Question 1

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

Accepted Answer

The chi-square goodness of fit test is used to determine if observed data follows a specified distribution, and one possible use is to test if specified multinomial probabilities in the null hypothesis are correct. Therefore, the statement is true.

Question 2

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, indicating that there is a relationship between them. Therefore, the correct answer is B.

Question 3

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 observed cell frequencies. The expected cell frequencies are calculated based on statistical assumptions about the distribution of variables and the independence of those variables.

Question 4

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

This is true. A large value of the chi-square statistic indicates a significant difference between the observed and expected frequencies, leading to rejection of the null hypothesis.

Question 5

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

Accepted Answer

The expected frequencies are calculated assuming that the null hypothesis is true. This means that if the null hypothesis is rejected, then the expected frequencies may not be a good estimate of the actual frequencies.

Question 6

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

Accepted Answer

Trials of a multinomial probability are assumed to be independent.

Question 7

When we carry out a chi-square test of independence,if r_i is 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 is true because the estimated expected cell frequency for a chi-square test of independence is calculated by multiplying the row total and the column total and then dividing it by the sample size (n), which is represented by the formula given in the statement.

Question 8

In a contingency table,when all the expected frequencies equal the observed frequencies the calculated $\chi$² statistic equals zero.

Accepted Answer

The formula to calculate chi-square statistic is based on the deviation between the expected and observed frequencies. When all the expected frequencies equal the observed frequencies, there is no deviation and hence the value of chi-square statistic is zero.

Question 9

The $\chi$² goodness of fit test requires nominative level of data.

Accepted Answer

The F1F1F1S1F1F1F10² goodness of fit test requires nominative (or nominal) level of data, which is a type of categorical data where each observation belongs to one of a set of distinct categories.

Question 10

When using 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 statement is true. The chi-square goodness of fit test is used to determine whether the observed frequencies of categorical data fit the expected frequencies based on a specified multinomial distribution. If the null hypothesis is rejected, it means that the observed frequencies are significantly different from the expected frequencies, indicating that at least one of the multinomial probabilities is not equal to the supposed value specified in the null hypothesis.

Question 11

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 n the contingency table.

Accepted Answer

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

Question 12

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 is correct.

Accepted Answer

This is true. In a chi-square goodness of fit test, the null hypothesis assumes that the observed frequencies match the expected frequencies based on the specified probabilities. If the difference between the observed and expected frequencies is small, then the calculated chi-square statistic will be small, which in turn will result in a higher probability of accepting the null hypothesis. Conversely, if the difference is large, the calculated chi-square statistic will be large, which suggests that the null hypothesis is not supported by the data.

Question 13

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 applied to test if a sample data comes from other distribution forms such as Poisson.

Accepted Answer

The statement is false. The chi-square goodness of fit test can be used to test whether a sample data comes from various distribution forms such as Poisson, binomial, and exponential, as long as certain assumptions are met. The test is not limited to normal distributions and multinomial probabilities.

Question 14

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

Accepted Answer

A multinomial probability distribution is used to describe data that is classified into two or more categories in a multinomial experiment, making the statement true.

Question 15

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

Accepted Answer

A contingency table, also known as a cross-tabulation or crosstab, displays the frequency distribution of variables that are classified on two dimensions or scales. It shows the number or percentage of cases falling into each combination of categories for the two variables being analyzed.

Question 16

In performing a chi-square test of independence,as the difference between the respective observed and expected frequencies decrease,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 decrease, the probability of concluding independence increases. When the difference is zero, it indicates perfect independence, and the chi-square value is zero. Therefore, it is incorrect to state that the probability of concluding independence decreases as the difference decreases.

Question 17

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.

Question 18

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, not dependence.

Question 19

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 is used to determine if an observed sample distribution fits a theoretical distribution, not if a population distribution is normal. Therefore, rejecting the null hypothesis does not necessarily mean the population is normally distributed.

Question 20

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

Accepted Answer

The chi-square distribution is skewed to the right, not to the left.

Quiz 12: Chi-Square Tests

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

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

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

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.

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

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

When we carry out a chi-square test of independence,if r_i is 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.

In a contingency table,when all the expected frequencies equal the observed frequencies the calculated $\chi$ ² statistic equals zero.

The $\chi$ ² goodness of fit test requires nominative level of data.

When using 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.

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 n the contingency table.

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 is correct.

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 applied to test if a sample data comes from other distribution forms such as Poisson.

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

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

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

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.

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

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

Quiz 12: Chi-Square Tests

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

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

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

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.

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

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

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

In a contingency table,when all the expected frequencies equal the observed frequencies the calculated χ\chiχ 2 statistic equals zero.

The χ\chiχ 2 goodness of fit test requires nominative level of data.

When using 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.

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 n the contingency table.

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 is correct.

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

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

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

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.

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

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

When we carry out a chi-square test of independence,if r_i is 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.

In a contingency table,when all the expected frequencies equal the observed frequencies the calculated $\chi$ ² statistic equals zero.

The $\chi$ ² goodness of fit test requires nominative level of data.