝑑 under 𝐼 0 Here 𝐺 df 1 ,df 2 is the 𝐺 -distribution with df 1 and df 2 degrees of freedom 𝑁𝑇𝐵 Reject for large values of 𝐺 = 𝑁𝑇𝑋 only because we only reject 𝐼 0 if variations between groups are larger than expected under 𝐼 0

TESTING SIGNIFICANCE OF ANOVA So step 3 becomes: 𝑁𝑇𝐵 ▪ under 𝐼 0 , 𝐺 = 𝑁𝑇𝑋 ∼ 𝐺 𝑑−1,𝑜−𝑑 ▪ under the assumption: 𝜁 𝑗𝑘 ∼ 𝑂 0, 𝜏 2 In other words, the assumptions of ANOVA are: ▪ the observations 𝑧 𝑗𝑘 should be independent ▪ the sub-populations should be normally distributed ▪ the sub-populations should have equal variances Fortunately, ANOVA is somewhat robust to ▪ departures from normality and ▪ the equal variance assumptions

TESTING SIGNIFICANCE OF ANOVA Five step procedure for ANOVA ▪ Step 1: ▪ 𝐼 0 : 𝛽 1 = 𝛽 2 = ⋯ = 𝛽 𝑑 = 0 ; 𝐼 1 : not 𝐼 0 ; 𝛽 = 0.05 ▪ Step 2: 𝑁𝑇𝐵 sample statistic: 𝐺 = 𝑁𝑇𝑋 ; reject for large values ▪ ▪ Step 3: 𝑁𝑇𝐵 under 𝐼 0 : 𝐺 = ▪ 𝑁𝑇𝑋 ∼ 𝐺 𝑑−1,𝑜−𝑑 requirement: normal populations with equal variance ▪ ▪ Step 4: calculate 𝐺 crit = 𝐺 upper;df 1 ,df 2 ,𝛽 ▪ or calculate 𝑞 −value = 𝑄 𝐺 ≥ 𝐺 calc ▪ ▪ Step 5 ▪ reject 𝐼 0 if 𝐺 calc > 𝐺 crit ▪ or reject 𝐼 0 if 𝑞 −value < 𝛽