Assume that there is no meaningful Interaction (even it exists) in the Hypothesi
ID: 3261584 • Letter: A
Question
Assume that there is no meaningful Interaction (even it exists) in the Hypothesis Test in part ‘b’ in Qu.#4 above. (In other words, all following sub-questions can be done without worrying if such analysis should be done in the first place!)
a. Conduct the 4 step Hypothesis Test if ShelfLevel used in this experiment has any effect in terms of the sales of cereal boxes. Calculate all the statistics manually and use the critical value of the statistic to reach your conclusion. You should use the appropriate “SS” from MiniTab.
b. Calculate the p-Value for the test in part ‘a’. c. Which factor explains a larger portion of the sum of squares? Give numbers. Also calculate the value of R2 manually.
d. Calculate the Bonferroni Margin of Error, ME. ME can be treated as the ±margin of error for the Bonferroni confidence intervals for pairs of treatment means. Specify which treatment means are significantly superior at the .05 level. Specifically comment if the Difference in Population Treatment Means exists when using (Barand-1 and Shelf Level-3) and (Brand-3 and Shelf Level-3). What can you say about the Difference in Population Treatment Means when comparing (Barand-1 and Shelf Level-3) and (Brand-2 and Shelf Level-4).
e. Can you justify the parametric analysis you have used above. Give specific reason/s with 4 specific plots. (Hint: Refer to your Week 8/9 Class Notes.)
Shelf_Level Brand1 Brand2 Brand3 Brand4 C6-T Boxes Brand ShelfLevel 1 335 300 450 270 385 275 435 335 Suggestion#1: Stack data in C2-C5 in C7 290 240 490 280 Suggestion#2: "ID" the Brand# in C8 2 445 350 500 463 Suggestion#3: Code the "Shelf_Level" in C9 390 335 488 445 460 410 463 455 3 550 413 520 475 575 425 550 488 588 400 508 470 4 300 238 375 263 313 275 388 295 338 250 400 273Explanation / Answer
We will conduct an ANOVA with effect as Shelf and Brand. We will assume tht there is no interaction between Shelf and Brand. Our anlaysis is to comment whether there is effect of brand and shelf separately on sales of cerslas boxes or not.
As I don't have exposure in Minitab we will conduct everything on R.
Firstly, here is the code to import the data properly in R.
Code:
box<-c(335, 300, 450, 270,
385, 275, 435, 335,
290, 240, 490, 280,
445, 350, 500, 463,
390, 335, 488, 445,
460, 410, 463, 455,
550, 413, 520, 475,
575, 425, 550, 488,
588, 400, 508, 470,
300, 238, 375, 263,
313, 275, 388, 295,
338, 250, 400, 273
)
shelf<-rep(c("1","2","3","4"),each=12)
brand<-rep(c("1","2","3","4"),12)
data<-data.frame(shelf,brand,box,stringsAsFactors = TRUE)
data
Output:
> box<-c(335, 300, 450, 270,
+ 385, 275, 435, 335,
+ 290, 240, 490, 280,
+ 445, 350, 500, 463,
+ 390, 335, 488, 445,
+ 460, 410, 463, 455,
+ 550, 413, 520, 475,
+ 575, 425, 550, 488,
+ 588, 400, 508, 470,
+ 300, 238, 375, 263,
+ 313, 275, 388, 295,
+ 338, 250, 400, 273
+ )
> shelf<-rep(c("1","2","3","4"),each=12)
> brand<-rep(c("1","2","3","4"),12)
> data<-data.frame(shelf,brand,box,stringsAsFactors = TRUE)
> data
shelf brand box
1 1 1 335
2 1 2 300
3 1 3 450
4 1 4 270
5 1 1 385
6 1 2 275
7 1 3 435
8 1 4 335
9 1 1 290
10 1 2 240
11 1 3 490
12 1 4 280
13 2 1 445
14 2 2 350
15 2 3 500
16 2 4 463
17 2 1 390
18 2 2 335
19 2 3 488
20 2 4 445
21 2 1 460
22 2 2 410
23 2 3 463
24 2 4 455
25 3 1 550
26 3 2 413
27 3 3 520
28 3 4 475
29 3 1 575
30 3 2 425
31 3 3 550
32 3 4 488
33 3 1 588
34 3 2 400
35 3 3 508
36 3 4 470
37 4 1 300
38 4 2 238
39 4 3 375
40 4 4 263
41 4 1 313
42 4 2 275
43 4 3 388
44 4 4 295
45 4 1 338
46 4 2 250
47 4 3 400
48 4 4 273
>
Now, following is the code to conduct the anova.
Code:
anova<-aov(box~shelf+brand,data=data)
summary(anova)
Output:
> anova<-aov(box~shelf+brand,data=data)
> summary(anova)
Df Sum Sq Mean Sq F value Pr(>F)
shelf 3 266886 88962 65.23 1.17e-15 ***
brand 3 122966 40989 30.05 1.93e-10 ***
Residuals 41 55915 1364
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
a) H0: All the treatment means for the 4 levels of shelfs are equal. Vs H1: Atleast one treatment mean is different from the other.
Test -Statistic = F = (SS(shelf)/d.f) / (SS(error)/d.f) (refer to the the results above. D.f = Degrees of freedom)
= (266886/3) / (55915/41) = 65.23
degrees of freedom of test statistic = (3,41)
Critical value at 5% significance level = 95 th percentile of F distribution with (3,41) d.f = qf(0.95,3,41) (code in R) = 2.833
As the test-statistic = 65.23 > critical value = 2.833, so at 5% level, we reject H0.
Conclusion: We conclude that there is enough significant statistical evidence to claim that shelflevels has a signifcant effect on sales of cereals boxes.
b) p-value = Pr(F > test-statistic) = 1- pf(65.23,3,41) = 0.0000
c) SS(shelf)/SS(total) = 266886/(266886+122966+55915) = 59.87%
SS(Brand)/SS(total) = 122966/(266886+122966+55915) = 27.59%
We see that, almost 60% of the total variability is explained by Shelf, where as 28% is by Brand. So, it is the factor Shelf which explain lastest porportion of Sum of squares.
R2 = ( SS(Shelf) + SS(brand) )/ SS(Total) = (266886+122966)/(266886+122966+55915) = 87.46% (ans)
d) Please upload the class notes. Otherwise it is very difficult to know your teacher have taught you. After that, I will answer (d)