In today's class, my goal is to show you how we analyze data,
from factorial experiments when three factors were used.
This example, is based on one from this textbook by Box, Hunter and Hunter,
called "Statistics for Experimenters".
The experiments described in that example, were run to find the combination of settings
that would reduce the amount of pollution discharged from the water treatment facility.
This is clearly a case where we would like to minimize the amount of pollutant.
So, minimizing our outcome variable would be the objective.
Three factors were considered.
The first one, factor C, was the chemical compound used.
Let's call that compound P, and compound Q. We don't really know their names.
Factor T, was the temperature of the treatment.
Whether we were treating the water at 72 or 100 Fahrenheit.
And factor S, was the stirring speed either a slow speed
of 200 revolutions per minute, or a high speed of 400.
Notice that every factor has two levels, and going back to that mathematical idea,
that two to the power "k", is the total number of experiments; "k" is equal to three
in this example, so we get a total of eight possible combinations.
Here's a short quiz to test that knowledge.
So let's take a look at the results.
We will always present our data, and analyze it using what we call "standard order".
Standard order, requires that we create a column for each of our factors.
So C, S, and T. Note that I could have used A, B, and C for the three factors,
but very often we'll switch to letters that actually match
of our factor names, but you don't have to.
So back to the standard order table.
And the rule is, we vary the first factor the fastest: - + - + - + - + The second factor,
temperature, is varied the next fastest, between its low and high levels.
So two minuses, two pluses, two minuses, two pluses.
And then the last factor S, is varied the slowest.
So four low levels (-) and four high levels (+).
Those make up our entire table.
Never run the experiments in the order of this table.
The order must be randomly selected.
So what we will do, is add a column to our table to keep track of the order
in which we actually ran the experiments.
Also add a column over here for the outcome variable.
In this case, the outcome was the pollutant amount, measured in pounds.
One thing that's so great about the standard order table, is that we can get a quick sense
of the factor's influence on the outcome variable.
Take a look, for example, at how the pollution amounts changes,
when we change the chemical compound, factor C. That factor goes low,
high, low, high, low, high, low, high.
We see that same pattern in the pollution amounts.
Take a look at the effect of factor S, the first four experiments, have a very high level
of pollution on average, while the last four experiments have a low level of pollution.
That also matches with factor S. We can already tell, just from this table, that factor C
and factor S are going to be really important to understanding the results.
Let's go back to our cube plot.
And this time, our cube plot is actually a cube.
We can draw it by showing the first factor along the horizontal axis.
The next factor on the vertical axis and the final factor, S, is shown in
and out of the page in this diagonal way.
Next, we transcribe the values onto this cube.
This is really easy when we follow the standard order sequence.
Take a look: 5, 30, 6, 33.
Then 4, 3, 5, and 4.
I love this visual representation of the experimental data.
It really helps us achieve our objective so quickly.
Take a few seconds and answer this question.
At what levels, should we set up three factors in order to achieve the lowest pollution amount?
That's right.
It's very clear we need to use chemical Q, operate at a low temperature,
and with high stirring speeds of 400 revolutions per minute.
Later on in the course, we're going
to start examining what happens when you move outside this cube.
And I want you to already start to think along those lines.
But let's come back to the data we have right here,
and analyze the main effects and the interactions.
Start with a first factor C. The choice
of either chemical P, or chemical Q at the high level.
If we look at the cube, we actually have four estimates of that main effect,
along each of the four horizontal edges.
At high temperature, and high stirring speed, in other words high T
and high S, that effect is equal to 4 - 5.
At high temperature and low speed, that's 33 - 6.
At T- and high speed, in other words S+, it is 3 - 4.
And finally, at low temperature and low speed, it's 30 - 5.
So four estimates of the effect of the chemical.
And the average of these four, is equal to 50 divided by 4, equal to 12.5.
Let's pause here.
I always tell my students, it's no good just calculating numbers.
What does this value of 12.5 really mean, in plain language?
How would you describe this value to your manager
who doesn't really understand any statistics?
What it says, is on average, we expect an increase in the pollution amount when we go
from using chemical P to using chemical Q. And remember,
by convention, we report half of that amount.
So report a value of 6.25.
One further thing to notice, is that the discrepancy of that chemical effect at high S,
and low S. Notice the very large difference there.
From the prior class number 2C, this should be alerting to you that there's an interaction
between factor C and factor S. But before we get to that, let's take a look at temperature.
When we examined the table earlier,
we didn't really notice anything special about temperature.
And we should be able to confirm that, numerically.
We've four estimates of the temperature effect along the vertical axis.
4 - 3 here; 33 - 30 up here; 5 - 4 up there; and 6 - 5 here up at the front.
So on average, we get a value of 1.5 as our difference.
Or if we report half of it, that's an effect of 0.75.
Lastly, let's take a look at the effect of stirring speed S. Along the four diagonal axis,
we have 4 - 33; up here, 3 - 30 down here; 5 - 6 here; and 4 - 5 over there.
The average of those differences, is -14.5.
And if we report half of it, that's -7.25.
The -14.5 tells us that we expect, on average, a reduction of 14.5 pounds of pollution,
when we go from a low stirring speed to a high stirring speed.
So clearly, it's in our favour to use high stirring speeds
in order to get that reduced pollution.
You should always step back at this point and make sure these results make sense.
Horizontally, we see going from chemical P to Q increases the pollution amounts.
That value of 6.25 looks about right.
A small value of 0.75 for temperature,
also looks right because it really has a very small effect.
And finally, increasing the stirring speed has the largest reduction
on pollution: a decrease of 7.25 units.
You noticed while I was reviewing these results, I started to build
up a numeric representation for you on the screen.
We did that in the class where we considered the ginger biscuits,
and I just followed the same idea here.
"y", represents the prediction of the pollution.
The "11.25" value here is the baseline.
It is the average of all eight of the outcome values,
(5 + 30 + 6 + 33 + 4 + 3 + 5 + 4) divided by eight.
The other three terms, are the separate effects of each factor.
Those are the main effects.
Let's see how we can use this model to make some predictions.
Consider the situation, where we were using chemical Q,
in other words x_C is coded as a value of +1.
Let's use low temperature.
So x_T, is coded as -1.
And also let's use low steering speed.
So x_S, is -1.
The predicted value, is 11.25 + 6.25 - 0.75 + 7.25.
That's a value of 24.
That's quite a bit different to the value of 30 pounds which was actually recorded.
There's something we haven't accounted for.
And that's the interaction between C and S. An interaction is
when you have one variable behaving very differently,
depending on the level of another variable.
We noticed earlier that the chemical effect, has a change from 30 to 5 and 33 to 6 over here
on the front face at low stirring speed.
Yet on the back face of the cube at high speeds, the effect is almost zero.
3 - 4; 4 - 5.
Very small amounts.
It's very clear, the stirring speed modifies the effect of the chemical.
There's an interaction between S and C. How do we quantify this?
Well, like we did in class 2C, we have to add a new term to our prediction model.
And that term, bCS, is multiplied by xC and xS.
But how do we go calculate the b_{CS} value?
Let's go follow the same ideas as did in class 2C.
We have two chances to calculate it.
One instance, at high temperature and one instance at low temperature.
We'll calculate both, and then average the answer.
And then, as we've always done, report half the value.
So at high temperature, the difference due to C, at high speed, is 4 - 5.
The difference due to C at low speed is much greater, 33 - 6.
As you remember, interactions are always reported
as half the difference, going from high to low.
In other words, that's -1 - 27, which equals -28, and half of that is -14.
Let's report that at the lower temperature.
The difference due to C at high speed is 3 - 4.
The difference due to C at low speed is much greater, 30 - 5.
Report half the value, from high to low, and that is -1 - 25 = -26.
Dividing that by two, gives -13.
So now we have two estimates of the interaction effect.
One estimate is -14, the other estimate is -13.
The average of those two numbers is -13.5.
And when we report it, let's put in, in our model a value of half that amount.
So in other words, -6.75.
That's the value for b_{CS} = -6.75.
Now, let me just pause here for a second,
and emphasize that this is all very tedious if you do it by hand.
And we're going to show some computerized ways to do this faster in the next few classes.
But, I always recommend, let's start with "by hand" and then see the advantage
of it later on, when we go to computers.
So let's take our predictions now again from the previous example and see if they are improved.
The predicted value earlier, was 11.25 + 6.25 - 0.75 + 7.25; but now with this interaction term,
we have an additional part "-6.75" times "1" times "-1".
What that means is we actually get an additional amount, of 6.75 due to the interaction.
Getting us a prediction of 30.75, much, much closer to the actual value.
Notice here that the interaction actually works against us.
That interaction has increased the amount of pollution.
We could also calculate CT interactions and TS interactions.
I've only shown you the CS interactions.
In fact, there's even a three factor interaction, a CTS interaction,
but all of this gets very tedious and error prone.
Coming up in the next module we can't wait to show you some computerized shortcuts
that will take care of all of this work for you.
Now, at the risk of this class going on a little bit too long, I want you to sit back,
and just think about that interaction for a second.
Don't just see it as a number, but let's try to interpret what's really going on over here.
Why does chemical Q appear to be less effective at lower speed,
but at high speed it works really well?
Maybe chemical Q, just takes a little bit longer to dissolve in water than chemical P does.
At low stirring speeds, chemical Q is not effective,
but at high speeds both chemicals are equally effective.
Now here's where experiments can be really powerful.
We saw that the lowest pollution was over here in this corner.
When we use chemical Q, with high speed, and low temperature.
But what if, the government's requirements was pollution had to be smaller than 10?
And imagine also, that chemical Q cost you double the amount
of chemical P. You can see where this is going.
We can see that any operating point on this plane, would be effective as long
as it's not the point with low speed and chemical Q. In fact,
it might be a whole lot more economically profitable, to operate at this point over here,
producing five pounds of pollution.
We still meet the requirements for safe operation,
because we're below the government level of 10 units, and we use less energy for stirring,
and a cheaper chemical, P. Actually what we've done here,
is considered an additional outcome in our mind: "profit".
Recognize that profits or costs often play a role in any system.
So you should always be aware of the economic impacts of every corner in your cube.
To end this class, and this example, I want you to consider this.
Does the fact that temperature having a small effect imply that temperature is meaningless?
The answer is no.
It is important to recognize that even effects
with small numbers have an important interpretation.
It means that over the range of temperatures selected, in this case between 72
and a 100 Fahrenheit, that temperature has a small
to negligible change on the pollution amount.
Now this is a key insight, because the engineer or operator can take this,
and select operating conditions which are the most economically advantageous.
Again this comes down to profit.
It is conceivable that when using lower temperature we will save energy.
And, because temperature has such a small effect on the system overall it means
that we will not significantly affect the pollution level
when we operate at low temperature.
That's a great result.
So I want to thank you for staying with me during these examples.
I know that they have been longer than normal, but I hope they have been insightful.
In the module coming up next, we're going to start looking
at how we can do fewer experiments, but still get a good amount
of information about our process.
Hope to see you over there.