Question 1

Chi-square tests of independence are always lower-tailed because a perfect fit between  and  makes  equal to zero.

Accepted Answer

Chi-square tests of independence can be either lower-tailed, upper-tailed, or two-tailed depending on the direction of the research hypothesis. The statement given in the question is not an accurate explanation of the directionality of the chi-square test.

Question 2

The chi-square test of a contingency table is used to determine if there is enough evidence to infer that two nominal variables are related, and to infer that differences exist among two or more populations of nominal variables.

Accepted Answer

The statement is true. The chi-square test is commonly used in statistical analysis to determine the independence between two categorical variables, or to test the hypothesis that there is an association or relationship between them. It can also be used to compare the distribution of multiple groups of categorical data.

Question 3

In a chi-square test of independence, the value of the test statistic was  15.652, and the critical value at  was 11.1433. Thus, we must reject the null hypothesis at  .

Accepted Answer

The test statistic value is greater than the critical value, which means there is sufficient evidence to reject the null hypothesis. Therefore, the statement is true.

Question 4

A contingency table classifies data with respect to two qualitative variables that are each divided into two or more categories.

Accepted Answer

This statement correctly describes what a contingency table does. It organizes data based on two categorical variables, allowing for comparison between different groups.

Question 5

Numbers in a contingency table show the frequency of occurrence of all possible combinations of categories.

Accepted Answer

This statement is true.

Question 6

A chi-square test for independence is applied to a contingency table with 4 rows and 4 columns for two qualitative variables. The degrees of freedom for this test must be 9.

Accepted Answer

The degrees of freedom for a chi-square test for independence in a contingency table is calculated as (number of rows - 1) * (number of columns - 1), which for a 4x4 table is (4-1)*(4-1) = 3*3 = 9.

Question 7

A contingency table classifies data according to two or more categories associated with each of two qualitative variables that are statistically independent of one another.

Accepted Answer

A contingency table classifies data according to two or more categories, but it is used to study the association between the variables, not to assume they are statistically independent.

Question 8

In a typical contingency-table test, we calculate each expected cell frequency as the product of row total times column total times sample size.

Accepted Answer

In a typical contingency-table test, expected cell frequencies are calculated using a specific formula that involves the row and column totals, but does not include the sample size. The formula is (row total × column total) / total sample size.

Question 9

The degrees of freedom associated with a chi-square test of independence where data are summarized in a contingency table with r rows and c columns equal the number of rows times the number of columns in the table; that is, rc.

Accepted Answer

The degrees of freedom associated with a chi-square test of independence where data are summarized in a contingency table with r rows and c columns equal the product of (r-1) and (c-1). So, the correct formula is (r-1) x (c-1).

Question 10

The chi-square test of independence is based upon three or more quantitative variables.

Accepted Answer

The chi-square test of independence is actually based upon two categorical variables.

Question 11

A chi-square test for independence with 6 degrees of freedom results in a test statistic of 13.25. Using the chi-square table, the most accurate statement that can be made about the p-value for this test is that p-value is greater than 0.025 but smaller than 0.05.

Accepted Answer

Looking at a chi-square distribution table with 6 degrees of freedom, a test statistic of 13.25 falls in the range of 0.025 < p-value < 0.05 (the intersection of row 6 and column for test statistic 13.25). Therefore, the most accurate statement is that the p-value is greater than 0.025 but smaller than 0.05.

Question 12

The degrees of freedom associated with a contingency-table test of independence equal the number of rows times the number of columns in the table.

Accepted Answer

The degrees of freedom associated with a contingency-table test of independence equal (number of rows - 1) times (number of columns - 1) in the table.

Question 13

A chi-square test for independence with 10 degrees of freedom results in a test statistic of 17.894. Using the chi-square table, the most accurate statement that can be made about the p-value for this test is that 0.05 < p-value < 0.10.

Accepted Answer

When using a chi-square table, the degrees of freedom and test statistic are used to determine the p-value. In this case, with 10 degrees of freedom and a test statistic of 17.894, the p-value falls between 0.05 and 0.10, which means that the statement "0.05 < p-value < 0.10" is the most accurate statement that can be made about the p-value for this test. Therefore, the answer is A.

Question 14

A chi-square test for independence is applied to a contingency table with 3 rows and 4 columns for two qualitative variables. The degrees of freedom for this test must be 12.

Accepted Answer

The degrees of freedom for a chi-square test for independence with a contingency table of 3 rows and 4 columns should be (number of rows - 1) x (number of columns - 1) = 2 x 3 = 6, not 12.

Question 15

To be valid, a chi-square test of independence requires that each expected frequency equal 30 or more.

Accepted Answer

The rule of thumb is that each expected frequency should be at least 5, not 30. However, if the sample size is very large, then expected frequencies of 1 or 2 may be acceptable.

Question 16

In a typical chi-square test of independence, we calculate each expected cell frequency (  ) as the product of row total (  ) times column total (  ) times sample size (n).

Accepted Answer

In a typical chi-square test of independence, each expected cell frequency is calculated as (row total x column total)/n. Sample size is not included in this calculation.

Question 17

A chi-square test for independence is applied to a contingency table with 3 rows and 5 columns for two qualitative variables. The degrees of freedom for this test is 8.

Accepted Answer

The degrees of freedom for a chi-square test for independence is calculated as (number of rows - 1) * (number of columns - 1), which in this case is (3 - 1) * (5 - 1) = 2 * 4 = 8.

Question 18

In a chi-square test of independence with 6 degrees of freedom and a level of significance of 0.05, the critical value from the chi-square table is 12.5916. The computed value of the test statistics is 11.264. This will lead us to reject the null hypothesis.

Accepted Answer

The computed value of the test statistic is less than the critical value, therefore we fail to reject the null hypothesis.

Question 19

Chi-square tests of independence are always lower-tailed because a perfect fit between  and  makes the test statistic  equal to zero.

Accepted Answer

Chi-square tests of independence can be one-tailed or two-tailed depending on the research question and hypotheses. A perfect fit between the variables in the contingency table would result in a chi-square value of zero, but this does not necessarily mean that the test is always lower-tailed.

Question 20

To be valid, a chi-square test of independence requires that the expected frequency for each cell in the contingency table equal 10 or more.

Accepted Answer

The rule of thumb for the chi-square test of independence is that the expected frequency for each cell in the contingency table should be 5 or more, not 10.

Quiz 14: Analysis of Categorical Data

Chi-square tests of independence are always lower-tailed because a perfect fit between and makes equal to zero.