1. Chi-Square X2

2. What is CHI-Square?
Chi-square is a statistical test commonly used to compare observed data with data we would expect to obtain according to a specific hypothesis
This test determines if deviations from expected results (differences between observed and expected) were the result of chance, or were they due to other factors

3. Chi Square
Below is the formula for the Chi-squared test

4. Defining the variables
x2: this is the chi-squared value
o: this is the value that was observed in the experiment
e: this is the expected value
Σ: this means that you take the sum of all the values

5. Null Hypothesis
The chi-square test is always testing what scientists call the null hypothesis, which states that there is no significant difference between the expected and observed result.
On the next page are the expected value for M&Ms by color.
Example of null hypothesis: The M&Ms will not vary from the values given by the company
NOTE: In the null hypothesis, it is a negative statement.

6. An Example
According to the Mars Company website, the colors of M&Ms are produced in certain percentages:
20% brown
20% yellow
20% red
20% blue
10% green
10 % orange

7. What we found… Color Observed Brown 76 Yellow 65 Red 70 Blue 85 Green28
Orange 26
total
350

8. What we expected… Color Expected Brown 20% of 350 = 70 Yellow Red Blue
Green
10% of 350 = 35
Orange
total
350

9. Brown
Yellow
Red
Blue
Green
Orange
Observed (o) 76 65 70 85 28 26
Expected (e) 35
o-e 6 -5 15 -7 -9
(o-e)2 36 25
225 49 81
(o-e)2/e
0.514
0.357
3.214
1.400
2.314 X2
7.799

10. What we know… X2 = 7.799 What we examine on the chart:
For the AP Exam, the significance that is examined to accept of invalidate the null hypothesis is: p < (FIND ON CHART)
Degrees of freedom: The degrees of freedom is the number of samples minus one
6 colors  (6-1) = 5 degrees of freedom

11. accept null hypothesis
reject null hypothesis

13. For Reals…

14. Sooo… Reject or fail to reject null hypothesis???
According to the Degrees of Freedom on the Chi-Squared Table, the value can be up to 11.07
Our value for x2 = 7.799
Since < 11.07, we will accept the null hypothesis:
The M&Ms do not vary significantly from the values given by the company. Therefore, they do not statistically vary from the expected values (NO OUTSIDE INFLUENCE)

15. Try this A card dealer deals out the following cards
22 clubs
32 spades
34 hearts
12 diamonds
Is the dealer cheating OR does random chance account for the distribution?

16. SOLUTION CHI-squared value = 12.32 Degrees of freedom = 3
P value = 7.82
X2 > p value
Therefore, the distribution is not due to chance (something fishy is going on)