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Chapter 12: Problem 32

If the expected count of a category is less than 1 , what can be done to the categories so that a goodness-of-fit test can still be performed?

Short Answer

Expert verified
Combine categories with expected counts less than 1 with adjacent categories to meet the requirements for the test.

Step by step solution

01

- Identify the Issue

Understand that the expected count of some categories is less than 1, which can make the chi-square goodness-of-fit test invalid.
02

- Combine Categories

Combine categories with expected counts less than 1 with adjacent categories until the combined expected count is at least 1.
03

- Update Expected Counts

Recalculate the expected counts for the combined categories to ensure they meet the requirement for the goodness-of-fit test.
04

- Perform the Goodness-of-Fit Test

Now, with all expected counts being at least 1, perform the chi-square goodness-of-fit test on the modified categories.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

expected count
In a chi-square goodness-of-fit test, the expected count is crucial. The expected count represents the frequency we expect in each category if the null hypothesis is true. This is calculated by multiplying the total sample size by the proportion of that category under the null hypothesis. For example, if we have 100 observations divided into 5 categories where each category is expected to have 20 observations, then each category's expected count is 20. Ensuring that these counts are correct is essential for the validity of the chi-square test. If an expected count is less than 1, it can lead to inaccurate test results.

By accurately calculating expected counts, we can compare observed frequencies to what we would expect under the null hypothesis and determine if there is a significant difference.
combining categories
Sometimes, certain categories in your data might have very low expected counts, less than 1. This can invalidate your chi-square goodness-of-fit test. A practical solution is to combine these categories with others until their combined expected count is greater than or equal to 1. This can involve merging adjacent categories or grouping them in a way that makes sense given the context of the data.

Combining categories helps maintain the test's assumptions, ensuring results are accurate and meaningful. For instance, if two categories have expected counts of 0.5 each, combining them results in a new category with an expected count of 1. This simple adjustment allows you to proceed with your chi-square test without compromising its validity.
recalculate expected counts
After combining categories, it's important to recalculate the expected counts. This ensures that the modified categories now meet the assumption of having expected counts of at least 1. Begin by assigning the new combined categories and then calculate the new expected counts based on how you pooled the data.

For example, if you combined two categories with expected counts of 0.5, the new combined category should have an expected count of 1. Recalculating these counts is a crucial step in preparing your data for the chi-square goodness-of-fit test. Once updated, you can proceed confidently with the test, knowing that all assumptions are met.

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Most popular questions from this chapter

Explain the differences between the chi-square test for independence and the chi-square test for homogeneity. What are the similarities?

Determine the expected counts for each outcome. $$ \begin{array}{lllll} \hline \boldsymbol{n}=\mathbf{5 0 0} & & & & \\ \hline p_{i} & 0.2 & 0.1 & 0.45 & 0.25 \\ \hline \text { Expected counts } & & & & \\ \hline \end{array} $$

Family Structure and Sexual Activity A sociologist wants to discover whether the sexual activity of females between the ages of 15 and 19 years and family structure are associated. She randomly selects 380 females between the ages of 15 and 19 years and asks each to disclose her family structure at age 14 and whether she has had sexual intercourse. The results are shown in the table. Data are based on information obtained from the National Center for Health Statistics. $$ \begin{array}{lcccc} &&{\text { Family Structure }} \\ \hline & \text { Both Biological } & & & \\ \text { Had Sexual } & \text { or Adoptive } & \text { Single } & \text { Parent and } & \text { Nonparental } \\ \text { Intercourse } & \text { Parents } & \text { Parent } & \text { Stepparent } & \text { Guardian } \\ \hline \text { Yes } & 64 & 59 & 44 & 32 \\ \hline \text { No } & 86 & 41 & 36 & 18 \\ \hline \end{array} $$ (a) Compute the expected values of each cell under the assumption of independence. (b) Verify that the requirements for performing a chi-square test of independence are satisfied. (c) Compute the chi-square test statistic. (d) Test whether family structure and sexual activity of 15 - to 19-year-old females are independent at the \(\alpha=0.05\) level of significance. (e) Compare the observed frequencies with the expected frequencies. Which cell contributed most to the test statistic? Was the expected frequency greater than or less than the observed frequency? What does this information tell you? (f) Construct a conditional distribution by family structure and draw a bar graph. Does this evidence support your conclusion in part (d)?

Explain why chi-square goodness-of-fit tests are always right tailed.

According to the manufacturer of M\&Ms, \(13 \%\) of the plain M\&Ms in a bag should be brown, \(14 \%\) yellow, \(13 \%\) red, \(24 \%\) blue \(, 20 \%\) orange, and \(16 \%\) green. A student randomly selected a bag of plain M\&Ms. He counted the number of \(\mathrm{M} \& \mathrm{Ms}\) that were each color and obtained the results shown in the table. Test whether plain M\&Ms follow the distribution stated by M\&M/Mars at the \(\alpha=0.05\) level of significance. $$ \begin{array}{lc} \text { Color } & \text { Frequency } \\ \hline \text { Brown } & 57 \\ \hline \text { Yellow } & 64 \\ \hline \text { Red } & 54 \\ \hline \text { Blue } & 75 \\ \hline \text { Orange } & 86 \\ \hline \text { Green } & 64\\\ \hline \end{array} $$
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