1. A marketing expert for a publishing house wants to measure reader preference

Question

1. A marketing expert for a publishing house wants to measure reader preference for 5 different covers of the same paperback novel. Five newsstands are selected at random and the novel is displayed at each newsstand for 5 weeks, one for each cover. One week is sufficient to determine the sales potential because a new cover makes its impact immediately followed by a pattern of diminishing returns. The number of sales are listed below with the cover given in parentheses. The file name is neus.txt. Week Newstand (D) 200 (C) 290 (A) 280 (E) 230 (B) 265 T(c)260TB)-280 TE)-245 TA)-285TD)245 | (A) 250 | (D) 245 | (C) 280 | (B) 250 | (E) 180 (B) 260 (E) 190 (D) 230 (C) 205 (A) 200 (E) 340 A 335 (B) 265(D) 270 (C) 230 Analyze the data from this experiment (use -0.2) and draw conclusions. T 111 IV

Explanation / Answer

This is a Latin square design.Analysis of latin square design in R as follows:

Newsstand= c(rep("NS1",1),rep("NS2",1),rep("NS3",1),rep("NS4",1),rep("NS5",1))
> Newsstand
[1] "NS1" "NS2" "NS3" "NS4" "NS5"
> Novel=c("D","C","A","B","E","C","B","D","E","A","A","E","C","D","B","E","A","B","C","D","B","D","E","A","C")
> Novel
[1] "D" "C" "A" "B" "E" "C" "B" "D" "E" "A" "A" "E" "C" "D" "B" "E" "A" "B" "C"
[20] "D" "B" "D" "E" "A" "C"
> Week=c(rep("week1",5),rep("week2",5),rep("week3",5),rep("week4",5),rep("week5",5)
+

+ > Week=c(rep("week1",5),rep("week2",5),rep("week3",5),rep("week4",5),rep("week5",5))
> Week
[1] "week1" "week1" "week1" "week1" "week1" "week2" "week2" "week2" "week2"
[10] "week2" "week3" "week3" "week3" "week3" "week3" "week4" "week4" "week4"
[19] "week4" "week4" "week5" "week5" "week5" "week5" "week5"
> freq=c(200,260,250,260,340,290,280,245,190,335,280,245,280,230,265,230,285,250,205,270,265,245,180,200,230)
> mydata=data.frame(Week,Newsstand,Novel,freq)

matrix(mydata$Novel,5,5)
[,1] [,2] [,3] [,4] [,5]
[1,] "D" "C" "A" "E" "B"
[2,] "C" "B" "E" "A" "D"
[3,] "A" "D" "C" "B" "E"
[4,] "B" "E" "D" "C" "A"
[5,] "E" "A" "B" "D" "C"
> matrix(mydata$freq,5,5)
[,1] [,2] [,3] [,4] [,5]
[1,] 200 290 280 230 265
[2,] 260 280 245 285 245
[3,] 250 245 280 250 180
[4,] 260 190 230 205 200
[5,] 340 335 265 270 230
> myfit=lm(freq~Newsstand+Week+Novel,mydata)
> myfit

Call:
lm(formula = freq ~ Newsstand + Week + Novel, data = mydata)

Coefficients:
(Intercept) NewsstandNS2 NewsstandNS3 NewsstandNS4 NewsstandNS5  
280.2 10.0 -12.0 -36.0 35.0  
Weekweek2 Weekweek3 Weekweek4 Weekweek5 NovelB  
6.0 -2.0 -14.0 -38.0 -6.0  
NovelC NovelD NovelE  
-17.0 -32.0 -33.0  

> anova(myfit)
Analysis of Variance Table

Response: freq
Df Sum Sq Mean Sq F value Pr(>F)  
Newsstand 4 13816 3454.0 2.9193 0.0671 .
Week 4 6096 1524.0 1.2881 0.3286  
Novel 4 4446 1111.5 0.9394 0.4741  
Residuals 12 14198 1183.2
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>

Conclusion:

For given level of significance that is alpha=0.02 ,

The difference between considering the group the newsstand is not significant.because p-value >0.02.

The difference between Week is not significant here pvalue >0.02.

The difference between Novel is not significant because p-value>0.02

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