Business Statistics CONTENTS Comparing the variance of two - PowerPoint PPT Presentation

TWO 𝜏 2 S: COMPARISONS Business Statistics

CONTENTS Comparing the variance of two populations The 𝐺 -distribution The 𝐺 -test Levene’s test Old exam question Further study

COMPARING THE VARIANCE OF TWO POPULATIONS ▪ So far, the emphasis was on differences in centrality ▪ There are also questions on differences in dispersion ▪ Context: ▪ you can choose between two drilling machines ▪ both make holes of the specified size ▪ but the precision of the two may be different ▪ so you do an experiment (intended hole size: 3 mm)

COMPARING THE VARIANCE OF TWO POPULATIONS ▪ A second case ▪ Recall that we can compare two population means under the assumption of equal population variances ▪ using the pooled variance ▪ Thus, we may need to check if the populations variances are indeed equal

COMPARING THE VARIANCE OF TWO POPULATIONS Test statistic to consider 2 and S 𝑍 2 ▪ A combination of S 𝑌 2 is problematic 2 − 𝑇 𝑍 ▪ but the null distribution of 𝑇 𝑌 2 𝑇 𝑌 ▪ Instead, 2 is possible 𝑇 𝑍 2 𝑇 𝑍 ▪ or 2 (sometimes easier) 𝑇 𝑌 These hypotheses are ▪ What is the hypothesis? 2 equivalent to 𝑇 𝑌 2 = 𝜏 𝑍 2 = 1 2 ▪ 𝐼 0 : 𝜏 𝑌 𝑇 𝑍 2 ≥ 𝜏 𝑍 (or ≥ 1 or ≤ 1 ) 2 ▪ 𝐼 0 : 𝜏 𝑌 2 ≤ 𝜏 𝑍 2 ▪ 𝐼 0 : 𝜏 𝑌 2 = 𝜏 𝑍 2 + 3 etc. is not possible! ▪ but 𝜏 𝑌

THE 𝐺 -DISTRIBUTION ▪ For normally distributed populations, it is known that: 2 = 𝜏 𝑍 ▪ under 𝐼 0 : 𝜏 𝑌 2 : 2 𝑇 𝑌 2 ~𝐺 𝑜 𝑌 −1,𝑜 𝑍 −1 𝑇 𝑍 ▪ where 𝐺 df 1 ,df 2 is the F -distribution ▪ with df 1 and df 2 degrees of freedom ▪ note: 𝐺 is a ratio of two variances, use df 1 for the numerator and df 2 for the denominator

THE 𝐺 -DISTRIBUTION ▪ So, we compute the value of a test statistic 2 𝐺 calc = 𝑡 𝑌 2 𝑡 𝑍 ▪ and expect it to be around 1 if 𝐼 0 is true ▪ and reject 𝐼 0 if 𝐺 calc is “too” small or “too” large ▪ we need to look up the critical values of the 𝐺 -distribution 𝐺 crit,lower 1 𝐺 𝑑𝑠𝑗𝑢,𝑣𝑞𝑞𝑓𝑠 Of course (!) you expect 𝐺 = 1 when 𝐼 0 is true

THE 𝐺 -DISTRIBUTION ▪ 𝐺 -distribution ▪ like 𝜓 2 not symmetrical and strictly positive ▪ need to find 𝐺 crit,lower and 𝐺 crit,upper

THE 𝐺 -DISTRIBUTION df 1 Looking up critical values for 𝐺 𝛽/2 Is this 𝐺 crit,lower or 𝐺 crit,upper ? df 2

THE 𝐺 -DISTRIBUTION ▪ Finding 𝐺 crit,lower when you know 𝐺 crit,upper 1 𝐺 crit,lower df 1 , df 2 = 𝐺 crit,upper df 2 , df 1 1 ▪ so 𝐺 crit,lower = 3.85 = 0.26 2 2 𝑇 1 2 > 𝑏 ⇔ 𝑇 2 2 < 1 𝑇 2 𝑇 1 𝑏

THE 𝐺 -TEST Step 1: 2 = 𝜏 2 2 ≠ 𝜏 2 2 ; 𝐼 1 : 𝜏 1 2 ; 𝛽 = 0.05 ▪ 𝐼 0 : 𝜏 1 Step 2: 2 𝑇 1 ▪ sample statistic: 𝐺 = 2 ; reject for “too small” and “too large” 𝑇 2 values Step 3: ▪ null distribution: 𝐺~𝐺 df 1 ,df 2 ▪ both populations must be normally distributed (no CLT here!) Step 4: 2 𝑡 1 2 ; 𝐺 crit,lower = ⋯ ; 𝐺 crit,upper = ⋯ (use 𝛽/2 ) ▪ 𝐺 calc = 𝑡 2 Step 5: ▪ reject 𝐼 0 if 𝐺 calc < 𝐺 crit,lower or 𝐺 calc > 𝐺 crit,upper

THE 𝐺 -TEST Example: 2 = 0.012 with 𝑜 1 = 6 ▪ machine 1 gives 𝑡 1 2 = 0.016 with 𝑜 2 = 7 ▪ machine 2 gives 𝑡 2 Computations: upper critical values 0.012 can be read in the 𝐺 - ▪ 𝐺 calc = 0.016 = 0.75 table, so easier to look up 𝐺 crit,upper ▪ swap the two machines 0.016 ▪ 𝐺 calc = 0.012 = 1.33 ▪ null distribution: 𝐺~𝐺 6,5 ▪ 𝐺 crit;upper = 𝐺 6,5;0.025 = 6.98 ▪ 𝐺 calc < 𝐺 crit;upper , do not reject 𝐼 0

THE 𝐺 -TEST 2 = 𝜏 2 Trick to avoid the use of 𝐺 crit,lower in 𝐼 0 : 𝜏 1 2 ▪ put largest sample variance in numerator 2 𝑡 1 ▪ so calculated value of test statistic is 𝐺 calc = 2 or 𝐺 calc = 𝑡 2 2 𝑡 2 2 𝑡 1 ▪ with this, 𝐺 𝑑𝑏𝑚𝑑 > 1 , so we need only consider 𝐺 𝑑𝑠𝑗𝑢,𝑣𝑞𝑞𝑓𝑠 ▪ because 𝐺 𝑑𝑠𝑗𝑢,𝑚𝑝𝑥𝑓𝑠 is always < 1 ▪ formally reject for “too large” and “too small” ▪ so keep using 𝛽/2 and not 𝛽 (it is still a two-sided test)

THE 𝐺 -TEST One-sided tests: what is different compared to two-sided? Step 1: 2 ≥ 𝜏 2 2 < 𝜏 2 2 ; 𝐼 1 : 𝜏 1 2 ; 𝛽 = 0.05 ▪ 𝐼 0 : 𝜏 1 2 ≤ 𝜏 1 2 > 𝜏 1 ▪ reformulate as 𝐼 0 : 𝜏 2 2 ; 𝐼 1 : 𝜏 2 2 ; 𝛽 = 0.05 Step 2: 2 𝑇 2 ▪ sample statistic: 𝐺 = 2 ; reject for “too large” values 𝑇 1 Step 3: ▪ null distribution: 𝐺~𝐺 df 2 ,df 1 ▪ both populations must be normally distributed Step 4: 2 𝑡 2 2 ; 𝐺 crit,upper = ⋯ (use 𝛽 ) ▪ 𝐺 calc = 𝑡 1 Step 5: ▪ reject 𝐼 0 if 𝐺 𝑑𝑏𝑚𝑑 > 𝐺 crit,upper

EXERCISE 1 Given a sample 1 with 𝑜 1 = 9 and 𝑡 1 = 4 and a sample 2 2 = 𝜏 2 2 . with 𝑜 2 = 7 and 𝑡 2 = 5 . We want to test 𝐼 0 : 𝜏 1 a. What is step 2? b. What is step 3?

LEVENE’S TEST Recall that SPSS also did a comparison of two variances when we asked for comparing two means ▪ Levene’s test

LEVENE’S TEST The Levene test ▪ is based on a different principle ▪ requires no normal populations (!) ▪ also yields an 𝐺 calc ▪ but with different values of df ▪ reject for large values only 2 = 𝜏 2 2 ▪ yields a 𝑞 -value for 𝐼 0 : 𝜏 1

LEVENE’S TEST ▪ When comparing two 𝜈 s with the 𝑢 -test, we needed to 2 and 𝜏 2 2 estimate 𝜏 1 2 by 𝑡 1 2 and 𝜏 2 2 by 𝑡 2 ▪ we could estimate 𝜏 1 2 2 to estimate both 2 and use the pooled 𝑡 P 2 = 𝜏 2 ▪ or assume 𝜏 1 ▪ The second is only allowed if the two sample variances 𝑡 1 2 2 are not “too unequal” and 𝑡 2 Usually, a low 𝑞 - ▪ Levene’s test tests this value means bingo, but not so here ... ▪ a low 𝑞 - value means: unequal, so don’t do it ▪ do not necessarily use the same 𝛽 Rule of thumb: ▪ Why Levene and not the “normal” 𝐺 -test? use 𝛽 = 0.1 Levene is a non-parametric test ...

OLD EXAM QUESTION 26 March 2015, Q1c

FURTHER STUDY Doane & Seward 5/E 10.7 Tutorial exercises week 3 Fisher 𝐺 -test