Question 1 Foundation
State the chi-squared test statistic.
Answer.
Each squared discrepancy is divided by its own expected frequency, so a gap of 5 counts for more when E is small.
The Chi-Squared Test
Original questions written at exam standard. Work each one before you open the solution.
Difficulty
Question 2 Foundation
How are expected frequencies found for a contingency table?
Answer. Row total × column total ÷ grand total
This is what independence would predict. Check that the expected values reproduce the original totals.
Question 3 Standard
How many degrees of freedom does a contingency table have?
Answer.
.
Question 4 Standard
Why must all expected frequencies be at least 5?
Answer. Otherwise the chi-squared approximation breaks down.
A small makes unstable and inflates the statistic. Combine categories until every expected value reaches 5.
Question 5 Stretch
A die rolled 60 times gives . Find .
Answer.
Each expected frequency is 10, and the squared differences are . Dividing the total 10 by 10 gives 1.
Question 6 Stretch
A chi-squared test rejects for a contingency table. What can be concluded?
Answer. There is evidence of an association between the two variables — not that one causes the other.
A third factor may drive both. Chi-squared detects association only, and the wording of the conclusion should say exactly that.