Chi-Square X2

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

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

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

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.

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

What we found… Color Observed Brown 76 Yellow 65 Red 70 Blue 85 Green 28 Orange 26 total 350

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

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

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 < 0.05 (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

accept null hypothesis reject null hypothesis

For Reals…

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 7.799 < 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)

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?

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)