AP Statistics 2012-13Name: ______
M&M'S® Summer Assignment--Due the 1st Day of School
Can We Really Trust M&M'S®
A Chi-Square Goodness of Fit Investigation
According to the Mars Company:
“...the mix of colors of M&M'S® Minis®(teeny-tiny candies full of M&M'S® chocolate candy flavor)is 13% brown, 13% yellow, 12% red, 23% orange, 15% green, and 25% blue.”
However, the mix of colors has been known to change every fewyears and these proportions are based on the company's large scale production process. Your task over the summer is to determine how well your sample mix of colors matches the above proportions. We want to see if there is sufficient evidence to reject the company’s color claim. To dothis, we’ll be using the Chi-Square Goodness of Fit Test.
The Chi-Square Goodness of Fit Test is applied when you have one categorical variable from a single population. It is used to determine whether sample data are consistent with a hypothesized distribution. Our approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.
At a minimum, Steps #1-#4 must be completed. The other steps may require some research on your part and are optional at this time but, if attempted, they will give you a great insight into the use of statistics. This will not be graded so take a shot. ( is a great resource for this project.)
1.State the null and alternate hypotheses to test if the distribution of colors in your sample is the same as the company's claim.
Ho:
Ha:
2.Open your tube of Minis® and carefully count how many of each color are in your sample. Discard any broken or mutant candies. Record your data in the "Observed (O)" row in the table below. After you have counted each color, eat the candy! Color of your Minis®tube: ______
3.Using the statement from the Mars Company, calculate how many of each color you should be expected to see. You’ll have to figure this out using the total number of M&M'S® in your tube and the percentages listed above. Enter these counts in the “Expected (E)” row below.
Brown / Yellow / Red / Orange / Green / Blue / TotalObserved (O)
Expected (E)
(O - E)
(O- E)2
(O - E)2/E / X2=
4. For each color, find the difference between your count and the expected number (O - E), then square this value and divide the result (O - E)2by E. Enter each value in the correct rows of the table. Add up all of these “component” values to find the value of X2.
If your tube reflects the distribution advertised by the Mars Company, there should be very little difference betweenthe observed and expected counts. To quantify the difference, we’ll calculate a total which we’llcall “Chi-Square” orX2. If this total value is small, we have little evidence to suggest a difference in color distributions. However, the largerX2 gets, the more evidence we have to suggest the company’s claim may not be correct.
Are the entries in the last row all about the same or do any of the quantities stand out because they are "significantly" larger?
Did you get more of a particular color than you expected?
Did you get fewer of a particular color than you expected?
5.The chi-square goodness of fit test is appropriate when the following conditions are met:
The sampling method is simple random sampling.
The population is at least 10 times as large as the sample.
The variable under study is categorical.
The expected value of the number of sample observations in each level of the variable is at least 5.
Does it appear that the conditions have been satisfied for this significance test?
6.X2 has its own unique distribution. Some of its features include:
It is not symmetric--it is skewed to the right.
Its shape depends upon the degrees of freedom.
Its values are always non-negative.
Chi-Squaredistributions are skewed right and specified by degrees of freedom. In a Goodness ofFit test, the degrees of freedom equal one less than the number of categories.
The number of degrees of freedom is the number of values in the final calculation of a statistic that are free to vary.
In the graph above, you will see four chi-square distributions, one with df = 3, another with df = 5, a third with df = 10, and the fourth with df = 20. Notice that all four are skewed right but the center and spread of each are different.
7.To determine the likelihood of observing a difference between observed and expected asextreme as the one we observed, we must look up the p-value on a Chi-square table or you can use your TI-Nspire calculator and compare this to a specified significance level, α. We usually use α = 0.05. If a test of significance gives a p-value lower than the significance level α, the null hypothesis is rejected. Such results are informally referred to as 'statistically significant'.
For example, if someone argues that "there's only one chance in a thousand this could have happened by coincidence," a 0.001 level of statistical significance is being implied. The lower the significance level, the stronger the evidence required. Choosing level of significance is a somewhat arbitrary task, but for many applications, a level of 5% is chosen, for no better reason than that it is conventional.
A p-value is something you calculate when you want to evaluate two competing hypotheses. The p-value you get from your data will give you an idea of how plausible the hypotheses you are evaluating are.
You can find the p-value for our test by looking up X2 for 5 degrees of freedom from the tablebelow or by using your calculator.
What is the df for your M&M'S® sample? ______.
Use the table above to locate the two p-values your X2 value falls between at the stated df level. ______and ______
You can find the exact p-value for your X2 on your TI-Nspire by following the directions at (Page 135)
8. Based on your p-value and α = 0.05, what conclusions can you make about your sample of M&M'S® Minis®? Is there a statistically significant difference in the distributions?
9.Should you accept or reject the null hypothesis?
Make sure to record and save your data as we will be using it later in the year.