## Friday, January 23, 2015

### M&M ANOVA

Khan Academy does a nice job of explaining ANOVA at this link.  This is in fact where I learned it.  Below I have a nice application of ANOVA using M&Ms that I would like to share.

There are numerous tables below which can be glided over without any loss of understanding.  Indeed, if you just read the prose in between the tables you will be far better off.

In the Fall of 2014, I assigned a series of activities to my Elementary Statistics involving M&Ms.  These activities begin here. There were six groups of students involved and each group took a sample of ten bags of M&Ms.  These samples are listed below.

 Group 1 Color/Bag 1 2 3 4 5 6 7 8 9 10 Red 2 5 2 4 3 7 3 1 7 5 Orange 7 3 6 5 6 7 5 7 6 9 Yellow 2 3 0 1 2 0 2 4 3 4 Green 3 0 6 6 3 2 1 3 4 2 Blue 3 5 4 2 4 7 4 5 4 5 Brown 1 1 1 0 1 3 1 1 1 3

 Group 2 Color/Bag 1 2 3 4 5 6 7 8 9 10 Red 2 2 3 2 2 3 1 6 3 3 Orange 3 4 0 4 4 4 3 2 2 3 Yellow 2 5 5 1 1 1 0 1 2 4 Green 3 1 5 4 4 3 4 2 4 2 Blue 3 2 2 4 3 3 6 1 4 4 Brown 3 2 1 1 2 2 2 4 1 1
 Group 3 Color/bag 1 2 3 4 5 6 7 8 9 10 Red 1 1 1 0 2 3 0 4 0 0 Orange 2 0 1 2 5 6 5 3 1 0 Yellow 1 1 2 0 3 5 0 7 1 3 Green 1 3 1 2 2 1 5 4 1 1 Blue 2 2 4 2 4 5 5 2 2 1 Brown 1 0 0 1 2 0 3 1 2 1
 Group 4 Color/Bag 1 2 3 4 5 6 7 8 9 10 Red 4 5 1 3 1 0 2 2 3 2 Orange 1 1 7 3 3 6 3 5 4 6 Yellow 1 4 6 0 3 2 3 0 3 4 Green 2 3 1 8 4 4 5 6 1 3 Blue 5 1 1 5 4 2 3 3 3 2 Brown 2 3 4 1 3 3 2 1 2 0
 Group 5 Color/Bag 1 2 3 4 5 6 7 8 9 10 Red 2 3 1 5 5 2 0 1 5 5 Orange 2 0 1 3 2 3 3 1 5 3 Yellow 6 2 5 2 3 4 7 7 1 5 Green 2 5 5 7 1 2 4 6 5 1 Blue 1 5 6 2 4 5 4 1 2 3 Brown 3 6 0 1 3 2 3 2 1 1
 Group 6 Color/Bag 1 2 3 4 5 6 7 8 9 10 Red 1 2 2 4 3 3 1 2 1 7 Orange 3 3 1 1 3 1 3 2 1 2 Yellow 3 5 3 3 2 2 3 6 4 4 Green 3 1 7 3 3 6 2 2 4 2 Blue 4 4 1 4 3 3 3 4 4 3 Brown 2 1 4 1 4 2 4 2 5 1
This is a nice collection of real data and my thought was to make the most of it.  As a sample size of ten is small, my thought was to pool the data, but before this can be legitimately done, it must be justified.  One might argue that since all of the samples were taken from M&M's that might be justification enough, but I had lingering doubts.  What if proportion of M&M color is not consistent from batch to batch? What if M&Ms are put out in a variety of Fun Sizes?  What if my students had just royally goofed?  In order to be careful, I decided that after having taught elementary statistics for twenty years it was time to learn ANOVA.

I first wanted to get an good confidence interval for the average number of M&Ms per bag. I calculated that using the data for each group, finding the bag by bag total.  I put that into the following table:

 Bag/Group 1 2 3 4 5 6 1 18 16 8 15 16 16 2 17 16 7 17 21 16 3 19 16 9 20 18 18 4 18 16 7 20 20 16 5 19 16 18 18 18 18 6 26 16 20 17 18 17 7 16 16 18 18 21 16 8 21 16 21 17 18 18 9 25 16 7 16 19 19 10 28 17 6 17 18 19
I then calculated the mean for each of the groups individually and the grand mean of the total pooled data.  Using this, I calculated the sum of the squares for differences within each of the groups and the sum of the squares for differences between the groups.  Those calculations are in the table below:

 DATA Bag/Group 1 2 3 4 5 6 1 18 16 8 15 16 16 2 17 16 7 17 21 16 3 19 16 9 20 18 18 4 18 16 7 20 20 16 5 19 16 18 18 18 18 6 26 16 20 17 18 17 7 16 16 18 18 21 16 8 21 16 21 17 18 18 9 25 16 7 16 19 19 10 28 17 6 17 18 19 GrandMean Means 17.07 20.7 16.1 12.1 17.5 18.7 17.3

 SSW 1 2 3 4 5 6 7.29 0.01 16.81 6.25 7.29 1.69 13.69 0.01 26.01 0.25 5.29 1.69 2.89 0.01 9.61 6.25 0.49 0.49 7.29 0.01 26.01 6.25 1.69 1.69 2.89 0.01 34.81 0.25 0.49 0.49 28.09 0.01 62.41 0.25 0.49 0.09 22.09 0.01 34.81 0.25 5.29 1.69 0.09 0.01 79.21 0.25 0.49 0.49 18.49 0.01 26.01 2.25 0.09 2.89 53.29 0.81 37.21 0.25 0.49 2.89 Sums 156.10 0.90 352.90 22.50 22.10 14.10

 SSB 1 2 3 4 5 6 13.20 0.93 24.67 0.19 2.67 0.05 13.20 0.93 24.67 0.19 2.67 0.05 13.20 0.93 24.67 0.19 2.67 0.05 13.20 0.93 24.67 0.19 2.67 0.05 13.20 0.93 24.67 0.19 2.67 0.05 13.20 0.93 24.67 0.19 2.67 0.05 13.20 0.93 24.67 0.19 2.67 0.05 13.20 0.93 24.67 0.19 2.67 0.05 13.20 0.93 24.67 0.19 2.67 0.05 13.20 0.93 24.67 0.19 2.67 0.05 Sums 132.01 9.34 246.68 1.88 26.68 0.54
The SSW sums to 568.6 and the SSB sums to 417.1.  The numerator has m-1=5 degrees of freedom as we are comparing m=6 groups. The denominator has m*(n-1)=6*(10-1)=54 degrees of freedom as each of those groups took a sample of size n=10.  This gives an F test-statistics of F=7.92.  The critical number for those degrees of freedom with a significance level of  alpha=0.10 is 1.957.  As 7.92 is greater than 1.957, we must conclude that these samples are not all drawn from the same population.

This came as something of surprise to me.  As an educator of over 30 years experience, I immediately suspected student error.  Looking at the SSW and SSB table above, I noted that the numbers from group 3 were considerably larger than the rest.  I was curious as whether and how they had erred.  Discerning this was easy because I had had the students document their process.  In looking at the documentation from group 3, I found the follow photograph:

The student had been told to use Fun Size M&Ms.  It was assumed that they would plain and that the bags would not be mixed.  We are well tutored in how one spells ass-u-me.

I would be remiss at this point, however, if I did not say that I had pushed this further. Elementating group 3 does not fix the problem.  The remaining groups are not sampling the same populations and an examination of the documentation of the other groups does not reveal a similar glaring error in methods.   Of all six groups, only 4 and 6 seem to be sampling the same population.

I will be having my class do a similar experiment this semester--with better instructions from the teacher--and after this I will conduct this study again.

## Wednesday, January 21, 2015

### M&M Binomial

This uses data gathered from the M&M Activity.

Let's begin this with a thought experiment.  Imagine you have the job of filling bags with M&Ms.  One might imagine that there is a huge bin that has been filled with M&Ms and that you are just parcelling them out into the bags.

There is a mountain of M&Ms and a certain proportion of these are red, orange, yellow, green, blue, and brown.  The number is so large that the act of choosing, say, a red M&M on one trial does not appreciably reduce the probability of getting one on the next trial.  All of this argues that, if one is interested in particular colors, each trial is a Bernoulli Trial with a fixed probability of success.

Each M&M is about the same weight and you are aiming to fill your bag to at least a certain weight but not more than one M&M more than that.  This results in the bags having a small variation in number of M&Ms per bag.  All of this means that we have a, more or less, fixed number of M&Ms per bag.

We combine these two observations, and what we have looks astonishingly like a Binomial Experiment.

Using a sample of 10 bags of Plain M&Ms, I investigated whether they did in fact follow the binomial distribution for red M&Ms.  This sample contained a total of 182 M&Ms of which 28 were red.  This yields a proportion p=0.154 of red M&Ms. The ten bags contained from 17 to 19 each. I therefore chose to set n=19.

Given those, I used my data to create a frequency table.  In the table below, the first column is the possible number of successes for a binomial experiment with 19 trials, that is to say the numbers from 0 to 19 inclusive; keeping with standard notation, that column is labeled x.  The second column is the number of bags that had that many red M&Ms.  As we are setting up do to a chi square test to see if the binomial model fits, we have labeled the frequency column with an O.

 x O 0 1 1 2 2 2 3 0 4 4 5 0 6 1 7 0 8 0 9 0 10 0 11 0 12 0 13 0 14 0 15 0 16 0 17 0 18 0 19 0

We ask the question of how well this fits with the expectations of the binomial distribution B(19,0.154).  I do the sample calculation for x=3 below:

Putting this calculation into a spread sheet, I obtain:

 n O E 0 1 0.42 1 2 1.45 2 2 2.36 3 0 2.44 4 4 1.77 5 0 0.97 6 1 0.41 7 0 0.14 8 0 0.04 9 0 0.01 10 0 0.00 11 0 0.00 12 0 0.00 13 0 0.00 14 0 0.00 15 0 0.00 16 0 0.00 17 0 0.00 18 0 0.00 19 0 0.00

We can then compare the values of the O and the E columns by using the (O-E)^2/E measure.  The results are in the table below:

 n O E (O-E)^3/E 0 1 0.42 0.81 1 2 1.45 0.21 2 2 2.36 0.06 3 0 2.44 2.44 4 4 1.77 2.80 5 0 0.97 0.97 6 1 0.41 0.85 7 0 0.14 0.14 8 0 0.04 0.04 9 0 0.01 0.01 10 0 0.00 0.00 11 0 0.00 0.00 12 0 0.00 0.00 13 0 0.00 0.00 14 0 0.00 0.00 15 0 0.00 0.00 16 0 0.00 0.00 17 0 0.00 0.00 18 0 0.00 0.00 19 0 0.00 0.00
Note that the sum of the (O-E)^2/E column is 8.32.  The critical number for the chi square test for this is 30.14.  This is the table look up with 19 degrees of freedom and a significance level of 0.05.

As 8.32 is not larger than 30.14, we cannot reject the null hypothesis of the chi square test.  The null hypothesis is that the model fits.  We've no proven that it is binomial, but we can say that our data is not inconsistent with the binomial distribution B(19, 0.154).

Do this test with the data you collected from the M&M Activity.

### M&M Hypothesis Test

This activity will make use of the data you gathered in M&M Activity.

As has been observed, there are six colors typically found in each bag of M&Ms. In the previous blog post, we tested whether the proportions of color within each bag remained consistent with differences due to no more than random variation.

Using data I had gather, I hypothesize that each bag of M&Ms contains 18 total and that each of the colors has the following average:
Red 3
Orange 3
Yellow 2.5
Green 4
Blue 4
Brown 1.5

In order to perform this test, we use a table of the following type.

There are blanks to be filled in depending the particular putative mean and the set level of significance.  I chose to test the green at a 0.05 level of significance.(The level of significance is chosen ahead of time so as not to effect the decision.)  This gave me the following table:

This is a small sample of size n=10 with 9 degrees of freedom, so I used the T-table* to give me a critical number of 2.262.  The number of greens from my sample was: 5,5,4, 3,3,3,3,4,3,1.  This has an average of 3.4 and a standard deviation of 1.2,  The calculation and test were completed as follows:
In this case, I could not prove that the average is not 4.

Follow these procedures yourself and test at a 0.10 level of significance. You will have to look up a new T-value, but as you sample size is n=10, it will still have 9 degrees of freedom. Also test whether the number of M&Ms per bag is 18.

*(In order to use the T-statistic, we have to assume that the data is normally distributed. This can be justified, but it is beyond the scope of the current activity.)

## Friday, January 16, 2015

### M&M Chi-square

The following activity will make use of data gathered in the M&M Activity.

Every regular bag of plain M&Ms contains six colors: red, orange, yellow, green, blue, and brown.  In handling numerous samples of M&Ms, the question arises as to whether there is always about the same proportion of any given color, i.e, is there always about the same proportion of green?

This activity will test that using the chi-square.  We will not be able to prove that the proportion is the same; instead we will able to determine one of two things:

1. The proportion is NOT the same.
2. The amount per bag is consistent with being the same proportion.
I will conduct my test using green M&Ms as they are reputed to have mystical properties the description of which are beyond the scope of our investigation.

I drew a new sample of ten bags of M&Ms which were not connected to the previous activity. The data was as below:

The first column is simple the bag number and does not figure into calculations.  The second column is the total number of M&Ms per bag. The third column is the number of green M&Ms in a given bag.

We will be conducting a chi-square test.  In the chi-square, the null hypothesis is that all of the proportions are the same.  So it cannot be proven that the proportions are the same by the use of this test.  However, it can be proven that they are not the same.  We will compute a chi-square test statistic in the course of this, and if that is too large we will be forced to decide the proportions are not the same.  Otherwise, we will say the data is consistent with having the same proportion per bag.

To do the calculations, I put my data into an Excel spreadsheet will columns as labeled below:
 Bag Number N O 1 17 3 2 17 2 3 19 5 4 18 6 5 19 6 6 18 1 7 18 2 8 18 6 9 19 2 10 19 6 182 39
The first column contains the bag number, the second column, headed by N, contains the total number of M&Ms per bag.  The third column, headed by O, contains the number of green M&Ms per bag. The last entry in the N and the O columns is the total of that column.

The null hypothesis of the chi-square test allows us to assume all the proportions are the same in each sample and, therefore, to pool data.  So pHat, the proportion of M&Ms that are green is 39/182, i.e. pHat is approximately 0.21.  Since the first bag contains 17 M&Ms we would expect it to contain 17*pHat=17*0.21=3.64 M7Ms.  We call this the expected number and denote it by E.  We do this calculation for every bag and obtain the following table:

 Bag Number N O E 1 17 3 3.64 2 17 2 3.64 3 19 5 4.07 4 18 6 3.86 5 19 6 4.07 6 18 1 3.86 7 18 2 3.86 8 18 6 3.86 9 19 2 4.07 10 19 6 4.07 182 39

Note that the expectations in the E column differ from the observed reality in the O column.  The question is whether this is too much to be accounted for by mere random variation.  This requires a measure.  The one we use is (O-E)^2/E.  We take the difference of O and E, square it, and then divide that by the expectation.  The division by E allows to deal with variation that, while perhaps large in absolute terms, is small relative to the expected value.  The completed table appears as below:
 Bag Number N O E (O-E)^2/E 1 17 3 3.64 0.11 2 17 2 3.64 0.74 3 19 5 4.07 0.21 4 18 6 3.86 1.19 5 19 6 4.07 0.91 6 18 1 3.86 2.12 7 18 2 3.86 0.89 8 18 6 3.86 1.19 9 19 2 4.07 1.05 10 19 6 4.07 0.91 182 39 9.34

The final entry in the (O-E)^2/E column is the total amount of error for observations verses expectations.  In this case that amount is 9.34.  The question is whether this is too much to be plausible.  To decide this, we need to look up the critical number on the chi-square table.  As there were 10 samples, this will require 9 degrees of freedom. For DF=9, this number is 16.92.  As 9.34, is not larger than 16.92, we cannot reject the null hypothesis.  We must therefore conclude that the data is consistent with there being approximately the same proportion of green M&Ms in each bag.