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10.2: Goodness-of-Fit Test

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    Goodness-of-Fit Test

    jkesler


    [latexpage]

    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:

    $$\sum_k \frac{(O-E)^2}{E}$$

    where:

    • O = observed values (data)
    • E = expected values (from theory)
    • k = the number of different data cells or categories

    The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true. There are k terms of the form $\frac{(O-E)^2}{E}$.

    The number of degrees of freedom is df = (number of categories – 1) = k – 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 expected value for each cell needs to be at least five in order for you to use this test.

    Example 10.1

    Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to Table 10.1.

    Number of absences per term Expected number of students
    0–2 50
    3–5 30
    6–8 12
    9–11 6
    12+ 2
    Table 10.1

    A random survey across all mathematics courses was then done to determine the actual number (observed) of absences in a course. The chart in Table 10.2 displays the results of that survey.

    Number of absences per term Actual number of students
    0–2 35
    3–5 40
    6–8 20
    9–11 1
    12+ 4
    Table 10.2


    Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test.

    H0: Student absenteeism fits faculty perception.


    The alternative hypothesis is the opposite of the null hypothesis.

    H1: Student absenteeism does not fit faculty perception.

    a. Can you use the information as it appears in the charts to conduct the goodness-of-fit test?

    b. What is the number of degrees of freedom (df)?

    Try It 10.1

    A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in Table 10.5.

    Number produced Number defective
    0–100 5
    101–200 6
    201–300 7
    301–400 8
    401–500 10
    Table 10.5

    A random sample was taken to determine the actual number of defects. Table 10.6 shows the results of the survey.

    Number produced Number defective
    0–100 5
    101–200 7
    201–300 8
    301–400 9
    401–500 11
    Table 10.6

    State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.

    In the following Example 10.2, our solution will use the p-value method where we find the p-value corresponding to our test statistic and compare it with the significance level given.

    In Example 10.3, our solution will use the critical value method where we find the critical value corresponding to the given significance level, and compare the critical value to the test statistic.

    The p-value method and the critical value method are equivalent when determining whether to reject or fail to reject the null hypothesis.

    Example 10.2

    Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in Table 10.7. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5% significance level.

    Monday Tuesday Wednesday Thursday Friday
    Number of Absences 15 12 9 9 15
    Table 10.7Day of the Week Employees were Most Absent

    Try It 10.2

    Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 56 students were asked on which night of the week they did the most homework. The results were distributed as in Table 10.8.

    Sunday Monday Tuesday Wednesday Thursday Friday Saturday
    Number of Students 11 8 10 7 10 5 5
    Table 10.8

    From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use?

    Example 10.3

    One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population) as in Table 10.9.

    Number of Televisions Percent
    0 10
    1 16
    2 55
    3 11
    4+ 8
    Table 10.9

    The table contains expected (E) percents.

    A random sample of 600 families in the far western United States resulted in the data in Table 10.10.

    Number of Televisions Frequency
    Total = 600
    0 66
    1 119
    2 340
    3 60
    4+ 15
    Table 10.10

    The table contains observed (O) frequency values.

    At the 1% significance level, does it appear that the distribution “number of televisions” of far western United States families is different from the distribution for the American population as a whole?

    Try It 10.3

    The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in Table 10.12.

    Number of Pets Percent
    0 18
    1 25
    2 30
    3 18
    4+ 9
    Table 10.12

    A random sample of 1,000 students from the Eastern United States resulted in the data in Table 10.13.

    Number of Pets Frequency
    0 210
    1 240
    2 320
    3 140
    4+ 90
    Table 10.13

    At the 1% significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the p-value?

    Example 10.4

    Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are the coins fair? Test at a 5% significance level.

    Try It 10.4

    Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. Table 10.14 shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom?

    MDG Region Adult Literacy Rate (%)
    Developed Regions 99.0
    Commonwealth of Independent States 99.5
    Northern Africa 67.3
    Sub-Saharan Africa 62.5
    Latin America and the Caribbean 91.0
    Eastern Asia 93.8
    Southern Asia 61.9
    South-Eastern Asia 91.9
    Western Asia 84.5
    Oceania 66.4
    Table 10.14

    10.2: Goodness-of-Fit Test is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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