Q: Which formula is used to calculate the expected cell frequency in a contingency table for the Chi-square test?

$$\text{Expected cell frequency} = \frac{(\text{row total}) \times (\text{column total})}{\text{overall total}}$$

Question 1

Which of the following best describes the purpose of a contingency table in statistics?

Accepted Answer

To display the relationship between two categorical variables by showing the frequency of observations for each combination of categories.

Question 2

Given the following data on credit defaults:

Previous late payment: Yes (Default: 347, No Default: 486, Total: 833)
Previous late payment: No (Default: 655, No Default: 13510, Total: 14165)

What is the estimated absolute risk of default for those with a previous late payment?

Accepted Answer

$$\frac{347}{833}$$

Question 3

Using the same credit default data, what is the estimated relative risk of default for those with a previous late payment compared to those with no previous late payment?

Accepted Answer

$$\frac{\frac{347}{833}}{\frac{655}{14165}}$$

Question 4

Which of the following statements about relative risk is correct?

Accepted Answer

Relative risk compares the likelihood of an event occurring in one group to another group.

Question 5

A study finds that 26.7% of tennis players who warmed up reported an injury, while 60% of those who did not warm up reported an injury. What is the relative risk of injury for those who did not warm up compared to those who did?

Accepted Answer

$$\frac{0.60}{0.267}$$

Question 6

Which formula is used to calculate the expected cell frequency in a contingency table for the Chi-square test?

Accepted Answer

$$	ext{Expected cell frequency} = \frac{(	ext{row total}) 	imes (	ext{column total})}{	ext{overall total}}$$

Question 7

In the tennis injury study, the observed frequency for 'Injury = Yes' and 'Warm-up = No' is 21. The expected frequency for this cell is 14. What is the contribution of this cell to the Chi-square test statistic?

Accepted Answer

$$\frac{(21-14)^2$}{14}$$

Question 8

A researcher tests the association between age group and trade union membership using a $$3 	imes 2$$ contingency table. What are the degrees of freedom for the Chi-square test?

Accepted Answer

$$(3-1) 	imes (2-1) = 2$$

Question 9

If the calculated Chi-square test statistic is 18.64 and the critical value at $$\alpha = 0.05$$ with 2 degrees of freedom is 5.991, what is the correct conclusion?

Accepted Answer

Reject $$H_0$$ and conclude there is a statistically significant association.

Question 10

Which of the following is the correct null hypothesis for a Chi-square test of association between two categorical variables?

Accepted Answer

$$H_0$$: There is no association between the two categorical variables.

Question 11

A company compares the quality of raw materials from two suppliers (A and B) classified as 'good', 'acceptable', or 'reject'. Which test is most appropriate to determine if there is an association between supplier and quality?

Accepted Answer

Chi-square test of association

Question 12

Why is it important to interpret relative risk within the context of absolute risk?

Accepted Answer

Because a high relative risk may still correspond to a very low absolute risk, especially for rare events.

Question 13

Which of the following best describes the critical value in a Chi-square test?

Accepted Answer

The value that the test statistic must exceed to reject the null hypothesis at a given significance level.

Question 14

In a $$2 	imes 2$$ contingency table, the calculated Chi-square statistic is 2.5. The critical value at $$\alpha = 0.05 is 3.841. $$What is the correct decision?

Accepted Answer

Fail to reject $$H_0$$; there is not enough evidence of an association.

Question 15

Which of the following is NOT an assumption of the Chi-square test of association?

Accepted Answer

The data must be normally distributed.

Question 16

A sample of 5441 customers is classified by type (A, B, C) and whether they bought a product. What is the relative risk of buying for Type A compared to Type C?

Accepted Answer

$$\frac{\frac{750}{1236}}{\frac{1223}{2619}}$$

Question 17

Which of the following best explains why we round expected frequencies down to the nearest whole number in the Chi-square test examples provided?

Accepted Answer

Because observed frequencies are always whole numbers, and expected frequencies should match this for calculation consistency.

L14 Chi-Square Tests, Risk, and Association in Categorical Data

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