Deck 19: Categorical Outcomes: Chi-Square and Loglinear Analysis

Question 1

Which of the following is the correct calculation for the degrees of freedom (df) when using a 2 × 2 contingency table?

A) (n - 1) × (n - 1) = 1
B) (n + 1) × (n + 1) = 9
C) (n - 1) + (n - 1) = 2
D) (n + 1) + (n + 1) = 6

Answer 1

A

Question 2

Which of the following is the most appropriate measure of central tendency when working with categorical data?

A) Mean
B) Mode
C) Median
D) None of the above are appropriate.

Answer 2

D

Question 3

Calculate the chi-square value for the below example.

A) 2.876
B) 2.600
C) 2.881
D) 0.586

Answer 3

A

Question 4

When samples sizes are small, the significance from values from a chi-square distribution can become inaccurate. What test can be applied to reduce this effect?

A) Fisher's exact test
B) Loglinear analysis
C) Seymour's test of precision
D) Multiple t-tests

Question 5

What is the minimum number of observations required for the expected frequencies?

A) 1
B) 5
C) 10
D) 15

Question 6

If Yates's continuity correction is applied, how is the chi-square value calculated earlier altered?

A) 2.86
B) 2.91
C) 2.79
D) 3.01

Question 7

What is the purpose of using Yates's continuity correction?

A) It reduces the probability of Type I error.
B) It reduces the probability of Type II error.
C) It makes the test more liberal.
D) It corrects for error in large samples.

Question 8

Calculate the expected value for the athletes who were inspired.

A) 142
B) 146
C) 150
D) 138

Question 9

Which of the following is not a measure used to modify the chi-square test value?

A) Phi
B) Contingency coefficient
C) Cramér's V
D) Hotelling's adjustment

Question 10

Before a chi-square test is performed, what should be inspected by the researcher?

A) Contingency table to inspect the expected frequencies.
B) Kolmogorov-Smirnov test to evaluate whether the data are normally distributed.
C) Shapiro-Wilks test to evaluate whether the data are normally distributed.
D) A histogram or box-whisker plot to inspect the distribution of the data.

Question 11

Which of the following could be considered a categorical variable? (You may select more than one option.)

A) The sex of each participant
B) The sport played by each participant
C) The intrinsic motivation score for each participant
D) The resting heart rate for each participant

Question 12

If a backward elimination in loglinear analysis was to be used, what would be the initial effect or interaction potentially eliminated?

A) Gender × athletic status × level of inspiration
B) Gender
C) Athletic status × level of inspiration
D) The order is not important so any effect or interaction could be potentially eliminated first.

Question 13

Based on the chi-square value calculated and the critical value identified above, what can be deduced from the association between athletic status and whether the London Olympic Games inspired them to become more physically active?

A) A significant difference exists between athletic status and the level of inspiration.
B) No significant difference exists between athletic status and the level of inspiration.
C) The relationship between athletic status and the level of inspiration is unclear.
D) A minor significant relationship exists between athletic status and the level of inspiration.

Question 14

If the example above was developed to incorporate gender, athletic status and whether they were inspired by the London Olympics, how many main effects and interactions would exist?

A) 5
B) 6
C) 7
D) 8

Question 15

What are the principal assumptions of analysing categorical data? (You may select more than one option.)

A) Independence of residuals.
B) No expected frequency should be less than 5.
C) Data should be normally distributed.
D) Sphericity must be assumed.

Question 16

What is the purpose of a chi-square test?

A) To measure the degree of association between categorical variables.
B) To distinguish the degree of separation between categorical variables.
C) To evaluate the degree of differentiation between continuous variables.
D) To quantify the degree of differentiation between categorical variables.

Question 17

Identify the critical value from a chi-square distribution table based on the df calculated above and a significance level of .05.

A) 3.84
B) 5.99
C) 7.81
D) 9.49

Question 18

How does Yates's continuity correction alter the chi-square equation?

A) It deducts 0.5 from the observed minus expected result.
B) It adds 0.5 to the observed minus expected result.
C) It adds 1.5 to the observed minus expected result.
D) It deducts 1.5 from the observed minus expected result.

Question 19

When should loglinear analysis be used?

A) When there are two or more categorical variables.
B) When there are three or more categorical variables.
C) When there are four or more categorical variables.
D) When the data are not normally distributed.

Question 20

Examine the table below. What do the z-scores (i.e. standardized residuals) for the two variables tell us?

A) Athletes who were inspired by the London Olympics had a significant z-score.
B) Non-athletes who were inspired by the London Olympics had a significant z-score.
C) None of the z-scores were significant.
D) Athletes who were not inspired by the London Olympics had a significant z-score.

Question 21

If a significant result is obtained, how might you interpret the findings more thoroughly?

A) Refer to percentage totals in the contingency table.
B) Refer back to the raw data.
C) Refer to Fisher's exact test statistic.
D) Refer to the continuity correction value.

Question 22

For a main effect or an interaction to be significant, what value should the z-score ideally exceed?

A) 1.96
B) 2.50
C) 0.92
D) 1.30

Question 23

Examine the results from the loglinear analysis. Which of the following interactions were significant? (You may select more than one option.)

A) Sex × activity × inspired
B) Sex × activity
C) Sex × inspired
D) Activity × inspired

Question 24

How would you report a chi-square result?

A) χ² (1) = 2.88, p = .09
B) p = .09
C) χ² (1) = p = .09, = 2.88
D) χ² = 2.88