Chapin Manufacturing Company operates 24 hours a day, five days a week. The work

Question

Chapin Manufacturing Company operates 24 hours a day, five days a week. The workers rotate shifts each week. Management is interested in whether there is a difference in the number of units produced when the employees work on various shifts. A sample of five workers is selected and their output recorded on each shift. At the 0.01 significance level, can we conclude there is a difference in the mean production rate by shift or by employee?

difference in the mean production rate.

Chapin Manufacturing Company operates 24 hours a day, five days a week. The workers rotate shifts each week. Management is interested in whether there is a difference in the number of units produced when the employees work on various shifts. A sample of five workers is selected and their output recorded on each shift. At the 0.01 significance level, can we conclude there is a difference in the mean production rate by shift or by employee?

Explanation / Answer

this is two way anova table without any replication with treatment as employess and block as the shift.

We will do the analysis in R. This is code by which you can frame the dataset and run the two way anova.

Code:

emp<-c(rep(c("Skaff","Lum","Clark","Treece","morgan"),3))
emp
shift<-c(rep("Day",5),rep("Afternoon",5),rep("Night",5))
shift
unit<-c(35,36,24,31,24,25,24,26,21,25,32,26,27,25,22)

mdata<-data.frame(emp,shift,unit)
mdata

mdata$shift<-as.factor(mdata$shift)
mdata$emp<-as.factor(mdata$emp)

anova(lm(unit ~ emp + shift,mdata))

Output:

>
> emp<-c(rep(c("Skaff","Lum","Clark","Treece","morgan"),3))
> emp
[1] "Skaff" "Lum" "Clark" "Treece" "morgan" "Skaff" "Lum" "Clark"
[9] "Treece" "morgan" "Skaff" "Lum" "Clark" "Treece" "morgan"
> shift<-c(rep("Day",5),rep("Afternoon",5),rep("Night",5))
> shift
[1] "Day" "Day" "Day" "Day" "Day" "Afternoon"
[7] "Afternoon" "Afternoon" "Afternoon" "Afternoon" "Night" "Night"
[13] "Night" "Night" "Night"
> unit<-c(35,36,24,31,24,25,24,26,21,25,32,26,27,25,22)
>
> mdata<-data.frame(emp,shift,unit)
> mdata
emp shift unit
1 Skaff Day 35
2 Lum Day 36
3 Clark Day 24
4 Treece Day 31
5 morgan Day 24
6 Skaff Afternoon 25
7 Lum Afternoon 24
8 Clark Afternoon 26
9 Treece Afternoon 21
10 morgan Afternoon 25
11 Skaff Night 32
12 Lum Night 26
13 Clark Night 27
14 Treece Night 25
15 morgan Night 22
>
> mdata$shift<-as.factor(mdata$shift)
> mdata$emp<-as.factor(mdata$emp)
>
> anova(lm(unit ~ emp + shift,mdata))
Analysis of Variance Table

Response: unit
Df Sum Sq Mean Sq F value Pr(>F)
emp 4 92.400 23.100 1.6861 0.24523
shift 2 85.733 42.867 3.1290 0.09911 .
Residuals 8 109.600 13.700
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>

For us treatments are the employess, look the anova table employee row , d.f of F-statistic = (4,8)

So for treatment reject H0 if F> F.INV(1-0.01,4,8) = 7.007 (ans)

For us blocks are the shift, look the anova table shift employee row , d.f of F-statistic = (2,8)

So for blocks reject H0 if F> F.INV(1-0.01,2,8) = 8.649 (ans)

As we are testing at 0.01 significance level, so 0.01 has been used.

As F-value for shift = 3.129 < 8.649 so, we fail to reject H0 to conclude that there is no difference in mean production rate across shifts.

As F-value for employee= 1.686 < 7.007 so, we fail to reject H0 to conclude that there is no difference in mean production rate across employees..

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