Question 1

If the expected frequency of a cell is less than 5, you should combine cells of the table.

Accepted Answer

When the expected frequency of a cell is less than 5, it is recommended to combine adjacent cells to get a minimum expected frequency of 5. This helps to ensure the validity of statistical tests and prevent unreliable results caused by small sample sizes.

Question 2

A chi-squared test is used to describe a population of nominal data.

Accepted Answer

A chi-squared test is commonly used to analyze nominal data to determine if there is a significant difference between observed and expected frequencies, which can provide insights into patterns in the population.

Question 3

A chi-squared goodness-of-fit test is always a two-tailed test.

Accepted Answer

A chi-squared goodness-of-fit test is not always a two-tailed test. The tail(s) of the test depend on the specific hypothesis being tested. For example, if the null hypothesis is that the observed data follows a specific distribution, the test may be one-tailed. If the null hypothesis is simply that the observed data does not differ significantly from the expected values, the test may be two-tailed.

Question 4

The only way the chi-squared test statistic can be zero is if the observed frequencies are all exactly the same as the expected frequencies.

Accepted Answer

The chi-squared test statistic measures the difference between observed and expected frequencies. If the observed frequencies are all exactly the same as the expected frequencies, the difference will be zero, resulting in a chi-squared test statistic of zero. Therefore, the statement is true.

Question 5

A multinomial experiment with two categories is identical to a binomial experiment.

Accepted Answer

In a multinomial experiment with two categories, there are only two possible outcomes for each trial (e.g. success or failure). This is the same as a binomial experiment, which also has two possible outcomes. Therefore, a multinomial experiment with two categories is identical to a binomial experiment.

Question 6

The alternative hypothesis in a goodness-of-fit test is that none of the p_i values are equal to their values specified in H₀.

Accepted Answer

The alternative hypothesis in a goodness-of-fit test typically states that at least one of the observed frequencies does not match the expected frequency under the null hypothesis, not that none of the observed frequencies match.

Question 7

In conducting a chi-squared goodness-of-fit test, an essential condition is that all expected frequencies are at least five.

Accepted Answer

This is true as a rule of thumb. It ensures that the sample size is large enough to reliably measure the distribution of the population. However, if some expected frequencies are less than five but the overall sample size is large enough, adjustments to the test can be made, such as combining categories or using a different test.

Question 8

For a chi-squared distributed random variable with 12 degrees of freedom and a level of significance of 0.05, the test statistics is 25.168. The chi-squared value from the table is 21.0261. These results will lead us to reject the null hypothesis.

Accepted Answer

The calculated test statistic (25.168) is greater than the critical value from the chi-squared table (21.0261) at a significance level of 0.05 with 12 degrees of freedom. Therefore, we reject the null hypothesis.

Question 9

You cannot use a chi-squared goodness-of-fit test when there are only two possible outcomes for each trial in your experiment.

Accepted Answer

A chi-squared goodness-of-fit test can be used even when there are only two possible outcomes (e.g., success or failure) for each trial in an experiment; it is not limited by the number of categories or outcomes.

Question 10

All of the expected frequencies in a chi-squared goodness-of-fit test must be equal to each other.

Accepted Answer

In a chi-squared goodness-of-fit test, the expected frequencies are not required to be equal to each other; they are determined based on the null hypothesis and the sample data.

Question 11

For a chi-squared distributed random variable with 10 degrees of freedom and a level of significance of 0.025, the chi-squared table value is 20.4831. Suppose the value of your test statistic is 16.857. This will lead you to reject the null hypothesis.

Accepted Answer

The value of the test statistic (16.857) is less than the table value (20.4831) at a level of significance of 0.025. Therefore, we fail to reject the null hypothesis.

Question 12

If the expected frequency of a cell is less than 5, you should increase the significance level.

Accepted Answer

If the expected frequency of a cell is less than 5, increasing the significance level is not appropriate. Instead, one should use techniques such as combining adjacent cells or using a different statistical test to address the issue of low expected frequencies in the data.

Question 13

A small chi-squared test statistic in a goodness-of-fit test supports the null hypothesis.

Accepted Answer

A small chi-squared test statistic indicates that the observed data does not deviate much from the expected data, which supports the null hypothesis.

Question 14

In a goodness-of-fit test, all of the proportions specified in the null hypothesis must be equal to each other.

Accepted Answer

In a goodness-of-fit test, the proportions specified in the null hypothesis do not have to be equal to each other; they can be any set of proportions that are hypothesized for the categories being tested.

Question 15

A chi-squared distribution is symmetric.

Accepted Answer

The chi-squared distribution is not symmetric, but rather it is right-skewed.

Question 16

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 17

If there are only two categories, the chi-squared goodness-of-fit test is the same as the z-test for p, the population proportion (as long as the sample/cell sizes meet the conditions).

Accepted Answer

When there are only two categories, the chi-squared goodness-of-fit test simplifies to a test for the difference between two proportions, which is equivalent to the z-test for p.

Question 18

The total of the observed frequencies in a multinomial experiment must equal nk where n is the number of trials and k is the number of categories.

Accepted Answer

The answer of The total of the observed frequencies in...

Question 19

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

For a left-tailed area in the chi-squared distribution, we need to look at the cumulative probability in the left tail of the distribution. Since the given area is 0.10, this means that 10% of the distribution lies in the left tail.

To find the corresponding table value for this area and degrees of freedom, we would need to consult a chi-squared distribution table. However, it is not necessarily true that the table value will be exactly 9.23635. The table value depends on the specific degrees of freedom and the confidence level (i.e. area) we are looking at.

Therefore, we cannot determine the correctness of the statement based on the information given.

Question 20

If the observed frequencies are all smaller than the expected frequencies in a goodness-of-fit test, the chi-squared test statistic will be negative.

Accepted Answer

The chi-squared test statistic can never be negative. It is always a non-negative value.

Quiz 15: Chi-Squared Tests