Question

A test is made of five different incentive-pay schemes for piece workers. Eight workers are assigned randomly to each plan. The total number of items produced by each worker over a 20-day period is recorded:

A	B	C	D	E
1106	1214	1010	1054	1210
1203	1186	1069	1101	1193
1064	1165	1047	1029	1169
1119	1177	1120	1066	1223
1087	1146	1084	1082	1161
1106	1099	1062	1067	1200
1101	1161	1051	1109	1189
1049	1153	1029	1083	1197

•Perform ANOVA to see if there is a difference among the five plans.

•Perform a comparison of means using Tukey’s method.

•Which plan(s) would you recommend to increase productivity?

•State the ANOVA assumptions and comment on whether you think they are satisfied.

•Perform a non-parametric test (Kruskal-Wallis) to see if there is a difference among the groups.

Answer 1

Here,

H₀: there is a difference among the five plans.

The R- code for ANOVA is as;

y=c(1106,1203,1064,1119,1087,1106,1101,1049,1214,1186,1165,1177,1146,1099,1161,1153,1010,1069,1047,1120,1084,1062,1051,1029,1054,1101,1029,1066,1082,1067,1109,1083,1210,1193,1169,1223,1161,1200,1189,1197)
x=rep(c(1,2,3,4,5),each=8)
d=data.frame(y,x)
a=aov(y~x,d);a
summary(aov(y~x,d))

And the output is ;

> a=aov(y~x,d);a
Call:
aov(formula = y ~ x, data = d)

Terms:
x Residuals
Sum of Squares 6195.2 137578.8
Deg. of Freedom 1 38

Residual standard error: 60.17054
Estimated effects may be unbalanced
> summary(aov(y~x,d))
Df Sum Sq Mean Sq F value Pr(>F)
x 1 6195 6195 1.711 0.199
Residuals 38 137579 3620

Here p-value is 0.199 which is greater than 0.05 thus we accept null hypothesis and conclude that

There is a no difference among the five plans.

The R-code for Tykey method is as;

y=c(1106,1203,1064,1119,1087,1106,1101,1049,1214,1186,1165,1177,1146,1099,1161,1153,1010,1069,1047,1120,1084,1062,1051,1029,1054,1101,1029,1066,1082,1067,1109,1083,1210,1193,1169,1223,1161,1200,1189,1197)
x=rep(c("l","m","n","o","p"),each=8)
d=data.frame(y,x)
a=aov(y~x)
TukeyHSD(a)

And the output is ;

> TukeyHSD(a)
Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = y ~ x)

$x diff lwr upr p adj

m-l 58.250 10.73122 105.768784 0.0099294

n-l -45.375 -92.89378 2.143784 0.0671590
o-l -30.500 -78.01878 17.018784 0.3649223
p-l 88.375 40.85622 135.893784 0.0000527
n-m -103.625 -151.14378 -56.106216 0.0000033
o-m -88.750 -136.26878 -41.231216 0.0000492
p-m 30.125 -17.39378 77.643784 0.3772209
o-n 14.875 -32.64378 62.393784 0.8949479
p-n 133.750 86.23122 181.268784 0.0000000
p-o 118.875 71.35622 166.393784 0.0000002

Here differences "m-l","p-l","n-m","o-m","p-n","p-o"

are significant while other differences are not significant.

The R-code for Kruskal -Wallis test is as;

y=c(1106,1203,1064,1119,1087,1106,1101,1049,1214,1186,1165,1177,1146,1099,1161,1153,1010,1069,1047,1120,1084,1062,1051,1029,1054,1101,1029,1066,1082,1067,1109,1083,1210,1193,1169,1223,1161,1200,1189,1197)
x=rep(c("l","m","n","o","p"),each=8)
d=data.frame(y,x)
kruskal.test(y~x,d)

And the output is :

> kruskal.test(y~x,d)

Kruskal-Wallis rank sum test

data: y by x
Kruskal-Wallis chi-squared = 26.383, df = 4,
p-value = 2.648e-05

Here p-value is less than 0.05 thus we reject null hypothesis as There is a no difference among the groups.