Hypothesis test with Chi Square – Independence
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Scenario: Crammer Nation University just completed their Greek Life rush. Delta Apple Pi is curious whether the hat that a recruit wore during recruitment had an impact on their bid across all fraternities. They conduct a random sample of 85 recruits. Utilizing an alpha level of 0.05, compute a hypothesis test and determine whether these two variables are independent of one another.
| Backwards hat | No hat | Bucket hat | Total | |
|---|---|---|---|---|
| Received bid | 21 | 10 | 8 | 39 |
| Did not receive bid | 13 | 26 | 7 | 46 |
| Total | 34 | 36 | 15 | 85 |
| Clue | Insight |
|---|---|
| The purple values are counts of data. | This is a Chi Square test. |
| We're being asked to "determine whether these two variables are independent of one another." | This means we'll run an Independence hypothesis test. |
| Assumption | Validate |
|---|---|
| The data is counted. | The data in purple is all counted values. ✅ |
| The counts must be randomly selected from the population. | The scenario details we're taking a "random sample". ✅ |
| Each count must be 5 or greater. | All our counts in purple are greater than 5! ✅ |
To be clear here: the "variables" in this case are "hat that a recruit wore" and "impact on their bid".
H0: The variables are independent.
Ha: The variables are not independent.
We're conducting a Chi Square test, so our test statistic will be X2!
Σ means the sum of each count.
obs means observed values.
exp means expected values (which we must calculate).
| Backwards hat | No hat | Bucket hat | Total | |
|---|---|---|---|---|
| Received bid | (34 x | (36 x | ( | 39 |
| Did not receive bid | (34 x | ( | ( | 46 |
| Total | 34 | 36 | 15 | 85 |
X2 = [(21 - 15.6)2 / 15.6] + [(10 - 16.518)2 / 16.518] + [(8 - 6.882)2 / 6.882] + [(13 - 18.4)2 / 18.4] + [(26 - 19.482)2 / 19.482] + [(7 - 8.118)2 / 8.118]
X2 = [(5.4)2 / 15.6] + [(-6.518)2 / 16.518] + [(1.118)2 / 6.882] + [(-5.4)2 / 18.4] + [(6.518)2 / 19.482] + [(-1.118)2 / 8.118]
X2 = [29.16 / 15.6] + [42.484 / 16.518] + [1.250 / 6.882] + [29.16 / 18.4] + [42.484 / 19.482] + [1.250 / 8.118]
X2 = [1.869] + [2.572] + [0.182] + [1.585] + [2.181] + [0.154]
X2 = 8.543
We've got 2 rows and 3 columns!
df = (# of rows - 1) x (# of columns - 1)
df = (2 - 1) x (3 - 1)
df = (1) x (2)
df = 2
0.025 < p-value < 0.05
Our X2 value sits somewhere in between 0.025 and 0.05!
0.025 < p-value < 0.05
α = 0.05
Considering our entire p-value range 0.025 to 0.05 is less than our alpha level of 0.05, this means we'll reject the null hypothesis!
Answer: Since our entire p-value range of 0.025 < p-value < 0.05 is less than our alpha level of 0.05, we reject the null hypothesis and do have enough evidence to support the alternative hypothesis, which states the hat that a recruit wore and whether or not they received a bid are not independent of each other at Crammer Nation University.
In other words... a recruit's hat selection may be associated with (and not independent of) their bid! (Remember... this is not a guaranteed cause-and-effect relationship, rather, it's an association!)