Question 1

When using the chi-square goodness-of-fit test,if the value of the chi-square statistic is large enough,we reject the null hypothesis.

Accepted Answer

When using the chi-square goodness-of-fit test, if the calculated value of the chi-square statistic is larger than the critical value from the table or p-value is less than the level of significance, we reject the null hypothesis, indicating that there is a significant difference between the observed and expected frequencies.

Question 2

Expected cell frequencies for a multinomial distribution are calculated by assuming statistical dependence.

Accepted Answer

Expected cell frequencies for a multinomial distribution are calculated by assuming statistical independence.

Question 3

When we carry out a chi-square test of independence,in the alternative hypothesis we state that the two classifications are statistically independent.

Accepted Answer

In the alternative hypothesis of a chi-square test of independence, we state that the two classifications are NOT statistically independent.

Question 4

The actual counts in the cells of a contingency table are referred to as the expected cell frequencies.

Accepted Answer

The actual counts in the cells of a contingency table are referred to as the observed cell frequencies, not expected. Expected cell frequencies are calculated based on the assumption of independence between the variables being studied.

Question 5

In performing a chi-square test of independence,as the difference between the respective observed and expected frequencies decreases,the probability of concluding that the row variable is independent of the column variable decreases.

Accepted Answer

As the difference between the observed and expected frequencies decreases, the probability of concluding that the row variable is independent of the column variable increases. This is because a smaller difference between observed and expected frequencies indicates a stronger agreement between the row and column variables, making it less likely that they are independent.

Question 6

When using a chi-square goodness-of-fit test with multinomial probabilities,the rejection of the null hypothesis indicates that at least one of the multinomial probabilities is not equal to the value stated in the null hypothesis.

Accepted Answer

This is true because the chi-square goodness-of-fit test is used to compare observed frequencies with expected frequencies based on a specified null hypothesis that assumes equal probabilities for each outcome. If the calculated chi-square value is greater than the critical value, the null hypothesis is rejected indicating that at least one of the multinomial probabilities is not equal to the expected value stated in the null hypothesis.

Question 7

One use of the chi-square goodness-of-fit test is to determine if specified multinomial probabilities in the null hypothesis are correct.

Accepted Answer

This is a correct use of the chi-square goodness-of-fit test. The test compares observed frequencies to expected frequencies based on the specified probabilities in the null hypothesis. If there is a significant difference between the observed and expected frequencies, it suggests that the specified probabilities are not accurate.

Question 8

A contingency table summarizes data that has been classified on two dimensions or scales.

Accepted Answer

A contingency table displays the frequency distribution of two categorical variables. It is also known as a cross-tabulation or a two-way table.

Question 9

When we carry out a chi-square test of independence,if r_i is the row total for row i and c_j is the column total for column j,then the estimated expected cell frequency corresponding to row i and column j equals (r_i)(c_j)/n.

Accepted Answer

This formula is correct and is used to calculate the expected cell frequency for the chi-square test of independence.

Question 10

The chi-square distribution is a continuous probability distribution that is skewed to the left.

Accepted Answer

The chi-square distribution is actually skewed to the right.

Question 11

A fastener manufacturing company uses a chi-square goodness-of-fit test to determine if a population of all lengths of ¼-inch bolts it manufactures is distributed according to a normal distribution.If we reject the null hypothesis,it is reasonable to assume that the population distribution is at least approximately normally distributed.

Accepted Answer

The chi-square goodness-of-fit test cannot prove that a population is normally distributed, only that it is not significantly different from a normally distributed population. Therefore, rejecting the null hypothesis does not necessarily mean that the population is normally distributed.

Question 12

The χ² goodness-of-fit test requires the nominative level of data.

Accepted Answer

The χ² goodness-of-fit test requires the nominal level of data, which means the data should be categorical with no inherent order or numerical value.

Question 13

When we carry out a chi-square test of independence,the chi-square statistic is based on (rc - 1)degrees of freedom,where r and c denote,respectively,the number of rows and columns in the contingency table.

Accepted Answer

The chi-square statistic is based on (r-1)(c-1) degrees of freedom, not (rc-1).

Question 14

The chi-square goodness-of-fit test can only be used to test whether a population has specified multinomial probabilities or to test if a sample has been selected from a normally distributed population.It cannot be used if sample data come from other distribution forms such as the Poisson.

Accepted Answer

The chi-square goodness-of-fit test can be used to test for goodness-of-fit to any distribution, not just the multinomial or normal distributions. It can also be used to test for goodness-of-fit to Poisson, exponential, or other distributions.

Question 15

The trials of a multinomial probability are assumed to be dependent.

Accepted Answer

Trials of a multinomial probability are assumed to be independent.

Question 16

In a contingency table,when all the expected frequencies equal the observed frequencies,the calculated χ² statistic equals zero.

Accepted Answer

When all the expected frequencies equal the observed frequencies, there is a perfect fit between the observed and expected data, resulting in a calculated χ² statistic of zero.

Question 17

In performing a chi-square goodness-of-fit test with multinomial probabilities,the smaller the difference between observed and expected frequencies,the higher the probability of concluding that the probabilities specified in the null hypothesis are correct.

Accepted Answer

This is true because a smaller difference between observed and expected frequencies means that the data is closer to what was expected under the null hypothesis, which makes it less likely that the differences are due to chance. As a result, the chi-square statistic will be smaller and the p-value will be larger, giving a higher probability of not rejecting the null hypothesis.

Question 18

A multinomial probability distribution describes data that are classified into two or more categories when a multinomial experiment is carried out.

Accepted Answer

This is the definition of a multinomial probability distribution.

Question 19

When we carry out a chi-square test of independence,the expected frequencies are based on the null hypothesis.

Accepted Answer

In a chi-square test of independence, the expected frequencies are calculated under the assumption that the null hypothesis is true. Therefore, the answer is true.

Question 20

In a contingency table,if all of the expected frequencies equal the observed frequencies,then we can conclude that there is a perfect association between rows and columns.

Accepted Answer

Conclusion is that the rows and columns are statistically independent.

Quiz 12: Chi-Square Tests