Question 1

The chi-square goodness-of-fit test is always one-tailed left.

Accepted Answer

The chi-square goodness-of-fit test can be one-tailed left, one-tailed right, or two-tailed, depending on the research question and hypothesis being tested. The choice of tail(s) is determined by the directionality and nature of the research question.

Question 2

The formula for finding the chi-square test value is __________.

Accepted Answer

The answer of The formula for finding the chi-square test...

Question 3

A random group of apartments was selected from a city to analyze the number of bedrooms they have. Is there evidence to reject the hypothesis that the apartments are equally distributed between 1-bedroom, 2-bedroom, and 3-bedroom apartments, at = 0.05? A) There is not evidence to reject the claim that the apartments are equally distributed between the three sizes because the test value 0.412 < 5.991 B) There is evidence to reject the claim that the apartments are equally distributed between the three sizes because the test value 5.991 > 0.412 C) There is not evidence to reject the claim that the apartments are equally distributed between the three sizes because the test value 0.412 < 7.815 D) There is evidence to reject the claim that the apartments are equally distributed between the three sizes because the test value 7.815 > 0.412

Accepted Answer

A

Question 4

A random group of oranges were selected from an orchard to analyze their ripeness. Based on the time of year, the orchard owner believes that 20% of the oranges are ready for picking now, 40% will be ready in three days, 30% will be ready in one week, and 10% will be ready in two weeks. Is there evidence to reject this hypothesis at = 0.05? A) There is evidence to reject the claim that the oranges are distributed as claimed because the test value 21.553 > 9.488 B) There is not evidence to reject the claim that the oranges are distributed as claimed because the test value 5.991 < 21.553 C) There is evidence to reject the claim that the oranges are distributed as claimed because the test value 21.553 > 7.815 D) There is not evidence to reject the claim that the oranges are distributed as claimed because the test value 9.488 < 21.553

Accepted Answer

We need to perform a chi-square goodness of fit test to determine if the observed frequencies (ripeness of the oranges) differ significantly from the expected frequencies (based on the claim of the orchard owner). The test value is calculated as 21.553 using the formula chi-square = Σ [(O - E)^2 / E], where O is the observed frequency and E is the expected frequency. We compare this test value to the critical value at a significance level of 0.05 and degrees of freedom equal to the number of categories minus one (df = 4-1 = 3). Using a chi-square distribution table or calculator, the critical value is 7.815. Since the test value is greater than the critical value (21.553 > 7.815), we reject the null hypothesis and conclude that there is evidence to reject the claim that the oranges are distributed as claimed. Therefore, choice C is the correct answer.

Question 5

A random group of students was selected from a large student conference to analyze their class in school. Is there evidence to reject the hypothesis that the number of students is equally distributed between the four classes, at = 0.05? A) There is not evidence to reject the claim that students are equally distributed between the four classes because the test value 7.815 < 8.897 B) There is evidence to reject the claim that the customers' preferences are distributed between the four classes because the test value 8.897 > 9.488 C) There is not evidence to reject the claim that students are equally distributed between the four classes because the test value 9.488 < 8.897 D) There is evidence to reject the claim that students are equally distributed between the four classes because the test value 8.897 > 7.815

Accepted Answer

The question is testing the hypothesis that the number of students is equally distributed between four classes. This is a chi-square goodness of fit test, and the calculated test statistic is 8.897. The degrees of freedom for this test are 3 (four classes minus one). Looking at a chi-square distribution table with 3 degrees of freedom, at a 0.05 significance level, the critical value is 7.815. Since the calculated test statistic (8.897) is greater than the critical value (7.815), there is evidence to reject the null hypothesis that the number of students is equally distributed between the four classes. Therefore, the answer is D.

Question 6

The null hypothesis should not be rejected for the data given in the figure below.

Accepted Answer

Without the figure or any additional information, it is impossible to determine the validity of the null hypothesis.

Question 7

A random group of customers at a fast food chain were asked whether they preferred hamburgers, chicken sandwiches, or fish sandwiches. The restaurant's marketing department claims that 45% of customers prefer hamburgers, 37% of the customers prefer chicken sandwiches, and 18% of the customers prefer fish sandwiches. Is there evidence to reject this hypothesis at = 0.05? A) There is evidence to reject the claim that the customers' preferences are distributed as claimed because the test value 7.815 > 2.843 B) There is not evidence to reject the claim that the customers' preferences are distributed as claimed because the test value 2.843 < 5.991 C) There is evidence to reject the claim that the customers' preferences are distributed as claimed because the test value 5.991 > 2.843 D) There is not evidence to reject the claim that the customers' preferences are distributed as claimed because the test value 2.843 < 7.815

Accepted Answer

The answer of A random group of customers at a...

Question 8

In a chi-square goodness-of-fit test when there is close agreement between the observed frequency and the expected frequency, the chi-square test value will be small.

Accepted Answer

The chi-square test value will be small when there is close agreement between the observed frequency and the expected frequency, indicating a good fit.

Question 9

The critical chi-square value for 15 degrees of freedom when  and the test is right-tailed is 19.675.

Accepted Answer

The answer of The critical chi-square value for 15 degrees...

Question 10

A shoe salesman wants to see if his female customers have a preference in the color of shoe purchased. Compute the test value for the following data. 
A) 28.31
B) 16.77
C) 26.80
D) 21.50

Accepted Answer

The answer of A shoe salesman wants to see if...

Question 11

The chi-square distribution is continuous, whereas the goodness-of-fit test is __________.

Accepted Answer

The answer of The chi-square distribution is continuous, whereas the...

Question 12

In a goodness of fit test between one observed frequency distribution and one expected frequency distribution, the degrees of freedom are equal to the number of categories minus two.

Accepted Answer

The degrees of freedom for a goodness of fit test between one observed frequency distribution and one expected frequency distribution are equal to the number of categories minus one.

Question 13

For four degrees of freedom, the value of  shown in the  distribution below is 0.05.

Accepted Answer

The answer of For four degrees of freedom, the value...

Question 14

An assumption for the goodness-of-fit test is that the __________ frequency for each category must be five or more.

Accepted Answer

The answer of An assumption for the goodness-of-fit test is...

Question 15

A chewing gum manufacturer wishes to determine if his customers prefer one flavor gum to any other flavor. He selects a random sample of customers and records the flavor of gum that is purchased. Compute the test value for the data given. 
A) 6.18
B) 29.5
C) 11.56
D) 18.0

Accepted Answer

In order to determine if customers have a preference for one flavor of gum over another, we can conduct a chi-square goodness of fit test. The null hypothesis is that there is no preference for any particular flavor, while the alternative hypothesis is that there is a preference.

To find the test value, we first need to calculate the expected frequencies under the null hypothesis. Let's say we have four flavors of gum: peppermint, spearmint, cinnamon, and fruit. If we assume that customers are equally likely to choose any flavor, then we would expect each flavor to have a frequency of 25% (or 0.25, since we have four flavors). We can multiply this by the sample size to get the expected frequency for each flavor.

For example, if our sample size is 100, we would expect 25 customers to choose peppermint, 25 to choose spearmint, 25 to choose cinnamon, and 25 to choose fruit. If we actually observed 30 customers choosing peppermint, 20 choosing spearmint, 25 choosing cinnamon, and 25 choosing fruit, we can calculate the chi-square test statistic as follows:

((30-25)^2/25) + ((20-25)^2/25) + ((25-25)^2/25) + ((25-25)^2/25) = 2.24 + 1 + 0 + 0 = 3.24

The degrees of freedom for a chi-square goodness of fit test are equal to the number of categories minus one (df = 4-1 = 3, in this case). Using a chi-square distribution table or calculator, we can find the p-value associated with a test statistic of 3.24 and 3 degrees of freedom. The p-value turns out to be approximately 0.36.

Since the p-value is greater than our chosen level of significance (usually 0.05), we fail to reject the null hypothesis and conclude that there is not enough evidence to say that customers have a preference for one flavor of gum over another.

Question 16

Before computing the test value, one should state the hypothesis.

Accepted Answer

It is important to state the hypothesis before computing a test so that the appropriate statistical test can be chosen and the results can be interpreted correctly.

Question 17

An anatomy teacher hypothesizes that the final grades in her class are distributed as 10% A's, 23% B's, 45% C's, 14% D's, and 8% F's. At the end of the semester, she has the following grades.  Find the critical value. Use  .
A) 9.236
B) 7.779
C) 11.070
D) 9.488

Accepted Answer

The answer of An anatomy teacher hypothesizes that the final...

Question 18

Frequencies obtained by calculation are called expected frequencies.

Accepted Answer

Frequencies obtained through calculation or mathematical models are known as expected frequencies.

Question 19

A random group of desktop computer sales was selected from an electronic discount chain to analyze the size of monitor purchased with the computer. Is there evidence to reject the hypothesis that the number of monitors is equally distributed between the five types, at = 0.05? A) There is evidence to reject the claim that the monitors are equally distributed between the five classes because the test value 14.286 > 11.070 B) There is not evidence to reject the claim that the monitors are equally distributed between the five classes because the test value 9.488 < 14.286 C) There is evidence to reject the claim that the monitors are equally distributed between the five classes because the test value 14.286 > 9.488 D) There is not evidence to reject the claim that the monitors are equally distributed between the five classes because the test value 11.070 < 14.286

Accepted Answer

The answer of A random group of desktop computer sales...

Question 20

A random group of seniors was selected from a university and asked about their plans for the following year. The school advising office claims that 60% of the students plan to work, 30% of the students plan to continue in school, and 10% of the students plan to take some time off. Is there evidence to reject this hypothesis at  = 0.05? 
A) There is evidence to reject the claim that the students' plans are distributed as claimed because the test value 6.778 > 5.991
B) There is not evidence to reject the claim that the students' plans are distributed as claimed because the test value 6.778 > 7.815
C) There is evidence to reject the claim that the students' plans are distributed as claimed because the test value 6.778 > 7.815
D) There is not evidence to reject the claim that the students' plans are distributed as claimed because the test value 6.778 > 5.991

Accepted Answer

The answer of A random group of seniors was selected...

Quiz 11: Other Chi-Square Tests

The chi-square goodness-of-fit test is always one-tailed left.

The formula for finding the chi-square test value is __________.

A random group of apartments was selected from a city to analyze the number of bedrooms they have. Is there evidence to reject the hypothesis that the apartments are equally distributed between 1-bedroom, 2-bedroom, and 3-bedroom apartments, at = 0.05?

A random group of students was selected from a large student conference to analyze their class in school. Is there evidence to reject the hypothesis that the number of students is equally distributed between the four classes, at = 0.05?

The null hypothesis should not be rejected for the data given in the figure below.

In a chi-square goodness-of-fit test when there is close agreement between the observed frequency and the expected frequency, the chi-square test value will be small.

The critical chi-square value for 15 degrees of freedom when and the test is right-tailed is 19.675.

A shoe salesman wants to see if his female customers have a preference in the color of shoe purchased. Compute the test value for the following data.

The chi-square distribution is continuous, whereas the goodness-of-fit test is __________.

In a goodness of fit test between one observed frequency distribution and one expected frequency distribution, the degrees of freedom are equal to the number of categories minus two.

For four degrees of freedom, the value of shown in the distribution below is 0.05.

An assumption for the goodness-of-fit test is that the __________ frequency for each category must be five or more.

A chewing gum manufacturer wishes to determine if his customers prefer one flavor gum to any other flavor. He selects a random sample of customers and records the flavor of gum that is purchased. Compute the test value for the data given.

Before computing the test value, one should state the hypothesis.

An anatomy teacher hypothesizes that the final grades in her class are distributed as 10% A's, 23% B's, 45% C's, 14% D's, and 8% F's. At the end of the semester, she has the following grades. Find the critical value. Use .

Frequencies obtained by calculation are called expected frequencies.

A random group of desktop computer sales was selected from an electronic discount chain to analyze the size of monitor purchased with the computer. Is there evidence to reject the hypothesis that the number of monitors is equally distributed between the five types, at = 0.05?