Lab 2: Chi-Square Test of Independence Quiz
The question sheet
Reveal any answer as you study-
What is the goal of the student in this Lab 2 exercise?
- Estimate a population mean
- Build a histogram of snacks
- Evaluate a snack–gender relationship
- Compare two sample variances
Reveal answer
Answer: Evaluate a snack–gender relationship
Source evidence
PDF page 673: • The student will evaluate if there is a significant relationship between favorite type of snack and gender.
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The lab conducts a hypothesis test to determine whether the factors are what?
- Equal in mean
- Independent
- Uniform
- Normally distributed
Reveal answer
Answer: Independent
Source evidence
PDF page 673: Conduct a hypothesis test to determine if the factors are independent:
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According to the note, why might you combine two food categories?
- To increase degrees of freedom
- To make the table symmetric
- So each cell has expected value ≥ 5
- To reduce the sample size
Reveal answer
Answer: So each cell has expected value ≥ 5
Source evidence
PDF page 673: total the results. NOTE You may need to combine two food categories so that each cell has an expected value of at least five.
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A contingency table displays sample values for how many factors?
- Two
- Three
- One
- Four
Reveal answer
Answer: Two
Source evidence
PDF page 675: contingency table a table that displays sample values for two different factors that may be dependent or contingent on each other; facilitates determining conditional probabilities
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What is the null hypothesis for a chi-square test of independence?
- The two factors are dependent
- The two factors are independent
- The data are normal
- The variances are equal
Reveal answer
Answer: The two factors are independent
Source evidence
PDF page 675: To assess whether two factors are independent, you can apply the test of independence that uses the chi-square distribution. The null hypothesis for this test states that the two factors are independent. The test compares observed values to expected values. The test is right-tailed. Each observation or cell category must have an expected value of at least five.
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The test of independence makes use of which structure?
- A dot plot
- A frequency polygon
- A contingency table
- A box plot
Reveal answer
Answer: A contingency table
Source evidence
PDF page 675: The goodness-of-fit test is typically used to determine if data fits a particular distribution. The test of independence makes use of a contingency table to determine the independence of two factors. The test for homogeneity determines whether two populations come from the same distribution, even if this distribution is unknown.
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A chi-square test of independence is described as which type of test?
- Left-tailed
- Two-tailed
- Nondirectional
- Right-tailed
Reveal answer
Answer: Right-tailed
Source evidence
PDF page 675: To assess whether two factors are independent, you can apply the test of independence that uses the chi-square distribution. The null hypothesis for this test states that the two factors are independent. The test compares observed values to expected values. The test is right-tailed. Each observation or cell category must have an expected value of at least five.
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How are the degrees of freedom found in a test of independence?
- (columns–1)(rows–1)
- columns × rows
- n – 1
- k – 1
Reveal answer
Answer: (columns–1)(rows–1)
Source evidence
PDF page 676: • The number of degrees of freedom is equal to (number
PDF page 676: of columns–1)(number of rows–1). 2
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How is the expected number in a cell computed for a test of independence?
- total surveyed / k
- (row total)(column total)/total surveyed
- column total × rows
- row total + column total
Reveal answer
Answer: (row total)(column total)/total surveyed
Source evidence
PDF page 676: • If the null hypothesis is true, the expected number
PDF page 676: (row total)(column total) E = . total surveyed
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The chi-square distribution curve is described as skewed in which direction?
- To the right
- Bimodal
- Symmetric
- To the left
Reveal answer
Answer: To the right
Source evidence
PDF page 675: The chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population. An important parameter in a chi-square distribution is the degrees of freedom df in a given problem. The random variable in the chi-square distribution is the sum of squares of df standard normal variables, which must be independent. The key characteristics of the chi-square distribution also depend directly on the degrees of freedom. The chi-square distribution curve is skewed to the right, and its shape depends on the degrees of freedom df. For df > 90, the curve approximates the normal distribution. Test statistics based on the chi-square distribution are always greater than or equal to zero. Such application tests are almost always right-tailed tests.
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For which df does the chi-square curve approximate the normal distribution?
- df = 30
- df > 90
- df < 5
- df = 1
Reveal answer
Answer: df > 90
Source evidence
PDF page 675: The chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population. An important parameter in a chi-square distribution is the degrees of freedom df in a given problem. The random variable in the chi-square distribution is the sum of squares of df standard normal variables, which must be independent. The key characteristics of the chi-square distribution also depend directly on the degrees of freedom. The chi-square distribution curve is skewed to the right, and its shape depends on the degrees of freedom df. For df > 90, the curve approximates the normal distribution. Test statistics based on the chi-square distribution are always greater than or equal to zero. Such application tests are almost always right-tailed tests.
PDF page 695: 3 when the number of degrees of freedom is greater than 90 5 df = 2 7 a goodness-of-fit test 9 3 11 2.04
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The Transit Railroads study asks whether ticket class choice is independent of what?
- Time of day
- Passenger age
- Travel distance
- Ticket price
Reveal answer
Answer: Travel distance
Source evidence
PDF page 679: Use the following information to answer the next seven exercises: Transit Railroads is interested in the relationship between travel distance and the ticket class purchased. A random sample of 200 passengers is taken. Table 11.31 shows the results. The railroad wants to know if a passenger’s choice in ticket class is independent of the distance the passenger must travel.
High School Statistics
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