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12.5: Test for Homogeneity

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    The goodness–of–fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity, can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence.

    The expected value for each cell needs to be at least five in order for you to use this test.

    Hypotheses

    • \(H_{0}\): The distributions of the two populations are the same.
    • \(H_{a}\): The distributions of the two populations are not the same.

    Test Statistic

    • Use a \(\chi^{2}\) test statistic. It is computed in the same way as the test for independence.

    Degrees of Freedom (\(df\))

    • \(df = \text{number of columns} - 1\)

    Requirements

    • All values in the table must be greater than or equal to five.

    Common Uses

    Comparing two populations. For example: men vs. women, before vs. after, east vs. west. The variable is categorical with more than two possible response values.

    Example \(\PageIndex{1}\)

    Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are shown in Table \(\PageIndex{1}\). Do male and female college students have the same distribution of living arrangements?

    Table \(\PageIndex{1}\): Distribution of Living Arragements for College Males and College Females
      Dormitory Apartment With Parents Other
    Males 72 84 49 45
    Females 91 86 88 35

    Answer

    • \(H_{0}\): The distribution of living arrangements for male college students is the same as the distribution of living arrangements for female college students.
    • \(H_{a}\): The distribution of living arrangements for male college students is not the same as the distribution of living arrangements for female college students.

    Degrees of Freedom (\(df\)):

    \(df = \text{number of columns} - 1 = 4 - 1 = 3\)

    Distribution for the test: \(\chi^{2}_{3}\)

    Calculate the test statistic: \(\chi^{2} = 10.1287\) (calculator or computer)

    Probability statement: \(p\text{-value} = P(\chi^{2} > 10.1287) = 0.0175\)

    Compare α and the p-value: Since no \(\alpha\) is given, assume \(\alpha = 0.05\). \(p\text{-value} = 0.0175\). \(\alpha > p\text{-value}\).

    Make a decision: Since \(\alpha > p\text{-value}\), reject \(H_{0}\). This means that the distributions are not the same.

    Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the distributions of living arrangements for male and female college students are not the same.

    Notice that the conclusion is only that the distributions are not the same. We cannot use the test for homogeneity to draw any conclusions about how they differ.

    Example 11.5.2

    Both before and after a recent earthquake, surveys were conducted asking voters which of the three candidates they planned on voting for in the upcoming city council election. Has there been a change since the earthquake? Use a level of significance of 0.05. Table shows the results of the survey. Has there been a change in the distribution of voter preferences since the earthquake?

      Perez Chung Stevens
    Before 167 128 135
    After 214 197 225

    Answer

    \(H_{0}\): The distribution of voter preferences was the same before and after the earthquake.

    \(H_{a}\): The distribution of voter preferences was not the same before and after the earthquake.

    Degrees of Freedom (df):

    \(df = \text{number of columns} - 1 = 3 - 1 = 2\)

    Distribution for the test: \(\chi^{2}_{2}\)

    Calculate the test statistic: \(\chi^{2} = 3.2603\) (calculator or computer)

    Probability statement: \(p\text{-value} = P(\chi^{2} > 3.2603) = 0.1959\)

    Compare \(\alpha\) and the \(p\text{-value}\): \(\alpha = 0.05\) and the \(p\text{-value} = 0.1959\). \(\alpha < p\text{-value}\).

    Make a decision: Since \(\alpha < p\text{-value}\), do not reject \(H_{0}\).

    Conclusion: At a 5% level of significance, from the data, there is insufficient evidence to conclude that the distribution of voter preferences was not the same before and after the earthquake.

    References

    1. Data from the Insurance Institute for Highway Safety, 2013. Available online at www.iihs.org/iihs/ratings (accessed May 24, 2013).
    2. “Energy use (kg of oil equivalent per capita).” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/...G.OE/countries (accessed May 24, 2013).
    3. “Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubsearch/pubsinf...?pubid=2009030 (accessed May 24, 2013).
    4. “Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubs2009/2009030_sup.pdf (accessed May 24, 2013).

    Review

    To assess whether two data sets are derived from the same distribution—which need not be known, you can apply the test for homogeneity that uses the chi-square distribution. The null hypothesis for this test states that the populations of the two data sets come from the same distribution. The test compares the observed values against the expected values if the two populations followed the same distribution. The test is right-tailed. Each observation or cell category must have an expected value of at least five.

    Formula Review

    \(\sum_{i \cdot j} \frac{(O-E)^{2}}{E}\) Homogeneity test statistic where: \(O =\) observed values

    \(E =\) expected values

    \(i =\) number of rows in data contingency table

    \(j =\) number of columns in data contingency table

    \(df = (i −1)(j −1)\) Degrees of freedom

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