The following table is copied from \"Assn3_QU#4&5_ADM2304X_SS17 \" data file The

Question

The following table is copied from "Assn3_QU#4&5_ADM2304X_SS17 " data file

The Management Team from the Head Office of a national chain of grocery-stores wants to investigate their Shelf-Level displays of cereal brands and the Number of the various Brands of cereal boxes they sell in a week. The data was gathered based on the weekly
sales of boxes in one of their large outlets. The data is given in the file: "Assn3_QU#4&5_ADM2304X_SS17".
N.B.: Remember, You MUST first input this raw Data properly. Use the Column-wise approach.

a. What is the name of the analysis you will have to do? In this experimental design, what is the number of replicates?
b. Conduct the 4 step Hypothesis Test and find out if there is an interaction between the Brand and the ShelfLevel used in this experiment. Calculate all the test statistics manually and use the critical value of the statistic to reach your conclusion. You may use the appropriate values of “SS” from the MiniTab output.
c. Plot the Interaction Diagram and indicate if it is consistent with the conclusion you have reached in part ‘c’. Explain what this diagram and the conclusion reached in part ‘b’ tell us in terms real life (not just statistical jargon!).

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 4 300 238 375 263 313 275 388 295 338 250 400 273

Explanation / Answer

We define the structural model as follows.

A factor is an independent variable. A k factor ANOVA addresses k factors.

In general, suppose we have two factors A and B. Factor A has r levels and factor B has c levels. We organize the levels for factor A as rows and the levels for factor B as columns. We use the index i for the rows (i.e. factor A) and the index j for the columns (i.e. factor B). Thus we use an r × c table where the entries in the table are

image1297

We use terms such as xi (or xi.) as an abbreviation for the mean of {xij: 1 j c}. Similarly, we use terms such as xj (or x.j) as an abbreviation for the mean of {xij: 1 i r}.

We estimate the level means from the total mean for factor A by i = + i where i denotes the effect of the ith level for factor A (i.e. the departure of the ith level mean i for factor A from the total mean ). We have a similar estimate for the sample of xi = x + ai.

Note that

Eiai=0

Similarly we estimate the level means from the total mean for factor B by j = + j where j denotes the effect of the jth level for factor B (i.e. the departure of the jth level mean j for factor B from the total mean ). We have a similar estimate for the sample of xj = x + bj.

As for factor A,

The two-way ANOVA will either test for the main effects of factor A or factor B, namely

H0: 1. = 2. == r. (Factor A)

or

H0: .1 = .2 == .c (Factor B)

If testing for factor A, the null hypothesis is equivalent to

H0: i = 0 for all i

If testing for factor B, the null hypothesis is equivalent to

H0: j = 0 for all j

Finally, we can represent each element in the sample as xij = + i + j + ij where ij denotes the error (or unexplained amount). As before we have the sample version xij = x + ai + bj + eij where eij is the counterpart to ij in the sample.

Observation: Since

xij=u+ai=bj+eij=u+(ui-u)+(uj-u)+ eij
eij= xij- ui- uj+u
Eieij=Ejeij=0

Part II

The first step is to specify the null hypothesis. For a two-tailed test, the null hypothesis is typically that a parameter equals zero although there are exceptions. A typical null hypothesis is 1 - 2 = 0 which is equivalent to 1 = 2. For a one-tailed test, the null hypothesis is either that a parameter is greater than or equal to zero or that a parameter is less than or equal to zero. If the prediction is that 1 is larger than 2, then the null hypothesis (the reverse of the prediction) is 2 - 1 0. This is equivalent to 1 2.
The second step is to specify the level which is also known as the significance level. Typical values are 0.05 and 0.01.
The third step is to compute the probability value (also known as the p value). This is the probability of obtaining a sample statistic as different or more different from the parameter specified in the null hypothesis given that the null hypothesis is true.
Finally, compare the probability value with the level. If the probability value is lower then you reject the null hypothesis. Keep in mind that rejecting the null hypothesis is not an all-or-none decision. The lower the probability value, the more confidence you can have that the null hypothesis is false. However, if your probability value is higher than the conventional level of 0.05, most scientists will consider your findings inconclusive. Failure to reject the null hypothesis does not constitute support for the null hypothesis. It just means you do not have sufficiently strong data to reject it.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.