Question 1

The distribution that is always greater than 0 and positively skewed is the distribution for A)Z . B)t . C)χ² . D)U .

Accepted Answer

The chi-square distribution (option C, represented with the symbol χ²) is always greater than 0 and positively skewed. This is because it is based on squared values, making negative values impossible, and it tends to skew to the right, especially with lower degrees of freedom.

Question 2

In a chi-square test,the variables are
A)rank-order.
B)equal-interval.
C)quantitative.
D)nominal.

Accepted Answer

In a chi-square test, the variables are typically categorical or nominal, meaning they have no inherent numerical value or ranking. Therefore, the best choice would be D - nominal.

Question 3

In a chi-square test for independence,the null hypothesis is that
A)the two population variances are independent.
B)the two variables are independent in the population.
C)the means of the populations are not equal.
D)the means of the populations are equal.

Accepted Answer

In a chi-square test for independence, the null hypothesis states that the two variables are independent in the population, meaning that there is no relationship between them. The alternative hypothesis would state that there is a relationship between the two variables. The test is used to determine if there is a significant association between two categorical variables. Therefore, choice B is the correct answer.

Question 4

"(N_Columns-1)(N_Rows-1)" is the formula for the degrees of freedom for A)the chi-square test for goodness of fit. B)the chi-square test for independence. C)the chi-square statistic estimated from a one-way analysis of variance. D)the chi-square statistic estimated from a two-way analysis of variance .

Accepted Answer

The formula "(NS1U1B1ColumnsS1U1B0-1)(NS1U1B1RowsS1U1B0-1)" is used to calculate the degrees of freedom for the chi-square test for independence, where NS1U1B1Columns is the number of columns in the contingency table and NS1U1B1Rows is the number of rows in the contingency table. This formula gives the total number of degrees of freedom for the test.

Question 5

In a chi-square test of independence,the term "expected frequency" generally refers to
A)the distribution of people over categories on the measured variables under the assumption of equal numbers of people in all cells.
B)the distribution of people over categories on one variable expected if the distribution of people over categories on the other variable is completely unrelated to it.
C)the examination of the spread of the grouping variable to see if it is truly an independent variable.
D)the population's distribution of the scores on the measured variable as estimated by the sample data.

Accepted Answer

In a chi-square test of independence, the expected frequency refers to the distribution of individuals over categories on one variable expected if the distribution of individuals over categories on the other variable is completely unrelated to it. It is calculated by taking the total number of individuals in each row/column and dividing by the total sample size, which gives us the expected number of individuals in each cell assuming independence. The expected frequencies are then compared with the observed frequencies to determine whether there is a significant association between the two variables. Option A pertains to a chi-square test of goodness of fit; option C refers to the examination of variance using ANOVA, and option D relates to the estimation of the population parameters from sample data through inferential statistics such as confidence intervals and hypothesis tests.

Question 6

When discussing the pattern in a contingency table,the word "independence" means
A)the distribution of one variable varies over different levels of the other.
B)the variables in the table should not be compared.
C)the observed frequencies are larger than the expected frequencies.
D)there is no relationship between the variables.

Accepted Answer

Independence means that there is no relationship between the variables in the contingency table. Option A describes an interaction, Option B is not a relevant consideration, and Option C describes a situation where the observed frequencies do not match the expected frequencies under a certain distribution assumption (such as a chi-square test). Independence assumes the null hypothesis of no association between the variables, with the alternative hypothesis suggesting some form of dependence or relationship.

Question 7

The degrees of freedom for the chi-square test for goodness of fit is
A)the number of categories minus one.
B)the mean number of individuals per category minus the number of categories.
C)the mean number of individuals per category minus one.
D)the total number of individuals minus the number of categories.

Accepted Answer

The degrees of freedom for the chi-square test for goodness of fit is calculated as the number of categories minus one. This is because one degree of freedom is lost in estimating the total sample size from the observed frequencies across categories.

Question 8

The degrees of freedom for the chi-square test for independence is
A)the number of categories minus one.
B)the number of columns minus one multiplied by the number of rows minus one.
C)the total number of category levels minus one.
D)the number of people minus the number of cells.

Accepted Answer

The degrees of freedom for the chi-square test for independence is calculated by multiplying the number of columns minus one by the number of rows minus one. This is because the test assesses whether two categorical variables are independent of each other, and the degrees of freedom reflect the number of independent elements in the data. The total number of category levels (option C) and the number of people minus the number of cells (option D) are not relevant to determining the degrees of freedom in this context. The number of categories minus one (option A) may apply in some cases, but not in the specific case of the chi-square test for independence.

Question 9

In a 3 × 4 contingency table,there are
A)four levels of each variable.
B)two levels of one variable,and three levels of the other.
C)three levels of one variable,and four levels of the other.
D)four levels of one variable,and five levels of the other.

Accepted Answer

In a contingency table, the rows and columns represent the levels of the variables being studied. In this case, there are three levels of one variable and four levels of the other, so the correct choice is C.

Question 10

When carrying out a chi-square test for independence,a good check on the arithmetic in figuring the expected frequencies is to make sure that
A)the expected frequency of each cell is no larger than the observed frequency.
B)the sum of all the expected frequencies times the degrees of freedom equals the sum of all the observed frequencies.
C)within any one row or column,the sum of the observed frequencies and the sum of the expected frequencies are the same.
D)the sum of all the expected frequencies for each row equals the sum of all the observed frequencies for each column minus 1.

Accepted Answer

In a chi-square test for independence, expected frequencies are calculated based on the assumption of independence between the two variables being compared. To check whether the expected frequencies have been calculated correctly, it is important to ensure that the sum of the observed frequencies and the sum of the expected frequencies for each row and column are equal. This is because the expected frequencies should reflect the same proportions as the observed frequencies, just assuming independence. Option A is not always true, as the expected frequency can be larger than the observed frequency in some cases. Option B is not a relevant check for the arithmetic for expected frequencies. Option D is incorrect, as the sum of expected frequencies for each row does not need to equal the sum of the observed frequencies for each column minus 1.

Question 11

Although similar in some ways,one difference between contingency tables and the tables used in factorial designs is
A)the factorial analysis of variance tables display marginal figures,whereas contingency tables do not.
B)the cells in contingency tables hold frequencies,whereas the cells in factorial analysis of variance tables represent means.
C)contingency tables,but not factorial analysis of variance tables,rely on marginal figures.
D)the numbers in the cells of factorial analysis of variance tables are means for the measured variable,whereas the numbers in the cells of contingency tables represent the means for grouping variables.

Accepted Answer

The cells in contingency tables hold frequencies, which represent the number of cases that fall into each category defined by the variables. In contrast, the cells in factorial analysis of variance (ANOVA) tables represent means of the dependent variable for each combination of levels of the independent variables. This distinction highlights the different purposes of these tables: contingency tables are used for analyzing categorical data, while factorial ANOVA tables are used for analyzing the effects of multiple factors on a continuous outcome.

Question 12

A way to determine the number of people expected to be in any one cell of a chi-square contingency table is to
A)multiply the number of people for the cell's column by its row proportion,figuring its row proportion as the number of people for the cell's row divided by the total number of people.
B)divide the total number of people by the number of people in the cell and multiply this result by a row proportion,figuring its row proportion as the number of people for the cell's row divided by the total number of people.
C)multiply the number of people in the cell's row by the number of people in the cell's column.
D)divide the number of people in the cell's row by the number of people in the cell's column.

Accepted Answer

To determine the expected number of people in any one cell of a chi-square contingency table, we need to calculate the row proportion and column proportion for that cell. We can then multiply the row proportion by the number of people in the column to get the expected number of people for that cell. Therefore, the correct method is to multiply the number of people for the cell's column by its row proportion, figuring its row proportion as the number of people for the cell's row divided by the total number of people. This method gives us an accurate estimate of the expected number of people in any one cell of the contingency table.

Question 13

The main difference between the test for goodness of fit and the test for independence is that
A)the goodness of fit test can be used only for 2 × 2 designs whereas the independence test is not limited to a particular design type.
B)expected frequencies are calculated for the independence test but not for the goodness of fit test.
C)the goodness of fit test is limited to one nominal variable whereas the independence test is not.
D)observed frequencies are compared to expected frequencies for the independence test but not for the goodness of fit test.

Accepted Answer

The goodness of fit test is limited to testing the fit of observed frequencies to an expected distribution for one nominal variable. The test for independence, on the other hand, can be used to assess the relationship between two nominal variables, where expected frequencies are calculated and compared to observed frequencies. Option A and D are incorrect as they do not accurately describe the differences between the two tests. Option B is incorrect as expected frequencies are calculated for both tests.

Question 14

Which calculation is necessary for figuring the chi-square statistic?
A)dividing the absolute differences between observed and expected frequencies by the expected frequency in each cell
B)dividing the absolute differences between observed and expected frequencies by the observed frequency in each cell
C)dividing the squared differences between observed and expected frequencies by the expected frequency in each cell
D)dividing the squared differences between observed and expected frequencies by the observed frequency in each cell

Accepted Answer

The chi-square statistic is calculated by dividing the squared differences between observed and expected frequencies by the expected frequency in each cell. This is necessary to determine if there is a significant difference between the observed and expected frequencies in a particular dataset.

Question 15

The main idea of a chi-square test is to
A)test the estimated degree of fit (proportion of variance accounted for)of one variable to the other variable.
B)compare the estimated population means,to see if they vary from each other by more than chance.
C)compare estimated population variances,to see if they vary from each other by more than chance.
D)test how well the pattern of observed frequencies fits some expected pattern of frequencies.

Accepted Answer

The main idea of a chi-square test is to test how well the pattern of observed frequencies fits some expected pattern of frequencies. It is commonly used to test the association between two categorical variables.

Question 16

The formula for the chi-squared statistic is A)Σ[(O-E)²/E] B)[Σ(O-E)²]/E C)O/[Σ(O-E)²] D)E/[Σ(O-E)²]

Accepted Answer

The chi-squared statistic is calculated by summing the squared differences between the observed and expected frequencies, divided by the expected frequencies. The formula for the chi-squared statistic is:χ² = Σ[(O-E)²/E]where:O is the observed frequencyE is the expected frequency

Question 17

The inventor of the chi-square test was
A)Karl Pearson.
B)W.S.Gossett.
C)Ronald Fisher.
D)Francis Galton.

Accepted Answer

Karl Pearson is credited with the development of the chi-square test, a statistical method used for testing hypotheses on categorical data.

Question 18

The values in a chi-square distribution are always greater than 0 and
A)are normally distributed.
B)less than 1.
C)can be quite large.
D)are negatively skewed.

Accepted Answer

The values in a chi-square distribution can be quite large and always greater than 0. They are not normally distributed, less than 1, or negatively skewed.

Question 19

A contingency table is a table in which A)the distributions of two nominal variables are laid out so that you have the frequencies of their combinations and totals can be seen. B)chi-squares for each category are displayed over each level of the predictor variable. C)χ² distributions are translated into t distributions. D)χ² distributions are translated into F distributions.

Accepted Answer

A contingency table is a table in which the distributions of two nominal variables are laid out so that you have the frequencies of their combinations and totals can be seen. Option B, C, and D are not correct as they do not define or relate to contingency tables.

Question 20

In a chi-square test for goodness of fit,the null hypothesis is that
A)the number of people in one category is no greater than the number of people in the other.
B)the variances of the populations of categories are the same.
C)the proportion of people over categories breaks down the same for the two populations.
D)the means of the two populations of categories are the same.

Accepted Answer

In a chi-square test for goodness of fit, the null hypothesis is that the proportion of people over categories breaks down the same for the two populations. The test determines if there is a significant difference between the observed and expected frequencies in each category.

Quiz 13: Chi-Square Tests

The distribution that is always greater than 0 and positively skewed is the distribution for

In a chi-square test,the variables are

In a chi-square test for independence,the null hypothesis is that

"(N_Columns-1) (N_Rows-1) " is the formula for the degrees of freedom for

In a chi-square test of independence,the term "expected frequency" generally refers to

When discussing the pattern in a contingency table,the word "independence" means

The degrees of freedom for the chi-square test for goodness of fit is

The degrees of freedom for the chi-square test for independence is

In a 3 × 4 contingency table,there are

When carrying out a chi-square test for independence,a good check on the arithmetic in figuring the expected frequencies is to make sure that

Although similar in some ways,one difference between contingency tables and the tables used in factorial designs is

A way to determine the number of people expected to be in any one cell of a chi-square contingency table is to

The main difference between the test for goodness of fit and the test for independence is that

Which calculation is necessary for figuring the chi-square statistic?

The main idea of a chi-square test is to

The formula for the chi-squared statistic is

The inventor of the chi-square test was

The values in a chi-square distribution are always greater than 0 and

A contingency table is a table in which

In a chi-square test for goodness of fit,the null hypothesis is that