Question 1

The number of degrees of freedom in a chi-square goodness-of-fit test is the number of categories minus the number of parameters estimated minus one.

Accepted Answer

This statement is true. The formula for calculating degrees of freedom in a chi-square goodness-of-fit test is df = (k - 1) - p, where k is the number of categories and p is the number of parameters estimated. Subtracting 1 adjusts for the constraint that the sum of the observed frequencies must equal the sum of the expected frequencies.

Question 2

The null hypothesis in a chi-square goodness-of-fit test is that the observed distribution is the same as the expected distribution.

Accepted Answer

The null hypothesis in a chi-square goodness-of-fit test is that the observed distribution is the same as the expected distribution.

Question 3

When using the chi-square goodness-of-fit test,the type of distribution (uniform,Poisson,normal)being used does not influence the hypothesis test.

Accepted Answer

The type of distribution being tested (uniform, Poisson, normal, etc.) directly influences the expected frequencies calculated for the chi-square goodness-of-fit test, which are crucial for determining how well the observed data fit the expected distribution under the null hypothesis.

Question 4

The decision rule in a chi-square goodness-of-fit test is to reject the null hypothesis if the computed chi-square is greater than the table chi-square.

Accepted Answer

This is correct. The decision rule in a chi-square goodness-of-fit test is to reject the null hypothesis if the computed chi-square is greater than the table chi-square, with degrees of freedom equal to the number of categories minus one. If the computed chi-square is less than or equal to the table chi-square, we fail to reject the null hypothesis.

Question 5

A goodness of fit test is to be performed to see if consumers prefer any of three package designs (A,B,and C)more than the other two. A sample of 60 consumers is used. What is the expected frequency for category A?
A) 1/3
B) 20
C) 60
D) 10
E) 30

Accepted Answer

Since there are three package designs and no preference is assumed, the expected frequency for each category is the total sample size divided by the number of categories. Therefore, for 60 consumers divided among 3 categories (A, B, and C), the expected frequency for each is 60/3 = 20.

Question 6

A variable contains five categories. It is expected that data are uniformly distributed across these five categories.To test this,a sample of observed data is gathered on this variable resulting in frequencies of 27,30,29,21,and 24. Using $\alpha$ = .01,the critical value of chi-square is _______.
A) 7.78
B) 15.09
C) 9.24
D) 13.28
E) 15.48

Accepted Answer

The answer of A variable contains five categories. It is...

Question 7

A chi-square goodness-of-fit test to determine if the observed frequencies in ten categories are Poisson distributed has nine degrees of freedom.

Accepted Answer

The answer of A chi-square goodness-of-fit test to determine if...

Question 8

A goodness of fit test is to be performed to see if Web Surfers prefer any of four Web sites (A,B,C and D)more than the other three. A sample of 60 consumers is used. What is the expected frequency for Web site A?
A) 1/4
B) 20
C) 15
D) 10
E) 30

Accepted Answer

Since there are four web sites and each one has an equal chance of being preferred, the expected frequency for each web site would be 1/4 of the total sample size. Thus, the expected frequency for Web site A would be (1/4)*60 = 15. Therefore, the best choice is C.

Question 9

A chi-square goodness-of-fit test is being used to test the goodness-of-fit of a uniform distribution for a dataset with "k" categories.This test has (k-3)degrees of freedom.

Accepted Answer

A chi-square goodness-of-fit test for a uniform distribution with "k" categories has (k-1) degrees of freedom, not (k-3).

Question 10

In a chi-square goodness-of-fit test,theoretical frequencies are also called expected frequencies.

Accepted Answer

Theoretical frequencies are the frequencies that are expected to occur based on the null hypothesis, so they are also referred to as expected frequencies.

Question 11

A variable contains five categories. It is expected that data are uniformly distributed across these five categories.To test this,a sample of observed data is gathered on this variable resulting in frequencies of 27,30,29,21,and 24. Using $\alpha$ = .01,the degrees of freedom for this test are _______.
A) 5
B) 4
C) 3
D) 2
E) 1

Accepted Answer

To test whether the observed frequencies are uniform across the categories, we will use a chi-square goodness-of-fit test.

The null hypothesis for this test is that the observed frequencies are uniformly distributed across the categories, and the alternative hypothesis is that they are not.

The test statistic for this test is:

χ2 = Σ [(O - E)2 / E]

where O is the observed frequency in each category, E is the expected frequency in each category (in this case, E = (27+30+29+21+24)/5 = 26.2), and the summation is over all categories.

The degrees of freedom for this test are (number of categories - 1). In this case, there are 5 categories, so the degrees of freedom are 4.

To determine whether to reject the null hypothesis, we compare the calculated value of χ2 to the critical value from the chi-square distribution with 4 degrees of freedom at the .01 level of significance.

Since the degrees of freedom are 4, the critical value is 13.28.

Calculating χ2 using the observed and expected frequencies, we get:

χ2 = [(27-26.2)2 / 26.2] + [(30-26.2)2 / 26.2] + [(29-26.2)2 / 26.2] + [(21-26.2)2 / 26.2] + [(24-26.2)2 / 26.2]
   = 0.31 + 1.75 + 0.46 + 1.70 + 0.34
   = 4.56

Since χ2 (4.56) is less than the critical value (13.28), we fail to reject the null hypothesis. Therefore, we conclude that the observed frequencies are not significantly different from a uniform distribution.

Question 12

In a chi-square goodness-of-fit test,actual frequencies are also called calculated frequencies.

Accepted Answer

Actual frequencies are the observed frequencies, while calculated frequencies are the expected frequencies under a specified hypothesis.

Question 13

A variable contains five categories. It is expected that data are uniformly distributed across these five categories.To test this,a sample of observed data is gathered on this variable resulting in frequencies of 27,30,29,21,and 24. Using $\alpha$ = .01,the observed value of chi-square is _______.
A) 12.09
B) 9.82
C) 13.28
D) 17.81
E) 2.09

Accepted Answer

The answer of A variable contains five categories. It is...

Question 14

A chi-square goodness-of-fit test to determine if the observed frequencies in seven categories are uniformly distributed has six degrees of freedom.

Accepted Answer

The formula for degrees of freedom in a chi-square goodness-of-fit test of uniform distribution is (number of categories - 1). In this case, there are seven categories, so the degrees of freedom would be 7-1=6. Therefore, the answer is A.

Question 15

The null hypothesis for a chi-square test of independence is that the two variables are not related.

Accepted Answer

The null hypothesis for a chi-square test of independence is that there is no relationship between the two variables being analyzed. This means that the two variables are not related to each other.

Question 16

A researcher is interested in using a chi-square test of independence to determine if age is independent of minutes spent reading.Age is divided into four categories while minutes spent reading is classified as high,medium,low.The number of degrees of freedom for this test is 12.

Accepted Answer

The number of degrees of freedom for a chi-square test of independence with r rows and c columns is given by (r-1)(c-1). In this case, there are 4 age categories and 3 levels of minutes spent reading, so there are actually 6 cells in the contingency table, and the degrees of freedom would be (4-1)(3-1) = 6. Therefore, the number of degrees of freedom stated in the question is incorrect.

Question 17

In a chi-square test of independence the contingency table has 4 rows and 3 columns.The number of degrees of freedom for this test is 7.

Accepted Answer

The number of degrees of freedom for a chi-square test of independence with a contingency table of 4 rows and 3 columns would be (4-1) * (3-1) = 6, not 7.

Question 18

In a chi-square test of independence the contingency table has 4 rows and 3 columns.The number of degrees of freedom for this test is 6.

Accepted Answer

The degrees of freedom for a chi-square test of independence are calculated as (number of rows - 1) * (number of columns - 1), which in this case is (4 - 1) * (3 - 1) = 3 * 2 = 6.

Question 19

The number of degrees of freedom in a chi-square goodness-of-fit test is the number of categories minus the number of parameters estimated.

Accepted Answer

The answer of The number of degrees of freedom in...

Question 20

A two-way table used for a test of independence is sometimes called a contingency table.

Accepted Answer

This is true. A contingency table is used to display the frequency distribution of two categorical variables and is often used to test for independence between the variables.

Quiz 16: Analysis of Categorical Data

The number of degrees of freedom in a chi-square goodness-of-fit test is the number of categories minus the number of parameters estimated minus one.

The null hypothesis in a chi-square goodness-of-fit test is that the observed distribution is the same as the expected distribution.

When using the chi-square goodness-of-fit test,the type of distribution (uniform,Poisson,normal)being used does not influence the hypothesis test.

The decision rule in a chi-square goodness-of-fit test is to reject the null hypothesis if the computed chi-square is greater than the table chi-square.

A goodness of fit test is to be performed to see if consumers prefer any of three package designs (A,B,and C) more than the other two. A sample of 60 consumers is used. What is the expected frequency for category A?

A variable contains five categories. It is expected that data are uniformly distributed across these five categories.To test this,a sample of observed data is gathered on this variable resulting in frequencies of 27,30,29,21,and 24. Using $\alpha$ = .01,the critical value of chi-square is _______.

A chi-square goodness-of-fit test to determine if the observed frequencies in ten categories are Poisson distributed has nine degrees of freedom.

A goodness of fit test is to be performed to see if Web Surfers prefer any of four Web sites (A,B,C and D) more than the other three. A sample of 60 consumers is used. What is the expected frequency for Web site A?

A chi-square goodness-of-fit test is being used to test the goodness-of-fit of a uniform distribution for a dataset with "k" categories.This test has (k-3)degrees of freedom.

In a chi-square goodness-of-fit test,theoretical frequencies are also called expected frequencies.

In a chi-square goodness-of-fit test,actual frequencies are also called calculated frequencies.

A variable contains five categories. It is expected that data are uniformly distributed across these five categories.To test this,a sample of observed data is gathered on this variable resulting in frequencies of 27,30,29,21,and 24. Using $\alpha$ = .01,the observed value of chi-square is _______.

A chi-square goodness-of-fit test to determine if the observed frequencies in seven categories are uniformly distributed has six degrees of freedom.

The null hypothesis for a chi-square test of independence is that the two variables are not related.

A researcher is interested in using a chi-square test of independence to determine if age is independent of minutes spent reading.Age is divided into four categories while minutes spent reading is classified as high,medium,low.The number of degrees of freedom for this test is 12.

In a chi-square test of independence the contingency table has 4 rows and 3 columns.The number of degrees of freedom for this test is 7.

In a chi-square test of independence the contingency table has 4 rows and 3 columns.The number of degrees of freedom for this test is 6.

The number of degrees of freedom in a chi-square goodness-of-fit test is the number of categories minus the number of parameters estimated.

A two-way table used for a test of independence is sometimes called a contingency table.