A test is made of five different incentive-pay schemes for piece workers. Eight

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:

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.

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

Explanation / Answer

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.

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