Question 1

A Chi square test has been conducted to assess the relationship between marital status and church attendance.The obtained Chi square is 23.45 and the critical Chi square is 9.488.What may be concluded?
A)reject the null hypothesis,church attendance and marital status are dependent
B)reject the null hypothesis,church attendance and marital status are independent
C)fail to reject the null hypothesis,church attendance and marital status are dependent
D)fail to reject the null hypothesis,church attendance and marital status are independent

Accepted Answer

The obtained Chi square value (23.45) is greater than the critical Chi square value (9.488), which indicates that there is a significant relationship between marital status and church attendance. Therefore, we reject the null hypothesis and conclude that church attendance and marital status are dependent.

Question 2

In the Chi square test,degrees of freedom are A)N - 1. B)N₁ + N₂ - 2. C)(r + 1)(c + 1). D)(r - 1)(c - 1).

Accepted Answer

The degrees of freedom (df) in a chi-square test is calculated by (r-1)(c-1), where r is the number of rows and c is the number of columns in the contingency table. Therefore, choice D is the correct answer. Choice A is incorrect because it assumes that the number of degrees of freedom is simply equal to the sample size minus one. Choices B and C are incorrect because they do not accurately account for the number of rows and columns in the contingency table.

Question 3

A researcher reports that her obtained chi square of -17.56 is significant because it exceeds the critical Chi square of -3.841 for an alpha of .05 with 2 degrees of freedom.What mistake has been made?
A)neither critical nor obtained Chi squares can have negative values
B)it is impossible to have only 2 degrees of freedom
C)it is impossible for obtained Chi square to exceed 10
D)a one-tailed test should have been used

Accepted Answer

Both critical and obtained Chi squares cannot have negative values. A negative value for the obtained Chi square suggests an error in calculation.

Question 4

Some potential difficulties arise in the chi square test when
A)sample size is small.
B)sample size is very large.
C)many cells have expected frequencies of five or less.
D)all of the above

Accepted Answer

A) When the sample size is small, the assumptions of the test may not be met and the test may not provide accurate results due to low power. 
B) When the sample size is very large, even small deviations from expected frequencies may lead to significant results, making the test overly sensitive. 
C) When many cells have expected frequencies of five or less, the assumptions of the chi-square test may not be met, and the test may not provide accurate results. Moreover, the results may be influenced by the choice of the grouping variable, such as combining cells or collapsing categories, leading to potential bias.

Therefore, all of the above are potential difficulties in the chi-square test, and the best choice is D.

Question 5

If Chi square obtained has a negative value,
A)an arithmetic error has been made.
B)a one-tailed test is called for.
C)the row and column variables must be switched.
D)the null hypothesis must be rejecteD.

Accepted Answer

The Chi square test always gives non-negative values. If a negative value is obtained, it indicates an arithmetic error in the calculation.

Question 6

For Chi square,a small sample is one in which
A)many expected frequencies are less than 5.
B)many observed frequencies are less than 5.
C)N is less than 100.
D)N is less than 50.

Accepted Answer

A small sample is one in which many expected frequencies are less than 5. This is because the validity of the Chi square test depends on the expected frequencies being greater than or equal to 5. When many expected frequencies are less than 5, the Chi square test may not be reliable and alternative tests may need to be used.

Question 7

The chi square goodness-of-fit test can be used when
A)degrees of freedom are less than one.
B)column percentages are equal.
C)sample size is very large.
D)the distribution of a single variable must be tested for significance.

Accepted Answer

The chi square goodness-of-fit test is used to test the significance of the distribution of a single variable. It is not used when degrees of freedom are less than one, when column percentages are equal, or when the sample size is very large.

Question 8

Chi square has been used to test the relationship between gender and support for abortion.The null hypothesis has been rejected.What may be concluded?
A)support for abortion depends on gender
B)gender and support for abortion are not related
C)females are more supportive of abortion
D)males are more supportive of abortion

Accepted Answer

The rejection of the null hypothesis indicates that there is a significant relationship between gender and support for abortion. Therefore, we can conclude that support for abortion depends on gender.

Question 9

When cell frequencies are small and 2x2 tables are used in the chi square test,an adjustment to the value of chi square can be made by applying
A)Yate's correction for continuity.
B)a different value to the expected frequencies.
C)a different value to the observed frequencies.
D)any of the above

Accepted Answer

Yates' correction for continuity should be applied when cell frequencies are small and 2x2 tables are used in the chi-square test. It adjusts the value of chi-square to account for the fact that the chi-square distribution is continuous, while the data is discrete. Using a different value for the expected or observed frequencies would not be appropriate in this situation.

Question 10

In a 2x2 table,if one cell frequency is known,
A)all other cell frequencies are determined.
B)two other cell frequencies are determined.
C)one other cell frequency is determined.
D)none of the other cell frequencies is determineD.

Accepted Answer

In a 2x2 table, if one cell frequency is known, the row and column totals can be determined. This, in turn, determines the other three cell frequencies.

Question 11

In a research study conducted to determine if arrests were related to the socioeconomic class of the offender,the chi square critical score was 9.488 and the chi square test statistic was 12.2.We can conclude that
A)the variables are independent.
B)being in a certain socioeconomic class causes arrests.
C)the variables are dependent.
D)the probability of getting these results by random chance alone is 0.5.

Accepted Answer

The chi square test statistic of 12.2 is greater than the critical score of 9.488, indicating that the variables (arrests and socioeconomic class) are dependent. This means that there is a relationship between the two variables, and being in a certain socioeconomic class may increase the likelihood of getting arrested. Option A is incorrect because if the variables were independent, the chi square test statistic would be lower than the critical score. Option B is a possible explanation, but we cannot conclude that it is the only explanation since correlation does not always imply causation. Option D is incorrect because the probability of getting these results by random chance alone is extremely low, given the values of the chi square test statistic and critical score.

Question 12

Yate's correction for small sample size A)reduces the value of the difference between f_o and f_e by .5 before squaring. B)increases the value of the difference between f_o and f_e by .5 before squaring. C)reduces the degrees of freedom by one half. D)increases the degrees of freedom by one half.

Accepted Answer

Yates correction for small sample size is used to adjust for the potential bias in chi-square tests when the sample size is small. It involves reducing the value of the difference between observed and expected frequencies in each cell by 0.5 before squaring and summing them up. This correction accounts for the fact that the chi-square distribution is discrete and tends to overestimate the significance of the test statistic with small sample sizes.

Question 13

For the chi square goodness-of-fit test,the null hypothesis would be that
A)the variables are independent.
B)the variables are dependent.
C)the distribution of the variable is determined by random chance.
D)the distribution of the variable is not determined by random chance.

Accepted Answer

The null hypothesis for the chi square goodness-of-fit test is that the distribution of the variable is determined by random chance. This means that there is no significant difference between the observed and expected values of the variable, and any deviations can be attributed to chance.

Question 14

How many degrees of freedom does a 3x3 bivariate table have?
A)8
B)6
C)4
D)3

Accepted Answer

A 3x3 bivariate table has 9 cells, and it is a table of counts, so the sum of all cells must equal the total sample size. Given this, the marginal totals for each variable fix their respective totals. The third marginal total, which is determined by adding up the counts in the third column, also fixes the third row total, which is determined by adding up the counts in the third row. Thus, there are 2 degrees of freedom for each marginal total, meaning that there are a total of 4 degrees of freedom in a 3x3 bivariate table.

Question 15

An obtained chi square of 10.78 has been calculated.Critical Chi square is 3.841.What should be concluded?
A)nothing,information is incomplete
B)reject the null hypothesis,the variables are not independent
C)reject the null hypothesis,the variables are independent
D)fail to reject the null hypothesis

Accepted Answer

The obtained chi square (10.78) is greater than the critical chi square (3.841), indicating that there is a significant relationship between the variables being studied. Therefore, we reject the null hypothesis and conclude that the variables are not independent.

Question 16

The sampling distribution for chi squares is
A)shaped like any other normal curve.
B)not normally distributed.
C)bimodal.
D)all of the above

Accepted Answer

The sampling distribution for chi squares is not normally distributed but instead skewed to the right.

Question 17

Unlike the chi square test of independence,in the chi square goodness of fit test,the expected frequencies are
A)not used.
B)computed.
C)determined from the null hypothesis.
D)the same as the observed frequencies.

Accepted Answer

In the chi square goodness of fit test, the expected frequencies are determined from the null hypothesis. The null hypothesis specifies the expected frequencies, which are then compared to the observed frequencies to determine if there is a significant difference.

Question 18

The value of the chi square test statistic is always
A)zero or a positive number.
B)greater than the number of cells in the bivariate table.
C)equal to or greater than the critical value.
D)greater than the degrees of freedom.

Accepted Answer

The chi square test statistic is calculated by summing the differences between observed and expected frequencies in each cell of a contingency table, squared and then dividing by the expected frequency. This results in a non-negative value, which means that the test statistic is always zero or a positive number.

Question 19

In the Chi square test for independence,the research hypothesis is
A)always one-tailed.
B)the variables are dependent.
C)the variables are independent.
D)expected frequencies are greater than observed frequencies.

Accepted Answer

The research hypothesis in the chi-square test for independence is that the two variables being tested are dependent on each other. Therefore, the best choice is B. The other options are not related to the research hypothesis in this test.

Question 20

In the Chi square test for independence,the null hypothesis and the research hypothesis
A)always contradict each other.
B)always agree with each other.
C)are never actually stated.
D)are usually both rejecteD.

Accepted Answer

The null hypothesis and research hypothesis in a Chi square test for independence are always contradicting each other. The null hypothesis assumes that there is no association between two variables being tested, while the research hypothesis suggests that there is a significant association between them. Therefore, they cannot agree with each other.

Quiz 11: Hypothesis Testing Iv: Chi Square

A Chi square test has been conducted to assess the relationship between marital status and church attendance.The obtained Chi square is 23.45 and the critical Chi square is 9.488.What may be concluded?

In the Chi square test,degrees of freedom are

A researcher reports that her obtained chi square of -17.56 is significant because it exceeds the critical Chi square of -3.841 for an alpha of .05 with 2 degrees of freedom.What mistake has been made?

Some potential difficulties arise in the chi square test when

If Chi square obtained has a negative value,

For Chi square,a small sample is one in which

The chi square goodness-of-fit test can be used when

Chi square has been used to test the relationship between gender and support for abortion.The null hypothesis has been rejected.What may be concluded?

When cell frequencies are small and 2x2 tables are used in the chi square test,an adjustment to the value of chi square can be made by applying

In a 2x2 table,if one cell frequency is known,

In a research study conducted to determine if arrests were related to the socioeconomic class of the offender,the chi square critical score was 9.488 and the chi square test statistic was 12.2.We can conclude that

Yate's correction for small sample size

For the chi square goodness-of-fit test,the null hypothesis would be that

How many degrees of freedom does a 3x3 bivariate table have?

An obtained chi square of 10.78 has been calculated.Critical Chi square is 3.841.What should be concluded?

The sampling distribution for chi squares is

Unlike the chi square test of independence,in the chi square goodness of fit test,the expected frequencies are

The value of the chi square test statistic is always

In the Chi square test for independence,the research hypothesis is

In the Chi square test for independence,the null hypothesis and the research hypothesis