Question 1

The chi-squared goodness-of-fit test is usually used as a test of multinomial parameters, but it can also be used to determine whether data were drawn from any distribution.

Accepted Answer

The chi-squared goodness-of-fit test can be used to test whether data were drawn from any distribution, not just multinomial parameters.

Question 2

A chi-squared goodness-of-fit test can be conducted either as a two-tail test or as a one-tail test.

Accepted Answer

A chi-squared goodness-of-fit test can only be conducted as a one-tail test, as it measures the degree of deviation from the expected distribution in one direction only. The test involves comparing the observed frequencies with the expected frequencies under a hypothesized distribution, and calculating the chi-square statistic. The p-value, which indicates the probability of observing the degree of deviation or greater under the null hypothesis, is then obtained from the chi-square distribution table with the appropriate degrees of freedom. As the p-value is calculated from one tail only, the test is considered a one-tail test.

Question 3

A multinomial experiment, where the outcome of each trial can be classified into one of two categories, is identical to the binomial experiment.

Accepted Answer

This statement is false. In a multinomial experiment, the outcome of each trial can be classified into multiple categories, whereas in a binomial experiment, the outcome can only be classified into two categories. Therefore, the two experiments are not identical.

Question 4

For a chi-squared distributed random variable with 12 degrees of freedom and a level of significance of 0.05, the chi-squared value from the table is 21.0261. The computed value of the test statistics is 25.1687. This will lead us to reject the null hypothesis.

Accepted Answer

The computed value of the test statistic (25.1687) is greater than the chi-squared value from the table (21.0261), which means that we can reject the null hypothesis at a significance level of 0.05. Therefore, the answer is A, true.

Question 5

In chi-squared tests, the conventional and conservative rule - known as the rule of five - is to require that difference between the observed and expected frequency for each cell be at least 5.

Accepted Answer

The rule of five in chi-squared tests actually states that no more than 20% of the cells should have expected frequencies less than 5, and all individual expected frequencies should be 1 or greater.

Question 6

Whenever the expected frequency of a cell is less than 5, one possible remedy for this condition is to combine it with one or more other cells.

Accepted Answer

This is true according to the chi-squared test assumption that the expected frequency of each cell should be at least 5. When this assumption is violated, combining the cell with one or more adjacent cells can help to increase the expected frequency and improve the validity of the analysis.

Question 7

A chi-squared test for independence 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 < 0.10.

Accepted Answer

According to the chi-squared table with 10 degrees of freedom, the critical value for a 0.05 level of significance is 18.31 and for a 0.10 level of significance is 14.68. Since the test statistic of 17.894 is between these two critical values, the p-value for the test must be between 0.05 and 0.10. Therefore, the most accurate statement is that 0.05 < p-value < 0.10.

Question 8

For a chi-squared distributed random variable with 10 degrees of freedom and a level of significance of 0.025, the chi-squared value from the table is 20.5. The computed value of the test statistic is 16.857. The decision is to retain Ho.

Accepted Answer

The computed value of the test statistic (16.857) is less than the critical value from the table (20.5) at a level of significance of 0.025. Since the computed value does not fall in the rejection region, the decision is to retain Ho.

Question 9

The area to the right of a chi-squared value is 0.01. For 8 degrees of freedom, the table value is 20.0902.

Accepted Answer

For a chi-squared distribution with 8 degrees of freedom, a table or calculator will show that the critical value corresponding to an area (or p-value) of 0.01 to the right (in the upper tail) is indeed 20.0902. This means that if the chi-squared statistic is greater than 20.0902, the result is in the top 1% of the distribution, indicating a very low probability (less than 0.01) that the observed distribution matches the expected distribution under the null hypothesis, assuming the null hypothesis is true.

Question 10

A left-tailed area in the chi-squared distribution equals 0.10. For 5 degrees of freedom, the table value equals 9.23635.

Accepted Answer

The answer of A left-tailed area in the chi-squared distribution...

Question 11

A left-tailed area in the chi-squared distribution equals 0.90. For 10 degrees of freedom, the table value equals 15.9871.

Accepted Answer

The answer of A left-tailed area in the chi-squared distribution...

Question 12

In a goodness-of-fit test, the null hypothesis states that the data came from a normally distributed population. The researcher estimated the population mean and population standard deviation from a sample of 200 observations. In addition, the researcher used 5 standardised intervals to test for normality. Using a 10% level of significance, the critical value for this test is 4.60517.

Accepted Answer

This is true based on the information given. The critical value for a goodness-of-fit test with 4 degrees of freedom and a 10% level of significance is 4.60517. However, it is not clear from the information given whether the test actually had 4 degrees of freedom.

Question 13

A test for independence is applied to a contingency table with 4 rows and 4 columns for two nominal variables. The number of degrees of freedom for this test will be 9.

Accepted Answer

The degrees of freedom for a test of independence in a contingency table is calculated as (number of rows - 1) * (number of columns - 1), which in this case is (4-1)*(4-1) = 3*3 = 9.

Question 14

The middle 0.95 portion of the chi-squared distribution with 9 degrees of freedom has table values of 3.32511 and 16.9190, respectively.

Accepted Answer

The answer of The middle 0.95 portion of the chi-squared...

Question 15

In a chi-squared test of independence, the value of the test statistic was $x ^ { 2 } =$ 15.652, and the critical value at $$\alpha = 0.025$$ was 11.1433. Thus we must reject the null hypothesis at $\alpha = 0.025$ .

Accepted Answer

Since the calculated test statistic value (15.652) is greater than the critical value (11.1433), we have enough evidence to reject the null hypothesis at a significance level of 0.025. Therefore, the answer is A) True.

Question 16

In a goodness-of-fit test, the null hypothesis states that the data came from a normally distributed population. The researcher estimated the population mean and population standard deviation from a sample of 100 observations. In addition, the researcher used 6 standardised intervals to test for normality. Using a 2.5% level of significance, the critical value for this test is 9.3484.

Accepted Answer

This question provides information about the null hypothesis, sample size, and significance level, but does not provide enough information to calculate the critical value for the test. However, the critical value for a goodness-of-fit test with 6 intervals and a 2.5% level of significance would typically be larger than 9.3484, so the statement is likely false.

Question 17

Whenever the expected frequency of a cell is less than 5, one remedy for this condition is to increase the size of the sample.

Accepted Answer

Increasing the sample size can lead to higher expected frequencies in each cell, thus addressing the issue of expected frequencies being less than 5, which is a common guideline for the validity of chi-square tests.

Question 18

The chi-squared test of independence is a Chi-squared test of a contingency table.

Accepted Answer

The chi-squared test of independence is used to determine if there is a significant association between two categorical variables in a contingency table. Therefore, it is a Chi-squared test of a contingency table.

Question 19

One type of Chi-squared goodness of fit test is a Chi-squared test for normality.

Accepted Answer

A Chi-squared test for normality is a type of Chi-squared goodness of fit test that tests whether a data set follows a normal distribution.

Question 20

The rule of five is used to ensure that the discrete distribution of the test statistic can be approximated by the continuous Chi-squared distribution.

Accepted Answer

The rule of five states that for the Chi-squared test to be valid, the expected count in each cell of the contingency table should be at least five. This ensures that the discrete distribution of the test statistic can be approximated by the continuous Chi-squared distribution.

Quiz 17: Additional Tests for Nominal Data: Chi-Squared Tests

The chi-squared goodness-of-fit test is usually used as a test of multinomial parameters, but it can also be used to determine whether data were drawn from any distribution.

A chi-squared goodness-of-fit test can be conducted either as a two-tail test or as a one-tail test.

A multinomial experiment, where the outcome of each trial can be classified into one of two categories, is identical to the binomial experiment.

For a chi-squared distributed random variable with 12 degrees of freedom and a level of significance of 0.05, the chi-squared value from the table is 21.0261. The computed value of the test statistics is 25.1687. This will lead us to reject the null hypothesis.

In chi-squared tests, the conventional and conservative rule - known as the rule of five - is to require that difference between the observed and expected frequency for each cell be at least 5.

Whenever the expected frequency of a cell is less than 5, one possible remedy for this condition is to combine it with one or more other cells.

A chi-squared test for independence 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 < 0.10.

For a chi-squared distributed random variable with 10 degrees of freedom and a level of significance of 0.025, the chi-squared value from the table is 20.5. The computed value of the test statistic is 16.857. The decision is to retain Ho.

The area to the right of a chi-squared value is 0.01. For 8 degrees of freedom, the table value is 20.0902.

A left-tailed area in the chi-squared distribution equals 0.10. For 5 degrees of freedom, the table value equals 9.23635.

A left-tailed area in the chi-squared distribution equals 0.90. For 10 degrees of freedom, the table value equals 15.9871.

A test for independence is applied to a contingency table with 4 rows and 4 columns for two nominal variables. The number of degrees of freedom for this test will be 9.

The middle 0.95 portion of the chi-squared distribution with 9 degrees of freedom has table values of 3.32511 and 16.9190, respectively.

In a chi-squared test of independence, the value of the test statistic was $x ^ { 2 } =$ 15.652, and the critical value at $\alpha = 0.025$ was 11.1433. Thus we must reject the null hypothesis at $\alpha = 0.025$ .

Whenever the expected frequency of a cell is less than 5, one remedy for this condition is to increase the size of the sample.

The chi-squared test of independence is a Chi-squared test of a contingency table.

One type of Chi-squared goodness of fit test is a Chi-squared test for normality.

The rule of five is used to ensure that the discrete distribution of the test statistic can be approximated by the continuous Chi-squared distribution.

Quiz 17: Additional Tests for Nominal Data: Chi-Squared Tests

The chi-squared goodness-of-fit test is usually used as a test of multinomial parameters, but it can also be used to determine whether data were drawn from any distribution.

A chi-squared goodness-of-fit test can be conducted either as a two-tail test or as a one-tail test.

A multinomial experiment, where the outcome of each trial can be classified into one of two categories, is identical to the binomial experiment.

For a chi-squared distributed random variable with 12 degrees of freedom and a level of significance of 0.05, the chi-squared value from the table is 21.0261. The computed value of the test statistics is 25.1687. This will lead us to reject the null hypothesis.

In chi-squared tests, the conventional and conservative rule - known as the rule of five - is to require that difference between the observed and expected frequency for each cell be at least 5.

Whenever the expected frequency of a cell is less than 5, one possible remedy for this condition is to combine it with one or more other cells.

A chi-squared test for independence 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 < 0.10.

For a chi-squared distributed random variable with 10 degrees of freedom and a level of significance of 0.025, the chi-squared value from the table is 20.5. The computed value of the test statistic is 16.857. The decision is to retain Ho.

The area to the right of a chi-squared value is 0.01. For 8 degrees of freedom, the table value is 20.0902.

A left-tailed area in the chi-squared distribution equals 0.10. For 5 degrees of freedom, the table value equals 9.23635.

A left-tailed area in the chi-squared distribution equals 0.90. For 10 degrees of freedom, the table value equals 15.9871.

A test for independence is applied to a contingency table with 4 rows and 4 columns for two nominal variables. The number of degrees of freedom for this test will be 9.

The middle 0.95 portion of the chi-squared distribution with 9 degrees of freedom has table values of 3.32511 and 16.9190, respectively.

In a chi-squared test of independence, the value of the test statistic was x2=x ^ { 2 } =x2= 15.652, and the critical value at α=0.025\alpha = 0.025α=0.025 was 11.1433. Thus we must reject the null hypothesis at α=0.025\alpha = 0.025α=0.025 .

Whenever the expected frequency of a cell is less than 5, one remedy for this condition is to increase the size of the sample.

The chi-squared test of independence is a Chi-squared test of a contingency table.

One type of Chi-squared goodness of fit test is a Chi-squared test for normality.

The rule of five is used to ensure that the discrete distribution of the test statistic can be approximated by the continuous Chi-squared distribution.

In a chi-squared test of independence, the value of the test statistic was $x ^ { 2 } =$ 15.652, and the critical value at $\alpha = 0.025$ was 11.1433. Thus we must reject the null hypothesis at $\alpha = 0.025$ .