Question 1

Consider a set of 50 measurements with mean 50.2 and standard deviation 18.7 and with the following observed frequencies.  It is desired to test whether these measurements came from a normal population. Calculate the expected frequency for the interval 60-79.99.

Accepted Answer

The answer of Consider a set of 50 measurements with...

Question 2

Consider a set of 50 measurements with mean 50.2 and standard deviation 18.7 and with the following observed frequencies.  It is desired to test whether these measurements came from a normal population. Calculate the expected frequency for the interval 40-59.99.

Accepted Answer

The answer of Consider a set of 50 measurements with...

Question 3

Consider the 3X2 contingency table below.  At  = .05,determine the tabular value of the chi-square statistic used to test for the independence of Factors A and B?

Accepted Answer

F1F1F1S1F1F1F10²_.05,2 = 5.99

Question 4

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

Accepted Answer

As the difference between observed and expected frequencies decreases, the chi-square statistic becomes smaller, indicating a higher probability that the variables are independent.

Question 5

An experiment consists of 400 observations,and four mutually exclusive groups.If the probability of a randomly selected item being classified into any of the four groups is equal,then the expected number of items that will be classified into group 1 is _____.

Accepted Answer

(.25)(400)= 100

Question 6

The number of degrees of freedom associated with a chi-square test for independence based upon a contingency table with 4 rows and 3 columns is _____.

Accepted Answer

(4 - 1)(3 - 1)= 6

Question 7

In performing a chi-square goodness of fit test for a normal distribution,if there are 7 intervals,then the degrees of freedom for the chi-square statistic is ______________.

Accepted Answer

7 - 3 = 4

Question 8

A special version of the chi-square goodness of fit test that involves testing the null hypothesis that all of the multinomial probabilities are equal is called the test for ___________.

Accepted Answer

The test for homogeneity involves comparing the proportions of one categorical variable across multiple groups or categories, to see if they are similar or different. It is a special version of the chi-square test and can be used to test the null hypothesis that all groups have the same proportions. Therefore, option D is the correct answer. Option A (goodness of fit) involves testing the fit of a single categorical variable to an expected distribution, option B (statistical independence) involves testing whether two categorical variables are independent, and option C (normality) involves testing whether continuous data follow a normal distribution. These concepts are not related to the test for homogeneity.

Question 9

A real estate company is analyzing the selling prices of residential homes in a given community.140 homes that have been solved in the past month are randomly selected and their selling prices are recorded.The statistician working on the project has stated that in order to perform various statistical tests,the data must be distributed according to normal distribution.In order to determine whether the selling prices of homes included in the random sample are normally distributed,the data is divided into 6 classes of equal size and the number of observations in each class is recorded.The chi-square goodness of fit test for normal distribution is performed and the results are summarized in the following table:  At a significance level of .05,we:

Accepted Answer

The answer of A real estate company is analyzing the...

Question 10

Consider a set of 50 measurements with mean 50.2 and standard deviation 18.7 and with the following observed frequencies.  It is desired to test whether these measurements came from a normal population. Calculate the expected frequency for the interval 0-39.99.

Accepted Answer

To test if the measurements came from a normal population, we need to use the normal distribution. We can estimate the expected frequency for the interval 0-39.99 by finding the area under the normal curve for this interval, using the mean and standard deviation of the data set. 
Using a z-score table, we can find the z-score for the lower bound of this interval as follows: 
z = (39.99 - 50.2) / 18.7 = -0.55
The area under the normal curve for this interval is equal to the area to the left of this z-score, which is 0.2912.
Therefore, the expected frequency for this interval is:
Expected frequency = 50 * 0.2912 = 14.56
So, the best choice is C.

Question 11

Consider the 3X2 contingency table below. At a significance level of .05,test H₀: the factors A and B are independent.

Accepted Answer

The answer of Consider the 3X2 contingency table below. ...

Question 12

Consider the 3X2 contingency table below.  How many degrees of freedom are associated with the chi-square test?

Accepted Answer

Degrees of freedom = (3 - 1)(2 - 1)= 2

Question 13

Consider the 3X2 contingency table below.  Compute the expected frequencies in row 1.

Accepted Answer

To calculate the expected frequencies in a contingency table, we use the formula:

Expected Frequency = (row total x column total) / grand total

The row total for row 1 is 30, and the grand total is 60. So, the expected frequency for row 1, column 1 is (30 x 20) / 60 = 10. The expected frequency for row 1, column 2 is (30 x 40) / 60 = 20. Therefore, the answer is C) 12,18, which are the closest values to the expected frequencies calculated above.

Question 14

In performing chi-square goodness fit test for a normal distribution,a researcher wants to make sure that all of the expected cell frequencies are at least five.The sample is divided into 7 intervals.The second through the sixth interval all have expected cell frequencies of at least five.The first and the last intervals have expected cell frequencies of 1.5 each.After adjusting the number of intervals,degrees of freedom for the chi-square statistic is ____.

Accepted Answer

The degrees of freedom for a chi-square goodness-of-fit test is equal to the number of intervals minus 1. In this case, the researcher has 7 intervals, but the first and last intervals have expected cell frequencies of less than 5, so they are combined into one interval. This leaves the researcher with 6 intervals, and the degrees of freedom is 6 - 1 = 5.

Question 15

The chi-square goodness of fit test for multinomial probabilities with 5 categories has _____ degrees of freedom.

Accepted Answer

5 - 1 = 4

Question 16

In performing a chi-square goodness of fit test with multinomial probabilities,the ___________ 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

The smaller the difference between observed and expected frequencies, the more consistent the data is with the null hypothesis. A larger difference would suggest that the observed frequencies are significantly different from the expected frequencies, which would lead to rejection of the null hypothesis.

Question 17

While a binomial distribution describes a count data that can be classified into one of two mutually exclusive categories,______________________ distribution describes count data that is classified into more than two mutually exclusive categories.

Accepted Answer

The multinomial distribution describes count data that is classified into more than two mutually exclusive categories. The other options (normal, skewed, uniform) do not describe count data with multiple categories.

Question 18

Consider the 3X2 contingency table below.  Compute the expected frequencies in row 3.

Accepted Answer

To calculate expected frequencies in row 3, we use the formula:
Expected Frequency = (row total * column total) / grand total
For the first cell in row 3:
Expected Frequency = (30 * 50) / 80 = 18.75
For the second cell in row 3:
Expected Frequency = (20 * 50) / 80 = 12.5
Rounding off to the nearest whole number, we get expected frequencies of 18 and 12 for the two cells in row 3, respectively. Therefore, the closest option is C, 12 and 18.

Question 19

Consider the 3X2 contingency table below.  Compute the expected frequencies in row 2.

Accepted Answer

The answer of Consider the 3X2 contingency table below. ...

Question 20

As the difference between observed frequency and expected frequency _______________,the probability of rejecting the null hypothesis increases.

Accepted Answer

The larger the difference between observed and expected frequency, the less likely the results occurred by chance and the more likely the null hypothesis is incorrect. Therefore, the probability of rejecting the null hypothesis increases as the difference between observed and expected frequency increases.

Quiz 12: Chi-Square Tests

Consider a set of 50 measurements with mean 50.2 and standard deviation 18.7 and with the following observed frequencies. It is desired to test whether these measurements came from a normal population. Calculate the expected frequency for the interval 60-79.99.

Consider a set of 50 measurements with mean 50.2 and standard deviation 18.7 and with the following observed frequencies. It is desired to test whether these measurements came from a normal population. Calculate the expected frequency for the interval 40-59.99.

Consider the 3X2 contingency table below. At = .05,determine the tabular value of the chi-square statistic used to test for the independence of Factors A and B?

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

An experiment consists of 400 observations,and four mutually exclusive groups.If the probability of a randomly selected item being classified into any of the four groups is equal,then the expected number of items that will be classified into group 1 is _____.

The number of degrees of freedom associated with a chi-square test for independence based upon a contingency table with 4 rows and 3 columns is _____.

In performing a chi-square goodness of fit test for a normal distribution,if there are 7 intervals,then the degrees of freedom for the chi-square statistic is ______________.

A special version of the chi-square goodness of fit test that involves testing the null hypothesis that all of the multinomial probabilities are equal is called the test for ___________.

Consider a set of 50 measurements with mean 50.2 and standard deviation 18.7 and with the following observed frequencies. It is desired to test whether these measurements came from a normal population. Calculate the expected frequency for the interval 0-39.99.

Consider the 3X2 contingency table below. At a significance level of .05,test H0: the factors A and B are independent.

Consider the 3X2 contingency table below. How many degrees of freedom are associated with the chi-square test?

Consider the 3X2 contingency table below. Compute the expected frequencies in row 1.

The chi-square goodness of fit test for multinomial probabilities with 5 categories has _____ degrees of freedom.

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

While a binomial distribution describes a count data that can be classified into one of two mutually exclusive categories,______________________ distribution describes count data that is classified into more than two mutually exclusive categories.

Consider the 3X2 contingency table below. Compute the expected frequencies in row 3.

Consider the 3X2 contingency table below. Compute the expected frequencies in row 2.

As the difference between observed frequency and expected frequency _______________,the probability of rejecting the null hypothesis increases.

Consider the 3X2 contingency table below. At a significance level of .05,test H₀: the factors A and B are independent.