Tapajyoti Ghosh Problem 6.7

8/9/2019 Tapajyoti Ghosh Problem 6.7

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TAPAJYOTI GHOSH CHBE 5779 HOMEWORK #9

Problem 6.7:

a) Solution:

α = 0.05

First we test the model with all the possible effects.Here the factors have been taken as categorical variables

because we have non numeric variables. Every factor has

only two levels. Thus this is 2^k factorial design. It is

used for screening factors.From the whole model test, we find that atleast one of the

factors is significant. So we go to effects test. The effects

shows us which factors are significant( p < alpha) and

which are not significant. (p > alpha). Now we determinehow to get to am adequate model by removing the factors

which are not significant.. However, we should alwaysremember that we should remove one factor at a time.

And refit it everytime rather than removing all the not

significant effects at once.So we start by moving the higher order effects. First

removing the ternary effect of C*D*A. And redoing the

analysis.



From the whole model test, we find that atleast one of

the factors is significant. So we go to effects test. The

effects shows us which factors are significant( p <alpha) and which are not significant. (p > alpha). Now

we determine how to get to an adequate model by

removing the factors which are not significant.So for the next analysis we remove the effect of

B*C*D and redo the entire analysis.

From the whole model test, we find that atleast one of

the factors is significant. So we go to effects test. The

effects shows us which factors are significant( p <alpha) and which are not significant. (p > alpha). Now

we determine how to get to an adequate model by

removing the factors which are not significant.So for the next analysis we remove the effect of B*D

and redo the entire analysis. The other two ternary

effects are now significant so we keep them in our

model.



From the whole model test, we find that

atleast one of the factors is significant. So we go to

effects test. The effects shows us which factors are

significant( p < alpha) and which are notsignificant. (p > alpha). Now we determine how to

get to an adequate model by removing the factorswhich are not significant.

So for the next analysis we remove the effect of

B*C and redo the entire analysis

From the whole model test, we findthat atleast one of the factors is

significant. So we go to effects

test. The effects shows us whichfactors are significant( p < alpha)

and which are not significant. (p >

alpha). Now we determine how toget to an adequate model by

removing the factors which are notsignificant.

So for the next analysis we remove

the effect of B*C and redo theentire analysis




atleast one of the factors is significant. So we go to

effects test. The effects shows us which factors aresignificant( p < alpha) and which are not

significant. (p > alpha). Now we determine how to


So for the next analysis we remove the effect ofA*C and redo the entire analysis


atleast one of the factors is significant. So we go toeffects test. The effects shows us which factors are

significant( p < alpha) and which are not

significant. (p > alpha). Now we determine how toget to an adequate model by removing the factorswhich are not significant.

So for the next analysis we remove the effect of

C*D and redo the entire analysis.



From the whole model test, we find thatatleast one of the factors is significant. So we go to

effects test. The effects shows us which factors are

significant( p < alpha) and which are not

significant. (p > alpha). Now we determine how to


Thus finally we get the best model which is the

most adequate. I believe this model is correct and

adequate fro the given data. I decide to keep the

main effect of B even thought it is not significant

as it is a main effect and we should include it in the

model. We might have decided to remove it too.

Thus now as we have the model, we can proceed

for residual analysis.

Oh No! OUTLIER. So we had an outlier in the experiment. I might have to do my entire analysis over

again. But I decide not to. Why? Because the number of outliers are really less. Only one. And we have

32 runs. So probably, or model would be robust enough such that even if we remove this outlier from

our analysis, out model which we have already derived will still work fine. And also, and we have two

replicates for a data point. So it is difficult to say which one is the outlier. I think the number of runs is

not adequate. We should have more runs to decide whether it is bad data or bad model.



Yes! My model works without the outlier data point

too. So it is robust and I have probably done right.

Hence, even if I had included in the process of

developing the model, the model was quite robust and

the presence of just a single outlier ensured that I

didn’t stray towards a very wrong direction. Hence, if I

had redone the entire analysis, I would have probably

got the same model.

Ah! So Finally I reach residual analysis. Very nice. Normaly distributed data. Goodness of fit test is

showing that is actually normally distributed. No outliers. Graph all good. Yup.Good to Go.



Plot of residuals vs predicted values shows equal variance at all levels. Hence our assumption was

correct. Thus our model is adequate.

A quick power analysis of the main effect shows that the main effect of B would might have been

rendered significant if we had done nearly 284 runs. TOO many runs! No need. Forget B. Power is low

though. High chance of type II error..hmmmm.

Tapajyoti Ghosh Problem 6.7

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