Question

Problem 2 R Officials of a large-food chain are experimenting with various promotional gimmicks. Forty franchises in the chain are randomly selected and split into subgroups. These treatments are then randomly assigned to subgroups no gimmick specials on selected items mes with food prizes games with money prizes IV. These data are obtained on sales during the 1-month experimental period (sales are in $100,000 units) 9 1.0 2 1.3 5 .7 .4 1.2 1.0 4 1.6 1.1 5 1.31.3 III 1.8 .3 2.0 2.8 5 1.3 2.91.7 IV 2.4 2.6 2.8 2.9 1.7 2.6 2.7 2.1 2.1 2.3 (a) Write a model in terms of treatment means Hi, and state the underlying assumptions (b) State the hypotheses to test whether the treatments differ with respect to mean sales (c) Construct an ANOVA table for these data. Perform hypothesis testing and write your (d) Use Tukey's procedure to compare the treatment means. Summarize your conclusion. (e) Select a contrast that can be used to determine the effect of games on sales. State the of the food chain. conclusion. Is there a best treatment? hypotheses, and test the contrast using Scheffe's test at the α-0.05 level. What practical conclusion can you draw based on the results of this test. Hint: Consider the contrast 0.5μι + 0.5μ-0.5μ3-0.5μ4

Really looking for confirmation on (d) and (e). Others I am confident in my answers. Thank you.

Answer 1

Answer:

Note: package DescTools loaded for ScheffeTest.

R code:

# ANOVA

y <- c(0.8,0.2,1.4,0.9,0.9,1,1.1,1.4,1.2,1,1.6,1.1,0.2,1.3,1.5,0.7,0.5,1.3,

1.3,0.4,1.8,0.3,1.1,2.0,2.8,1.5,1.3,2.9,1.7,0.8,2.4,2.6,2.8,2.9,1.7,2.6,2.7,

2.1,2.1,2.3)

group <- factor(c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,

       3,3,3,3,3,4,4,4,4,4,4,4,4,4,4))

r.aov <- aov(y ~ group)

summary(r.aov)

post1<-TukeyHSD(r.aov);post1

post2<-ScheffeTest(r.aov,contrasts=matrix( c(0.5,0.5,-0.5,-0.5)) )

post2

R output:

# ANOVA

> y <- c(0.8,0.2,1.4,0.9,0.9,1,1.1,1.4,1.2,1,1.6,1.1,0.2,1.3,1.5,0.7,0.5,1.3,

+ 1.3,0.4,1.8,0.3,1.1,2.0,2.8,1.5,1.3,2.9,1.7,0.8,2.4,2.6,2.8,2.9,1.7,2.6,2.7,

+ 2.1,2.1,2.3)

> group <- factor(c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,

+        3,3,3,3,3,4,4,4,4,4,4,4,4,4,4))

>

> r.aov <- aov(y ~ group)

> summary(r.aov)

            Df Sum Sq Mean Sq F value Pr(>F)   

group        3 13.81   4.603   15.68 1.1e-06 ***

Residuals   36 10.57   0.294                   

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

>

> post1<-TukeyHSD(r.aov);post1

Tukey multiple comparisons of means

    95% family-wise confidence level

Fit: aov(formula = y ~ group)

$`group`

             diff        lwr         upr       p adj

2-1 -1.665335e-15 -0.65264081 0.6526408 1.0000000

3-1 6.300000e-01 -0.02264081 1.2826408 0.0617704

4-1 1.430000e+00 0.77735919 2.0826408 0.0000055

3-2 6.300000e-01 -0.02264081 1.2826408 0.0617704

4-2 1.430000e+00 0.77735919 2.0826408 0.0000055

4-3 8.000000e-01 0.14735919 1.4526408 0.0111902

>

> post2<-ScheffeTest(r.aov,contrasts=matrix( c(0.5,0.5,-0.5,-0.5)) )

>

> post2

Posthoc multiple comparisons of means : Scheffe Test

    95% family-wise confidence level

$`group`

               diff lwr.ci       upr.ci            pval   

1,2-3,4 -1.03 -1.532464 -0.5275358 1.3e-05 ***

---

Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

d).

From Tukey test, group 4 is significantly different from groups 1,2,3. Other pairs of groups are not significant.

e).

Scheffe Test shows that the contrast (0.5,0.5,-0.5,-0.5) is significant. Average of groups 1 and 2 is significant from average of groups 3 and 4.