A. Intro to Chi-Square Tests for Independence
A chi-square test for independence is a type of significance test where we want to find convincing evidence for a certain claim, more specifically the alternative hypothesis Ha, regarding an effect of one factor on another. These types of hypotheses commonly involve using 2-way tables, and analyzing if there is convincing evidence for an association between the 2 factors. For the sake of practice, we will be using a made-up 2 way table that I made:
*MADE UP* 2 way table of Averaged PPR Points per Game and Height for ALL starting WRs in the last 3 years
B. Method for Chi-Square Test for Independence
Is there truly convincing evidence of an association between WR's height and their Averaged PPR Fantasy Points per Game in the population of all starting WRs in the last 3 years?
Likewise, we have to perform a methodical procedure that consists of 4 main steps.
1. The first step is "State." Here, we state 2 hypotheses, the null hypothesis H, H0, and the alternative hypothesis, Ha. The Ha is what we are testing, which is that there is an association between these two categories. So of course, Ho is that there is no association.
State:
H0: There is no association between height and averaged PPR fantasy points per game in the population of all starting WRs in the last 3 years.Ha: There is an association between height and averaged PPR fantasy points per game in the population of all starting WRs in the last 3 years.
2. The second step is "Plan." Similar to what we have previously done, this step involves 2 things. What type of test we are conducting, and whether or not our conditions are met. Our conditions slightly differ from our previous examples. However, the "random" condition and the "Independence" condition are the same, but remember again that these only apply if we sampled from the population. If we took absolutely every subject from a population, then these two conditions can't apply and are thus irrelevant. For our 3rd, the condition is still called "Large Counts," but except that it now requires that all expected values in the table are ≥ 5. But how do we calculate expected values? The formula is expected count = (row total x column total) / table total. Here's an example on our graph:
1. The first step is "State." Here, we state 2 hypotheses, the null hypothesis H, H0, and the alternative hypothesis, Ha. The Ha is what we are testing, which is that there is an association between these two categories. So of course, Ho is that there is no association.
State:
H0: There is no association between height and averaged PPR fantasy points per game in the population of all starting WRs in the last 3 years.Ha: There is an association between height and averaged PPR fantasy points per game in the population of all starting WRs in the last 3 years.
2. The second step is "Plan." Similar to what we have previously done, this step involves 2 things. What type of test we are conducting, and whether or not our conditions are met. Our conditions slightly differ from our previous examples. However, the "random" condition and the "Independence" condition are the same, but remember again that these only apply if we sampled from the population. If we took absolutely every subject from a population, then these two conditions can't apply and are thus irrelevant. For our 3rd, the condition is still called "Large Counts," but except that it now requires that all expected values in the table are ≥ 5. But how do we calculate expected values? The formula is expected count = (row total x column total) / table total. Here's an example on our graph:
Now, when repeat that calculation for every cell, we get these numbers:
We see that all of the expected counts are ≥ 5.
Plan: We will be doing a Chi-Square Test for Independence
Random condition: We did not sample any WRs. We instead looked at the whole population of starting WRs. Therefore, this condition does not apply. Independence condition: We did not sample any WRs. We instead looked at the whole population of starting WRs. Therefore, this condition does not apply. Large Counts: All expected counts are greater than 5. This is met
All conditions are met.
3. The third step is called "Do." We now calculate our standardized test statistic, x^2, to ultimately get our p-value. There is a calculator method and a hand-written method we can use to solve, but I will explain the handwritten method.
Random condition: We did not sample any WRs. We instead looked at the whole population of starting WRs. Therefore, this condition does not apply. Independence condition: We did not sample any WRs. We instead looked at the whole population of starting WRs. Therefore, this condition does not apply. Large Counts: All expected counts are greater than 5. This is met
All conditions are met.
3. The third step is called "Do." We now calculate our standardized test statistic, x^2, to ultimately get our p-value. There is a calculator method and a hand-written method we can use to solve, but I will explain the handwritten method.
This is the formula for finding our x^2 statistic, and we will use every cell included in our 2-way table for this calculation.
Do:
Nice! So we found our X^2 statistic, which is 8.32. But now what? Now, we are going to use this value, along with our df value, to help us find our p-value. We can do this with two methods, which are using a reference sheet or our calculators if it is able to.
Note: Unlike earlier, figuring out what our df (degrees of freedom) value is not as simple as n-1.
Here, we do the (number of columns - 1) x (number of rows - 1 ) to find it. We have 3 columns and 5 rows in this table, so we would do 2 x 4.
This gives us a df of 8.
#1. The first method of finding a p-value given that you have your X^2 value and df is to use a reference sheet provided by Collegeboard. Collegeboard will provide you with a reference sheet during the AP Exam. And anywhere else, you can search it online. We go down the column labeled "df" and would go to 8. Then, we go towards the right to find our x^2 value, which is 8.32. Unfortunately, there is no 8.32 on the sheet as the smallest number on this row starts at 10.22. So, we can estimate. At last, we go up to find our p-value. At the df of 8, an X^2 value of 10.22 is .25, so we can estimate that an X^2 value of 8,23 would be around 0.40. Regardless, anything above a p-value of 0.05 would be considered not statistically significant in this problem, so the exact pin-pointed value does not matter much.
#2. Alternatively, we can use a calculator. For instance, with a Ti-84 Plus CE, we would press "2nd vars" to go into the distribution tab. Then, we scroll down to "X^2 cdf."
Here, we input our values.
lower: 8.23
upper: 1000000
df: 8
=
x^2cdf (8.23, 100000, 8) = .411.
This gives us a p-value of 0.411.
4. The last step is called "Conclude" Here, like previous conclusions, we use our newly calculated p-value, to answer the question, which is if there is convincing evidence of an association between the two factors.
Conclude: Because the p-value of 0.411 > a = 0.05, we do not have convincing evidence to reject the null hypothesis, H0. This means that we cannot prove or state that there is an association between Averaged PPR Points per Game and Height for our fake population of all starting WRs in the last 3 years.
Note: Remember, this example used made-up data by me for practical sakes of demonstrating the process. Real data will be different from mine, which will grant different results.
2 Way Table of Round Projections and Actual Averaged PPR Points per Game for the first 50 WRs in the 2023 NFL Season (excluding those who played less than 7 games)
- Round projections according to Fantasy Pros ADP
- "First 50 WRs" as in the projected WR1 to WR50, adjustments were made if a player was excluded from the population
Prompt: Using the data above, is there convincing evidence that there is an association between Projected Round Numbers and Actual Averaged PPR Points per Game for the first 50 WRs in 2023?
Work: We will conduct our 4-step procedure below.
State:
H0: There is no association between Projected Round # and Averaged PPR Points per Game for the first 50 WRs in 2023.
Ha: There is an association between Projected Round # and Averaged PPR Points per Game for the first 50 WRs in 2023.
Plan:
We will be doing a Chi-Square Test for Independence
Random condition: We did not sample any WRs. We instead looked at the whole population of the first 50 WRs. Therefore, this condition does not apply.Independence condition: We did not sample any WRs. We instead looked at the whole population of the first 50 WRs. Therefore, this condition does not apply.Large Counts: All expected counts are greater than 5. (28 x 21) / 50, (28 x 29)/50, (22 x 21)/50, (22 x 29)/50, > 5. This is met
All conditions are met. Do: X^2 = ∑ (Observed - Expected)^2 / Expected X^2 = ((6 - 11.76)^2 / 11.76) + ((22 - 16.24)^2 / 16.24) + (( 15 - 9.24)^2 / 9.24) + ((7 - 12.76)^2 / 12.76) X^2 = 11.05 df = (2-1) (2-1) = 1 Using a calculator to calculate p-value --> X^2cdf (Lower: 11.05, Upper: 10000000, df: 1) = 0.00089 Conclude: Because the p-value of 0.00089 < a = 0.05, we do have convincing evidence to reject the null hypothesis, H0. This means that there is an association between Projected Round Numbers (again, projections according to Fantasy Pros) and Actual Averaged PPR Points per Game for our population of the first 50 WRs in the 2023 NFL Season, excluding those who played less than 7 games. Answer: There indeed is convincing evidence that there is an association between Projected Round Numbers and Actual Averaged PPR Points per Game for the first 50 WRs in the 2023 NFL Season.
Random condition: We did not sample any WRs. We instead looked at the whole population of the first 50 WRs. Therefore, this condition does not apply.Independence condition: We did not sample any WRs. We instead looked at the whole population of the first 50 WRs. Therefore, this condition does not apply.Large Counts: All expected counts are greater than 5. (28 x 21) / 50, (28 x 29)/50, (22 x 21)/50, (22 x 29)/50, > 5. This is met
All conditions are met. Do: X^2 = ∑ (Observed - Expected)^2 / Expected X^2 = ((6 - 11.76)^2 / 11.76) + ((22 - 16.24)^2 / 16.24) + (( 15 - 9.24)^2 / 9.24) + ((7 - 12.76)^2 / 12.76) X^2 = 11.05 df = (2-1) (2-1) = 1 Using a calculator to calculate p-value --> X^2cdf (Lower: 11.05, Upper: 10000000, df: 1) = 0.00089 Conclude: Because the p-value of 0.00089 < a = 0.05, we do have convincing evidence to reject the null hypothesis, H0. This means that there is an association between Projected Round Numbers (again, projections according to Fantasy Pros) and Actual Averaged PPR Points per Game for our population of the first 50 WRs in the 2023 NFL Season, excluding those who played less than 7 games. Answer: There indeed is convincing evidence that there is an association between Projected Round Numbers and Actual Averaged PPR Points per Game for the first 50 WRs in the 2023 NFL Season.