Question

16.11 The investigator mentioned in Review Question 14.14 wishes to conduct a more exten- sive test of the effect of alcohol consumption on the performance of automobile drivers, possibly to gain more information about the legal maximum for DUI arrests. Before the driving test, subjects drink a glass of orange juice laced with controlled amounts of vodka. Their performance is measured by the number of errors on a driving simulator. Five subjects are randomly assigned to each of five groups receiving different amounts of vodka (either 0, 1, 2, 4, or 6 ounces), and the following results were obtained: 1 DRIVING ERRORS AS A FUNCTION OF ALCOHOL CONSUMPTION (OUNCES) ZERO ONE TWO FOUR SIX 1 4 6 15 20 3 1 6 25 3 1 2 9 10 6 7 10 17 10 4 5 7 9 9 T 15 20 26 56 74 EX = G = 191 EX2 = 2371 (a) Summarize the results with an ANOVA table. (Note: Save these results for use with Review Question 17.7.) (b) If appropriate, estimate the effect size with ? (C) If appropriate, use Tukey's HSD test to pinpoint pairs of means that contribute to the significant F, given that Xo = 3, X, = 4, X, = 5.2, X4 = 11.2, and X6 = 14.8. Furthermore, if appropriate, estimate effect sizes with Cohen's d.

Answer 1

A)

treatment	G1	G2	G3	G4	G5
count, ni =	5	5	5	5	5
mean , x̅ i =	3.000	4.00	5.20	11.20	14.800
std. dev., si =	2.1	2.2	3.7	4.6	7.3
sample variances, si^2 =	4.500	5.000	13.700	21.200	52.700
total sum	15	20	26	56	74		191	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	7.64
( x̅ - x̅̅ )²	21.530	13.250	5.954	12.674	51.266
							TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	107.648	66.248	29.768	63.368	256.328		523.36
SS(within ) = SSW = Σ(n-1)s² =	18.000	20.000	54.800	84.800	210.800		388.4000

no. of treatment , k =   5  
df between = k-1 =    4  
N = Σn =   25  
df within = N-k =   20  
      
mean square between groups , MSB = SSB/k-1 =    523.36/4=   130.8400
mean square within groups , MSW = SSW/N-k =    388.4/20=   19.4200
      
F-stat = MSB/MSW =    130.84/19.42=   6.74

anova table
	SS	df	MS	F	p-value
Between:	523.4	4	130.8	6.74	0.001
Within:	388.4	20	19.4
Total:	911.8	24

Decision:   p-value<α , reject null hypothesis    

................

B))

eta square ,effect size = SSbet/SST=   0.5740

...............

C)

TUKEY HSD/KRAMER
Level of significance	0.05
no. of treatments,k	5
DF error =N-k=	20
MSE =	19.4200
q-critical value(0.05,5,20)=	4.2319

critical value = q*√(MSE/2*(1/ni+1/nj))
=8.34

................

					confidence interval
	mean difference	absolute mean difference		critical value	lower limit	upper limit			result
µ1-µ2	-1.00	1.00		8.34	-9.34	7.34			means are not different
µ1-µ3	-2.20	2.20		8.34	-10.54	6.14			means are not different
µ1-µ4	-8.20	8.20		8.34	-16.54	0.14			means are not different
µ1-µ5	-11.80	11.80		8.34	-20.14	-3.46			means are different
µ2-µ3	-1.20	1.20		8.34	-9.54	7.14			means are not different
µ2-µ4	-7.20	7.20		8.34	-15.54	1.14			means are not different
µ2-µ5	-10.8	10.80		8.34	-19.14	-2.46			means are different
µ3-µ4	-6.00	6.00		8.34	-14.34	2.34			means are not different
µ3-µ5	-9.6	9.60		8.34	-17.94	-1.26			means are different
µ4-µ5	-3.6	3.60		8.34	-11.94	4.74			means are not different

.................

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