Look along each row labeled by color. Notice that there is variation. In the Red row, for example, the values vary from 1 on the low end up to 4 on the high end. I've take this separated this data in the table below.
While there is variation, there is not all that much. We calculated in the last activity that the sample standard deviation is only 0.9 which is relatively small. The range of the data (high minus low) is 3. While there are rigorous techniques to justify this, it is not unreasonable to assume by just looking at it that each bag contains approximately the same proportion of red as any other.
We will now examine this assumption in more depth.
Notice that at the bottom of the Reds table, I've calculated a quantity that I've called Expected Proportion and have labeled it with the variable p-hat. Note that p-hat is 17/177 which rounds off to 0.096. This means that 9.6 percent of the M&Ms we counted are red. The 17 is obtained as the total of the red row and the 177 is the total number of M&Ms.
The question we ask now, are 9.6 percent of the M&Ms in each bag red? How do we figure that because, as you may have noticed from our first table, the number of M&Ms in each bag varies from 16 to 19? Well, if a bag contains 16 M&Ms and 9.6 percent of them are red, then 16 times 0.096 of them (1.7) are red. I've done this calculation in the table below:
I've carried out this same activity for all of the colors. The results are below. In carrying out the calculations for Red to three decimals, I noticed they weren't all that much different than using two decimals, so in the subsequent examples, I've only gone to two decimal places.
- Make a table of your original data, adding a total column and a total row as I have.
- You will need a separate table for each color of M&M.
- You will need to calculate a value of p-hat corresponding to each color. It is suggested that at least two people, working independently of each other, perform the calculations involved.
- You will need to submit your data in electronic form.