Multiple Comparisons Examples STAT 314

Transcription

1 Multiple Comparisos Examples STAT 31 Problem umbers match those from the ANOVA Examples hadout. 8. Four brads of flashlight batteries are to be compared by testig each brad i five flashlights. Twety flashlights are radomly selected ad divided radomly ito four groups of five flashlights each. The each group of flashlights uses a differet brad of battery. The lifetimes of the batteries, to the earest hour, are as follows. Brad A Brad B Brad C Brad D Perform a Tukey multiple cofidece iterval compariso usig a family cofidece level of 0.9. Let the subscripts 1,, 3, ad refer to Brad A, Brad B, Brad C, ad Brad D, respectively. Each sample size is, ad the total umber of pieces of data is 0. Step 1: Family Cofidece Level Family Cofidece Level is 0.9. Step : Critical Value = q( k =,df Error =16 =.0 Step 3: Obtai the Cofidece Iterval Edpoits I ANOVA Example #8, we calculated to be i 1 3 y i i k The umber of itervals will be! =! (! The edpoits for µ 1 µ y ± q( k,df Error ( ± ± from 7.37 to The edpoits for µ 1 ± ( ± ± from 6.37 to

2 The edpoits for µ 1 µ y ± q( k,df Error ( ± ± from.037 to The edpoits for µ ( y ± q( k,dferror ( ± ± from to The edpoits for µ µ ( y y ± ( ± ± from to 1.37 Step : Step : The edpoits for µ 3 µ ( y 3 y ± q( k,dferror ( ± ± from to Decide Which Populatio Meas are Differet Sice all of the itervals cotai zero, oe of them are cocluded to b e differet (i.e., we thik all of the meas are approximately the same. Costruct a Compariso Diagram Brad D Brad C Brad B Brad A ( (3 ( ( We ca be 9% cofidet that o two of the brad meas differ. These results are cosistet with our ANOVA coclusio that there was ot eough evidece (α = 0.0 to coclude that the mea lifetimes of the brads of batteries differ.

3 9. Maufacturers of golf balls always seem to be claimig that their ball goes the farthest. A writer for a sports magazie decided to coduct a impartial test. She radomly selected 0 golf professioals ad the radomly assiged four golfers to each of five brads. Each golfer drove the assiged brad of ball. The drivig distaces, i yards, are displayed i the followig table. Brad 1 Brad Brad 3 Brad Brad Coduct a Tukey multiple cofidece iterval compariso at the 9% family cofidece level. Let the subscripts 1,, 3,, ad refer to Brad 1, Brad, Brad 3, Brad, ad Brad respectively. Each sample size is, ad the total umber of data pieces is 0. Step 1: Family Cofidece Level Family Cofidece Level is 0.9. Step : Critical Value = q( k =,df Error =1 =.37 Step 3: Obtai the Cofidece Iterval Edpoits I ANOVA Example #9, we calculated to be 6.3. i 1 3 y i i k The umber of itervals will be! =! (! The edpoits for µ 1 µ y ± q( k,dferror ( ± ±1.877 from to 1.17 The edpoits for µ 1 ± q( k,dferror ( ± ±1.877 from to 1.877

4 The edpoits for µ 1 µ y ± ( ± ±1.877 from to The edpoits for µ 1 µ y ± q( k,dferror ( ± ±1.877 from 1.17 to 8.67 The edpoits for µ ( y ± q( k,df Error ( ± ± from.17 to.67 The edpoits for µ µ ( y y ± q( k,dferror ( ± ±1.877 from 7.67 to.17 The edpoits for µ µ ( y y ± q( k,df Error ( ± ±1.877 from to 11.37

5 The edpoits for µ 3 µ ( y 3 y ± q( k,dferror ( ± ±1.877 from to The edpoits for µ 3 µ ( y 3 y ± q( k,df Error ( ± ±1.877 from 8.17 to 1.67 The edpoits for µ µ ( y y ± q( k,dferror ( ± ±1.877 from.67 to.17 Step : Decide Which Populatio Meas are Differet Sice all of the itervals cotai zero, oe of them are cocluded to b e differet (i.e., we thik all of the meas are approximately the same. Step : Costruct a Compariso Diagram Brad 3 (3 Brad ( Brad 1 (1 Brad ( Brad ( We ca be 9% cofidet that o two of the brad meas differ. These results are cosistet with our ANOVA coclusio that there was ot eough evidece (α = 0.0 to coclude that the mea drivig distaces of the brads of golf balls differ.

6 10. The U.S. Bureau of Prisos publishes data i Statistical Report o the times served by prisoers released from federal istitutios for the first time. Idepedet radom samples of released prisoers for five differet offese categories yielded the followig iformatio o time served, i moths. Major i y i s i Couterfeitig Drug Laws Firearms Forger Fraud Usig a family cofidece level of 99%, perform a Tukey multiple cofidece iterval compariso. Let the subscripts 1,, 3,, ad refer to Couterfeitig, Drug Laws, Firearms, Forgery, ad Fraud, respectively. The total umber of pieces of data is 6. Step 1: Family Cofidece Level Family Cofidece Level is Step : Critical Value = q( k =,df Error = 60 =.9 Step 3: Obtai the Cofidece Iterval Edpoits I ANOVA Example #10, we calculated to be i 1 3 y i i k The umber of itervals will be! =! (! The edpoits for µ 1 µ y ± q( k,df Error 1 ( ± ±.8663 from to The edpoits for µ 1 ± 1 3 ( ± ±.303 from to 1.603

7 The edpoits for µ 1 µ y ± 1 ( ± ±.6081 from to.081 The edpoits for µ 1 µ y ± q( k,df Error 1 ( ± ±.30 from.30 to 8.30 The edpoits for µ ( y ± q( k,df Error 3 ( ± ±.179 from.979 to.379 The edpoits for µ µ ( y y ± q( k,df Error + 1 ( ± ±.76 from.676 to 8.76 The edpoits for µ µ ( y y ± ( ± ±.316 from 1.8 to 1.16

8 The edpoits for µ 3 µ ( y 3 y ± 3 ( ± ±.8818 from to The edpoits for µ 3 µ ( y 3 y ± q( k,df Error 3 ( ± ±.731 from to 1.31 The edpoits for µ µ ( y y ± q( k,df Error ( ± ± from to Step : Decide Which Populatio Meas are Differet Sice the itervals for µ µ ad µ 3 µ do ot cotai zero, we declare that µ ad µ are differet ad that µ 3 ad µ ; all other meas are ot differet. Step : Costruct a Compariso Diagram Fraud Couterfeitig Forgery Firearms Drug Laws ( (1 ( (3 ( We ca be 99% cofidet that the mea time served for firearms ad drug law offeses differs from that for fraud offeses; o other meas ca b e declared differet. These results are cosistet with our ANOVA coclusio that there exists eough evidece (α = 0.01 to coclude that the mea times served for the crime groups do differ. I fact, we ow kow which times served actually differ (with 99% cofidece.

9 1. Three sets of five mice were radomly selected to be placed i a stadard maze but with differet color doors. The respose is the time required to complete the maze as see below. Perform the appropriate aalysis to test if there is a effect due to door color. (Use α = 0.01 Color Time Red Gree Black Previous results yielded : Source df SS MS = SS/df F-statistic p-value Treatmets p-value < Error Total Sice it was determied that there is a effect due to door color o the mea time to complete the maze, we must perform multiple comparisos to fid the differeces. (Use α = 0.01 For this example, we ll use Tukey s W. TUKEY STUDENTIZED RANGE Step 1 : Step : Step 3 : Step : Step : Family Cofidece Level, 1 a For α = 0.0, use 9% cofidece. Critical Value = q( k = 3,df Error =1 =.0 Compute the Margi(s of Error W = = N k = = 8.90 Decide Which Populatio Meas are Differet = = 10.8 y = y 3 = = 8 y = = (differet = =.8 < 8.90 (ot differet y =. 8 = (differet Costruct a Compariso Diagram Black Red Gree (3 (1 ( =. We ca be 99% cofidet that mazes with gree doors seem to take loger tha the other mazes. These results are cosistet with our ANOVA coclusio that there exists eough evidece (α = 0.01 to coclude that the mea times to complete the mazes with differet door colors do differ. I fact, we ow kow which times actually differ (with 99% cofidece.

10 1. A study of firefighters i a large urba area cetered o the physical fitess of the egieers employed by the Fire Departmet. To measure the fitess, a physical therapist sampled five egieers each with, 10, 1, ad 0 years experiece with the departmet. She the recorded the umber of pushups that each perso could do i 60 secods. The results are listed below. Perform a aalysis of variace to determie if there are differeces i the physical fitess of egieers by time with departmet group. Use α = 0.0. Time with Departmet, years Previous results yielded : Source df S S MS = F-statistic p-value SS/df Treatmets p-value < Error Total Sice it was determied that there is a effect due to years of experiece o the physical fitess of the egieers, we must perform multiple comparisos to fid the differeces. (Use α = 0.0 For this example, we ll use Tukey s W. TUKEY STUDENTIZED RANGE Step 1 : Step : Step 3 : Step : Step : Family Cofidece Level, 1 a For α = 0.0, use 9% cofidece. Critical Value = q( k =,df Error =16 =.0 Compute the Margi(s of Error W = = N k = = 6.9 Decide Which Populatio Meas are Differet = = 8. y = = y 3 = = 3.6 y = = y = =1 < 6.9 (ot differet = = (differet y = 8. = (differet y = = (differet y y = 7. = (differet y 3 y = 3.6 =1.6 < 6.9 (ot differet Costruct a Compariso Diagram ( (3 ( ( We ca be 9% cofidet that a 0-year vetera has differet fitess from that of a vetera of or 10 years, ad a teure of 1-year vetera has differet fitess from that of a vetera of or 10 years. These results are cosistet with our ANOVA coclusio that there exists eough evidece (α = 0.0 to coclude that the mea physical fitess levels of the egieers do differ by time with the departmet group. I fact, we ow kow which times with the departmet group actually differ (with 9% cofidece.

11 1. A local bak has three brach offices. The bak has a liberal sick leave policy, ad a vice-presidet was cocered with employees takig advatage of this policy. She thought that the tedecy to take advatage depeded o the brach at which the employee worked. To see if there were differeces i the time employees took for sick leave, she asked each brach maager to sample employees radomly ad record the umber of days of sick leave take durig Te employees were chose, ad the data are listed below. Do this data idicate a differece i braches? Use a level of sigificace of 0.0. Previous results yielded : Brach Brach Brach Source df S S MS = SS/df F-statistic p-value Betwee < p-value < 0.0 Withi Total Sice it was determied that there is a differece i the braches regardig sick leave take durig 1990, we must perform multiple comparisos to fid the differeces. (Use α = 0.0 For this example, we ll use Tukey s W. TUKEY STUDENTIZED RANGE Step 1 : Step : Step 3 : Step : Step : Family Cofidece Level, 1 a For α = 0.0, use 9% cofidece. Critical Value = q( k = 3,df Error =7 =.16 Compute the Margi(s of Error W 1, = q( k,df Error = N k W 1,3 = q( k,df Error = N k W,3 = =N k Decide Which Populatio Meas are Differet 1 = = = = = = = = 17 y = = y 3 3 = y = =.6667 < 6.03 (ot differet = 17 0 = 3 < 6.03 (ot differet y = = (differet Costruct a Compariso Diagram Brach Brach 1 Brach 3 ( (1 ( = 0 We ca be 9% cofidet that Brach 3 has sigificatly more sick days of sick leave take i 1990 tha does Brach. These results are cosistet with our ANOVA coclusio that there exists eough evidece (α = 0.0 to coclude that the mea umber of days of sick leave take durig 1990 differs amog the three braches. I fact, we ow kow which braches actually differ (with 9% cofidece.