From the Decision to a Contextual Conclusion
In “Making a Decision in a Chi-Square Test,” you learned to compare a chi-square test’s p-value with the chosen significance level, \(\alpha\). This tutorial focuses on the next step: explaining what that decision says about the two categorical variables in the setting of a test of independence.
A test of independence uses one sample in which each individual is classified by two categorical variables. As explained in “Stating Hypotheses for a Test of Independence,” the null hypothesis says the variables are independent in the population, and the alternative says they are associated. Your conclusion should translate the decision about these hypotheses into a statement about the population and the variables named in the question.
The word association is central. When the variables are associated, knowing the category of one variable is useful for understanding the distribution of the other. In the sample, this can appear as different conditional distributions across categories, as covered in “Interpreting Association Versus Independence in a Table.” The chi-square test evaluates whether the overall pattern of counts is convincing evidence against independence; the conclusion need not identify a particular cell or category.
A good conclusion names both variables and the population. For example, “There is convincing evidence that workshop attendance and compost-bin use are associated among members of this cooperative” is more informative than “The result is significant.” The first statement explains what the test indicates and where the claim applies.
Build the Conclusion From the Test
Use the p-value and \(\alpha\) to choose the appropriate conclusion. Reject \(H_0\) when \(p\le\alpha\); fail to reject \(H_0\) when \(p>\alpha\). As in the previous tutorial, do not say “accept \(H_0\).” A large p-value means the data do not provide convincing evidence against independence; it does not prove that the variables are independent.
A useful conclusion has three connected parts: the decision, the strength of evidence, and the contextual claim. Include the comparison that supports your decision, then state what the data indicate about the named variables in the population. This makes the reasoning visible instead of leaving the reader to infer what “reject” means.
Compare the p-value with the \(\alpha\) selected for the test. Use the unrounded calculator value if the comparison is close.
Say “reject \(H_0\)” if \(p\le\alpha\), or “fail to reject \(H_0\)” if \(p>\alpha\).
For rejection, state that there is convincing evidence of an association between the variables in the population. For failure to reject, state that the sample does not provide convincing evidence of an association.
Do not turn a nonsignificant result into proof of independence, or an association into a claim of causation.
Before writing the final statement, check that the hypotheses, p-value, decision, and context all refer to the same two variables and population. Also make sure the conditions for the test were addressed, as in “Random Condition for Chi-Square Procedures,” “Independence and 10 Percent Condition for Chi-Square,” and “Checking the Expected Count Condition for Chi-Square.” A careful conclusion cannot compensate for a design or condition problem.
Worked Example: Convincing Evidence of an Association
Worked Example: Convincing Evidence of an Association
Suppose an invented random sample of 120 members from a gardening cooperative with 2,400 members is classified by whether each person attended a soil workshop and whether each person uses a compost bin. The observed counts are:
| Uses compost bin | Does not use compost bin | Total | |
|---|---|---|---|
| Attended workshop | 40 | 20 | 60 |
| Did not attend workshop | 20 | 40 | 60 |
| Total | 60 | 60 | 120 |
State. The population is all members of this gardening cooperative. The null hypothesis is that workshop attendance and compost-bin use are independent in this population. The alternative hypothesis is that these variables are associated.
Plan. Use a chi-square test of independence because one sample of cooperative members is classified by two categorical variables. The sample is random, and each member is counted in exactly one cell. The 10% condition holds because \(120\le0.10(2400)=240\). Under independence, each expected count is at least 5. For example, the expected count for members who attended the workshop and use a compost bin is \(60(60)/120=30\); the same row and column totals give 30 for each of the other three cells.
Do. Each of the four observed counts differs from its expected count of 30 by 10 in absolute value. Thus:
As a check, four equal contributions of \(100/30\) sum to \(400/30=13.3333\). The degrees of freedom are \((2-1)(2-1)=1\). The calculator gives an upper-tail p-value of approximately \(0.0003\), rounded to four decimal places.
Conclude. At \(\alpha=0.05\), \(0.0003<0.05\), so reject \(H_0\). The random sample provides convincing evidence that soil-workshop attendance and compost-bin use are associated among members of this gardening cooperative.
Worked Example: No Convincing Evidence of an Association
Worked Example: No Convincing Evidence of an Association
Suppose an invented random sample of 120 subscribers from a service with 5,000 subscribers is classified by preferred newsletter format and whether the subscriber reads the newsletter on a phone, tablet, or computer. The observed counts are:
| Preferred format | Phone | Tablet | Computer | Total |
|---|---|---|---|---|
| Brief summary | 22 | 19 | 19 | 60 |
| Detailed edition | 18 | 21 | 21 | 60 |
| Total | 40 | 40 | 40 | 120 |
State. The question is whether preferred newsletter format and device used to read the newsletter are associated among all subscribers to this service. The null hypothesis states that the variables are independent; the alternative states that they are associated.
Plan. Use a chi-square test of independence. The sample is random, each subscriber appears once in the table, and the 10% condition holds because \(120\le0.10(5000)=500\). The expected count in every cell is \(60(40)/120=20\), so every expected count is at least 5.
Do. The first row has deviations \(2,-1,-1\) from its expected counts, and the second row has deviations \(-2,1,1\). Therefore:
The degrees of freedom are \((2-1)(3-1)=2\). The upper-tail p-value is approximately \(0.7408\), rounded to four decimal places. As a check for 2 degrees of freedom, the upper-tail probability is \(e^{-0.60/2}=e^{-0.30}\approx0.7408\).
Conclude. At \(\alpha=0.05\), \(0.7408>0.05\), so fail to reject \(H_0\). The sample does not provide convincing evidence that preferred newsletter format and reading device are associated among the service’s subscribers. This does not prove that the variables are independent.
Worked Example: Evidence of Association Is Not Evidence of Cause
Worked Example: Evidence of Association Is Not Evidence of Cause
Suppose an invented random sample of 120 residents from a town of 4,000 is classified by housing type and whether the resident composts food scraps. The observed table is:
| Housing type | Composts food scraps | Does not compost | Total |
|---|---|---|---|
| Apartment | 28 | 12 | 40 |
| Townhouse | 20 | 20 | 40 |
| Detached home | 12 | 28 | 40 |
| Total | 60 | 60 | 120 |
State. The population is the town’s residents. The null hypothesis is that housing type and composting food scraps are independent in this population; the alternative is that they are associated.
Plan. A chi-square test of independence is appropriate because one random sample is classified by two categorical variables. Each resident contributes to exactly one cell. The 10% condition holds because \(120\le0.10(4000)=400\). Each expected count is \(40(60)/120=20\), so the expected-count condition is met.
Do. In the first and third rows, the observed counts differ from the expected counts by 8 in absolute value. The middle row matches its expected counts:
A check is \(4(64/20)=4(3.2)=12.8\). The degrees of freedom are \((3-1)(2-1)=2\), and the upper-tail p-value is approximately \(0.0017\), rounded to four decimal places.
Conclude. At \(\alpha=0.05\), \(0.0017<0.05\), so reject \(H_0\). The random sample provides convincing evidence that housing type and composting food scraps are associated among residents of this town. Because this is an observational sample, the conclusion does not show that housing type causes residents to compost, or that composting causes people to live in a particular type of home.
Common Mistakes and AP Exam Tip
- Writing only “the result is significant”: Name the two variables and the population. A full-credit conclusion explains what the evidence is about, not just that a decision was made.
- Claiming independence after failing to reject: Say that the sample does not provide convincing evidence of an association. Do not say the variables are independent or that the null hypothesis has been proven.
- Using “association” without context: Identify both variables and the population to which the conclusion refers. Avoid an unsupported claim about a broader population than the sampling design represents.
- Confusing association with causation: A test of independence can provide evidence that variables are associated. It does not establish that one variable causes the other.
- Making a more specific claim than the test supports: Rejection does not by itself tell you which category or cell is responsible for the result, or how strong or practically important the association is.
- Leaving out the decision comparison: State the p-value, \(\alpha\), and whether the p-value is below or above \(\alpha\). Then give the matching decision and contextual interpretation.
For a clear AP-style response, connect the decision directly to the alternative hypothesis. For rejection, write that there is convincing evidence that the named variables are associated in the population. For failure to reject, write that there is not convincing evidence of an association in that population. Avoid certainty, causal claims, and vague phrases that do not identify the variables.
Check Your Understanding
Use the decision and study context to write an accurate conclusion for each situation.
- A test of independence has \(p=0.018\) and \(\alpha=0.05\). What decision should be made, and what should the contextual conclusion say?
- A test has \(p=0.22\) and \(\alpha=0.05\). Write a correct conclusion and identify one claim that should be avoided.
- Why should a conclusion name the population as well as both categorical variables?
- A test rejects independence between preferred exercise setting and age group in a random survey. Does this result show that age group causes a person to prefer a particular setting? Explain.
- A test rejects the null hypothesis. Does that decision alone identify which cells have the largest departures from expected counts? Explain.