Question 1

Student Absenteeism 
Consider a multinomial experiment involving n = 200 students of a large high school.The attendance department recorded the number of students who were absent during the weekdays.The null hypothesis to be tested is: HS1U1B10S1U1B0: pS1U1B11S1U1B0 = .10,pS1U1B12S1U1B0 = .25,pS1U1B13S1U1B0 = .30,pS1U1B14S1U1B0 = .20,pS1U1B15S1U1B0 = .15.
-{Student Absenteeism Narrative} Test the hypothesis at the 5% level of significance with the following frequencies: (n = 100)

Accepted Answer

The answer of Student Absenteeism 
Consider a multinomial experiment involving...

Question 2

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

Accepted Answer

Rejecting the null hypothesis in a test of a contingency table indicates that there is a significant association between two variables, indicating that they are not independent. Therefore, the statement is true.

Question 3

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 values in a contingency table are calculated by multiplying the estimated joint probabilities for each cell by the total sample size, n.

Question 4

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

Accepted Answer

In a chi-square test for a contingency table, each expected cell frequency should be at least 5 to ensure the validity of the test results.

Question 5

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

Accepted Answer

To calculate expected values in a test of a contingency table, you assume that the null hypothesis is true. This is done by using the marginal totals of the table to calculate the expected frequencies under the assumption that the two variables in the table are independent.

Question 6

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 .025 but smaller than .05.

Accepted Answer

When using the chi-squared table with 6 degrees of freedom, the critical value for a 95% confidence level is 12.59 and for a 97.5% confidence level is 14.45. Therefore, the test statistic of 13.25 falls between them and corresponds to a p-value greater than 0.025 but smaller than 0.05, making option A true.

Question 7

Sales Volumes 
A telemarketer makes five calls per day.A sample of 200 days gives the frequencies of sales volumes listed below:  Assume the population is binomial distribution with a probability of purchase p equal to .50.
-{Sales Volumes Narrative} Should the assumption of a binomial distribution be rejected at the 5% significance level?

Accepted Answer

The answer of Sales Volumes 
A telemarketer makes five calls...

Question 8

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 row i total times the column j total, all divided by n.

Question 9

In a goodness-of-fit test,H₀ lists specific values for proportions and the test of a contingency table does not.

Accepted Answer

The statement is true. Goodness-of-fit test involves comparing observed frequencies to expected frequencies based on a specific distribution or set of proportions. H₀ lists specific values for proportions, while a test of a contingency table examines the association between two categorical variables and does not involve testing specific values for proportions.

Question 10

Sales Volumes 
A telemarketer makes five calls per day.A sample of 200 days gives the frequencies of sales volumes listed below:  Assume the population is binomial distribution with a probability of purchase p equal to .50.
-{Sales Volumes Narrative} Compute the expected frequencies for x = 0,1,2,3,4,and 5 by using the binomial probability function or the binomial tables.Combine categories if necessary to satisfy the rule of five.

Accepted Answer

The answer of Sales Volumes 
A telemarketer makes five calls...

Question 11

Student Absenteeism 
Consider a multinomial experiment involving n = 200 students of a large high school.The attendance department recorded the number of students who were absent during the weekdays.The null hypothesis to be tested is: HS1U1B10S1U1B0: pS1U1B11S1U1B0 = .10,pS1U1B12S1U1B0 = .25,pS1U1B13S1U1B0 = .30,pS1U1B14S1U1B0 = .20,pS1U1B15S1U1B0 = .15.
-{Student Absenteeism Narrative} Review the previous results.What is the effect of decreasing the sample size?

Accepted Answer

The answer of Student Absenteeism 
Consider a multinomial experiment involving...

Question 12

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 .05 < p-value < .10.

Accepted Answer

Looking at the chi-squared table with 10 degrees of freedom, we find that the critical value for a significance level of 0.05 is 18.31 and for a significance level of 0.10 is 15.99. Since the test statistic of 17.894 falls between these two critical values, we can conclude that the p-value is greater than 0.05 but less than 0.10. Therefore, the statement "0.05 < p-value < 0.10" is the most accurate statement that can be made about the p-value for this test.

Question 13

Student Absenteeism 
Consider a multinomial experiment involving n = 200 students of a large high school.The attendance department recorded the number of students who were absent during the weekdays.The null hypothesis to be tested is: HS1U1B10S1U1B0: pS1U1B11S1U1B0 = .10,pS1U1B12S1U1B0 = .25,pS1U1B13S1U1B0 = .30,pS1U1B14S1U1B0 = .20,pS1U1B15S1U1B0 = .15.
-{Student Absenteeism Narrative} Test the hypothesis at the 5% level of significance with the following frequencies: (n = 50)

Accepted Answer

The answer of Student Absenteeism 
Consider a multinomial experiment involving...

Question 14

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 15

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 in a contingency table, every expected cell frequency must be at least 5 in order to use the chi-square test for independence. There is no requirement on the observed cell frequencies. However, if any of the expected cell frequencies are less than 5, then the chi-square test may not be valid and other methods of analysis should be considered.

Question 16

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

Accepted Answer

The correct formula for the intersection of two independent events A and B is P(A and B) = P(A) x P(B). Therefore, the statement given in the question is false.

Question 17

Student Absenteeism 
Consider a multinomial experiment involving n = 200 students of a large high school.The attendance department recorded the number of students who were absent during the weekdays.The null hypothesis to be tested is: HS1U1B10S1U1B0: pS1U1B11S1U1B0 = .10,pS1U1B12S1U1B0 = .25,pS1U1B13S1U1B0 = .30,pS1U1B14S1U1B0 = .20,pS1U1B15S1U1B0 = .15.
-{Student Absenteeism Narrative} Test the hypothesis at the 5% level of significance with the following frequencies:

Accepted Answer

The answer of Student Absenteeism 
Consider a multinomial experiment involving...

Question 18

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 by (number of rows - 1) x (number of columns - 1). Therefore, in this case, the degrees of freedom would be (3-1) x (4-1) = 6, not 12.

Question 19

In a chi-squared test of a contingency table,the value of the test statistic was χ² = 15.652,and the critical value at α = .025 was 11.1433.Thus,we must reject the null hypothesis at α = .025.

Accepted Answer

The given test statistic value is greater than the critical value, which means that we reject the null hypothesis. Therefore, the answer is A - True.

Question 20

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

Both the chi-squared test for a contingency table and the goodness-of-fit test use the same formula for calculating the test statistic, which involves summing the squared differences between observed and expected frequencies, divided by the expected frequencies.

Quiz 15: Chi-Squared Tests

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

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.

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

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

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 .025 but smaller than .05.

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 a goodness-of-fit test,H0 lists specific values for proportions and the test of a contingency table does not.

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 .05 < p-value < .10.

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.

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

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

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 a chi-squared test of a contingency table,the value of the test statistic was χ2 = 15.652,and the critical value at α = .025 was 11.1433.Thus,we must reject the null hypothesis at α = .025.

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.

In a goodness-of-fit test,H₀ lists specific values for proportions and the test of a contingency table does not.

In a chi-squared test of a contingency table,the value of the test statistic was χ² = 15.652,and the critical value at α = .025 was 11.1433.Thus,we must reject the null hypothesis at α = .025.