Question 1

Suppose the value of your chi-squared test statistic in a goodness-of-fit test is equal to 0. What do you conclude?

Accepted Answer

The answer of Suppose the value of your chi-squared test...

Question 2

A chi-squared test of a contingency table with 6 degrees of freedom results in a test statistic of 13.25. Using the chi-squared 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 the chi-squared distribution table with 6 degrees of freedom, a test statistic of 13.25 corresponds to a p-value between 0.025 and 0.05, but not exactly at either value. Therefore, the most accurate statement that can be made is that the p-value is greater than 0.025 but smaller than 0.05.

Question 3

The degrees of freedom for the test statistic in a test of a contingency table is ( r - 1 )( c - 1 )where r is the number of rows in the table, and c is the number of columns.

Accepted Answer

This is the correct formula for calculating the degrees of freedom for a test of a contingency table.

Question 4

To calculate the expected values in a test of a contingency table, you assume that the null hypothesis is true.

Accepted Answer

When calculating expected values in a test of a contingency table, the null hypothesis is assumed to be true.

Question 5

In a test of a contingency table, rejecting the null hypothesis concludes the variables are not independent.

Accepted Answer

This is true. Rejecting the null hypothesis in a contingency table test means that there is a significant association between the variables and they are not independent.

Question 6

In a chi-squared test of a contingency table, the value of the test statistic was c ² = 15.652, and the critical value at a = 0.025 was 11.1433. Thus, we must reject the null hypothesis at a = 0.025.

Accepted Answer

Since the test statistic (15.652) is greater than the critical value (11.1433) at the significance level of α = 0.025, we reject the null hypothesis.

Question 7

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

Accepted Answer

For a chi-squared test with 10 degrees of freedom, a test statistic of 17.894 falls between the critical values for p-values of 0.05 and 0.01, indicating that the p-value is less than 0.05 but greater than 0.01.

Question 8

If each element in a population is classified into one and only one of several categories, the population is:

Accepted Answer

The population is described as being classified into multiple categories, which is a characteristic of a multinomial distribution.

Question 9

In a goodness-of-fit test, H ₀ lists specific values for proportions and the test of a contingency table does not.

Accepted Answer

In a goodness-of-fit test, the null hypothesis specifies expected proportions for each category, while in a test of contingency table, the proportions are not pre-specified and must be calculated from the observed data in the table.

Question 10

To determine whether data were drawn from a multinomial distribution with certain proportions, you use a:

Accepted Answer

A chi-squared goodness-of-fit test is specifically used to test if observed data fits a certain distribution, such as a multinomial distribution with certain proportions. The other options listed (B and C) are not applicable in this situation.

Question 11

A chi-squared test of a contingency table 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-squared test of a contingency table are calculated as (number of rows - 1) * (number of columns - 1), so for a 4x4 table, it would be (4-1)*(4-1) = 3*3 = 9.

Question 12

The test statistic for the chi-squared test of a contingency table is the same as the test statistic for the goodness-of-fit test.

Accepted Answer

The test statistic for the chi-squared test of a contingency table and the goodness-of-fit test are both calculated using the chi-squared formula, therefore they are the same. However, the way the formula is applied and the assumptions made for each test are different.

Question 13

If two events A and B are independent, the P ( A and B )= P ( A )+ P ( B ).

Accepted Answer

The correct formula for the probability of two independent events A and B occurring is P(A and B) = P(A) x P(B).

Question 14

To determine the critical values in the chi-squared distribution table, you need to know the:

Accepted Answer

The critical values in the chi-squared distribution table are determined by the degrees of freedom. The degrees of freedom is a parameter that depends on the sample size and the number of variables being analyzed, but it is not the same thing as the sample size itself. The probability of Type II error is not used to determine the critical values in the chi-squared distribution table.

Question 15

A chi-squared goodness-of-fit test is always conducted as a(n):

Accepted Answer

A chi-squared goodness-of-fit test is always conducted as an upper-tail test because the test statistic measures the extent to which the observed frequencies deviate from the expected frequencies, and significant deviations are detected in the upper tail of the chi-squared distribution.

Question 16

A chi-squared test of a contingency table 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-squared test of a contingency table are calculated as (# of rows - 1) x (# of columns - 1). Therefore, in this case, the degrees of freedom would be (3-1) x (4-1) = 6.

Question 17

In the test of a contingency table, the observed cell frequencies must satisfy the rule of 5.

Accepted Answer

The rule of 5 states that each cell in a contingency table must have an expected frequency of at least 5. However, the observed frequencies do not have to satisfy this rule. It is the expected frequencies that are important in determining if a test is appropriate for the data.

Question 18

To produce expected values for a test of a contingency table, you multiply estimated joint probabilities for each cell by the total sample size, n .

Accepted Answer

This statement is true. The expected value for each cell in a contingency table can be calculated by multiplying the estimated joint probability of that cell by the total sample size, n.

Question 19

The expected frequency for the cell in row i and column j is the row i total plus the row j total, all divided by n .

Accepted Answer

The expected frequency for the cell in row i and column j is the product of the row i total and the column j total, all divided by n.

Question 20

In the test of a contingency table, the expected cell frequencies must satisfy the rule of 5.

Accepted Answer

The rule of 5 states that all expected cell frequencies should be greater than or equal to 5. This ensures that the chi-squared test statistic is approximately normally distributed, and improves the accuracy of the test results.

Quiz 15: Chi-Squared Tests

Suppose the value of your chi-squared test statistic in a goodness-of-fit test is equal to 0. What do you conclude?

A chi-squared test of a contingency table with 6 degrees of freedom results in a test statistic of 13.25. Using the chi-squared 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.

The degrees of freedom for the test statistic in a test of a contingency table is ( r - 1 )( c - 1 )where r is the number of rows in the table, and c is the number of columns.

To calculate the expected values in a test of a contingency table, you assume that the null hypothesis is true.

In a test of a contingency table, rejecting the null hypothesis concludes the variables are not independent.

In a chi-squared test of a contingency table, the value of the test statistic was c ² = 15.652, and the critical value at a = 0.025 was 11.1433. Thus, we must reject the null hypothesis at a = 0.025.

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

If each element in a population is classified into one and only one of several categories, the population is:

In a goodness-of-fit test, H ₀ lists specific values for proportions and the test of a contingency table does not.

To determine whether data were drawn from a multinomial distribution with certain proportions, you use a:

A chi-squared test of a contingency table 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.

The test statistic for the chi-squared test of a contingency table is the same as the test statistic for the goodness-of-fit test.

If two events A and B are independent, the P ( A and B )= P ( A )+ P ( B ).

To determine the critical values in the chi-squared distribution table, you need to know the:

A chi-squared goodness-of-fit test is always conducted as a(n) :

A chi-squared test of a contingency table 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.

In the test of a contingency table, the observed cell frequencies must satisfy the rule of 5.

To produce expected values for a test of a contingency table, you multiply estimated joint probabilities for each cell by the total sample size, n .

The expected frequency for the cell in row i and column j is the row i total plus the row j total, all divided by n .

In the test of a contingency table, the expected cell frequencies must satisfy the rule of 5.

Quiz 15: Chi-Squared Tests

Suppose the value of your chi-squared test statistic in a goodness-of-fit test is equal to 0. What do you conclude?

A chi-squared test of a contingency table with 6 degrees of freedom results in a test statistic of 13.25. Using the chi-squared 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.

The degrees of freedom for the test statistic in a test of a contingency table is ( r - 1 )( c - 1 )where r is the number of rows in the table, and c is the number of columns.

To calculate the expected values in a test of a contingency table, you assume that the null hypothesis is true.

In a test of a contingency table, rejecting the null hypothesis concludes the variables are not independent.

In a chi-squared test of a contingency table, the value of the test statistic was c 2 = 15.652, and the critical value at a = 0.025 was 11.1433. Thus, we must reject the null hypothesis at a = 0.025.

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

If each element in a population is classified into one and only one of several categories, the population is:

In a goodness-of-fit test, H 0 lists specific values for proportions and the test of a contingency table does not.

To determine whether data were drawn from a multinomial distribution with certain proportions, you use a:

A chi-squared test of a contingency table 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.

The test statistic for the chi-squared test of a contingency table is the same as the test statistic for the goodness-of-fit test.

If two events A and B are independent, the P ( A and B )= P ( A )+ P ( B ).

To determine the critical values in the chi-squared distribution table, you need to know the:

A chi-squared goodness-of-fit test is always conducted as a(n) :

A chi-squared test of a contingency table 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.

In the test of a contingency table, the observed cell frequencies must satisfy the rule of 5.

To produce expected values for a test of a contingency table, you multiply estimated joint probabilities for each cell by the total sample size, n .

The expected frequency for the cell in row i and column j is the row i total plus the row j total, all divided by n .

In the test of a contingency table, the expected cell frequencies must satisfy the rule of 5.

In a chi-squared test of a contingency table, the value of the test statistic was c ² = 15.652, and the critical value at a = 0.025 was 11.1433. Thus, we must reject the null hypothesis at a = 0.025.

In a goodness-of-fit test, H ₀ lists specific values for proportions and the test of a contingency table does not.