Goodness-of-Fit Test Quiz
The question sheet
Reveal any answer as you study-
Which distribution is used for the hypothesis test that determines whether data fit a particular distribution?
- Binomial
- Chi-square
- Normal
- Student's t
Reveal answer
Answer: Chi-square
Source evidence
PDF page 478: In this type of hypothesis test, you determine whether the data "fit" a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities. The test statistic for a goodness-of-fit test is: 2
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In the goodness-of-fit test statistic, what does O represent?
- Observed values (data)
- Number of categories
- Degrees of freedom
- Expected values from theory
Reveal answer
Answer: Observed values (data)
Source evidence
PDF page 478: • O = observed values (data)
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How are the degrees of freedom found for a goodness-of-fit test?
- Number of categories
- Number of categories minus 1
- Sample size minus 2
- Sample size minus 1
Reveal answer
Answer: Number of categories minus 1
Source evidence
PDF page 478: (O − E) . hypothesis were true. There are n terms of the form E The number of degrees of freedom is df = (number of categories – 1). The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve. NOTE The number of expected values inside each cell needs to be at least five in order to use this test.
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The goodness-of-fit test is almost always which type of test?
- Right-tailed
- Two-tailed
- Left-tailed
- Non-directional
Reveal answer
Answer: Right-tailed
Source evidence
PDF page 478: (O − E) . hypothesis were true. There are n terms of the form E The number of degrees of freedom is df = (number of categories – 1). The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve. NOTE The number of expected values inside each cell needs to be at least five in order to use this test.
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To use the goodness-of-fit test, how many expected values should be in each cell?
- At least two
- At least ten
- At least five
- At least one
Reveal answer
Answer: At least five
Source evidence
PDF page 478: (O − E) . hypothesis were true. There are n terms of the form E The number of degrees of freedom is df = (number of categories – 1). The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve. NOTE The number of expected values inside each cell needs to be at least five in order to use this test.
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What are the expected values in a goodness-of-fit test?
- Values expected if null is false
- Values expected if null is true
- The actual survey data
- Randomly chosen numbers
Reveal answer
Answer: Values expected if null is true
Source evidence
PDF page 478: The observed values are the data values and the expected values are the values you would expect to get if the null 2
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In the absenteeism example, why could the original charts not be used directly?
- Expected number for 12+ was under five
- Observed values were missing
- Sample too small overall
- Too many categories
Reveal answer
Answer: Expected number for 12+ was under five
Source evidence
PDF page 479: a. No. Notice that the expected number of absences for the "12+" entry is less than five (it is two). Combine that
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After combining cells in the absenteeism example, how many degrees of freedom were there?
- 5
- 2
- 3
- 4
Reveal answer
Answer: 3
Source evidence
PDF page 479: b. There are four "cells" or categories in each of the new tables.
PDF page 479: df = number of cells – 1 = 4 – 1 = 3
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In the employee absence study, the null hypothesis was that absent days fit which distribution?
- Uniform distribution
- Normal distribution
- Chi-square distribution
- Binomial distribution
Reveal answer
Answer: Uniform distribution
Source evidence
PDF page 480: • H0: The absent days occur with equal frequencies, that is, they fit a uniform distribution.
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In the 60-manager absence study, what was the expected number of absences per day?
- 10
- 12
- 9
- 15
Reveal answer
Answer: 12
Source evidence
PDF page 481: If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: 15 + 12 + 9 + 9 + 15 = 60), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the expected (E) values. The values in the table are the observed (O) values or data.
PDF page 481: • Expected (E) values (12, 12, 12, 12, 12)
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What was the calculated chi-square test statistic in the employee absence study?
- 29.65
- 3
- 9.48
- 5.99
Reveal answer
Answer: 3
Source evidence
PDF page 481: Now add (sum) the last column. The sum is three. This is the χ test statistic.
PDF page 481: The calculated test statistics is 3 and the critical value of the χ distribution at 4 degrees of freedom the 0.05 level
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In the employee absence study, what was the critical value at 4 df and 0.05 level?
- 5.991
- 13.277
- 3
- 9.48
Reveal answer
Answer: 9.48
Source evidence
PDF page 481: The calculated test statistics is 3 and the critical value of the χ distribution at 4 degrees of freedom the 0.05 level
PDF page 481: of confidence is 9.48. This value is found in the χ table at the 0.05 column on the degrees of freedom row 4.
Introductory Business Statistics
Introductory Business Statistics by OpenStax, used under CC BY 4.0. Changes made by Stratacademy.
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