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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

In Thomas Pynchon's book Gravity Rainbow, the characters discuss whether the Poisson probabilistic model can be used to describe the locations that Germany's feared \(\mathrm{V}-2\) rocket would land in. They divided London into \(0.25-\mathrm{km}^{2}\) regions. They then counted the number of rockets that landed in each region, with the following results: $$ \begin{array}{l} \begin{array}{l} \text { Number of } \\ \text { rocket hits } \end{array} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \begin{array}{l} \text { Observed number } \\ \text { of regions } \end{array} & 229 & 211 & 93 & 35 & 7 & 0 & 0 & 1 \\ \hline \end{array} $$ (a) Estimate the mean number of rocket hits in a region by computing \(\mu=\sum x P(x) .\) Round your answer to four decimal places. (b) Explain why the requirements for conducting a goodness-offit test are not satisfied. (c) After consolidating the table, we obtain the following distribution for rocket hits. Using the Poisson probability model, \(P(x)=\frac{\mu^{x}}{x !} e^{-\mu},\) where \(\mu\) is the mean from part (a), we can obtain the probability distribution for the number of rocket hits. Find the probability of 0 hits in a region. Then find the probability of 1 hit, 2 hits, 3 hits, and 4 or more hits. $$ \begin{array}{lcccc} \hline \begin{array}{l} \text { Number of } \\ \text { rocket hits } \end{array} & 0 & 1 & 2 & 3 & 4 \text { or more } \\ \hline \begin{array}{l} \text { Observed number } \\ \text { of regions } \end{array} & 229 & 211 & 93 & 35 & 8 \\ \hline \end{array} $$ (d) A total of \(n=576\) rockets was fired. Determine the expected number of rocket hits by computing "expected number of rockets" \(=n p,\) where \(p\) is the probability of observing that particular number of hits in the region. (e) Conduct a goodness-of-fit test for the distribution using the \(\alpha=0.05\) level of significance. Do the rocket hits appear to be modeled by a Poisson random variable?

The General Social Survey asked a random sample of adult Americans two questions: (1) Would you favor or oppose a law which would require a person to obtain a police permit before he or she could buy a gun? (2) Do you favor or oppose the death penalty for persons convicted of murder? Results of the survey are below. Does the sample evidence suggest the proportion of adult Americans who favor permits for guns is different from the proportion of adult Americans who favor the death penalty for murder convictions? Use the \(\alpha=0.05\) level of significance. $$ \begin{array}{llcc} & & {\text { Death Penalty }} \\ & & \text { Favor } & \text { Oppose } \\ \hline{\text { Gun Laws }} & \text { Favor } & 176 & 69 \\ & \text { Oppose } & 58 & 17 \\ \hline \end{array} $$

The National Highway Traffic Safety Administration publishes reports about motorcycle fatalities and helmet use. The distribution shows the proportion of fatalities by location of injury for motorcycle accidents. $$ \begin{array}{lccccc} \hline \begin{array}{l} \text { Location } \\ \text { of injury } \end{array} & \begin{array}{l} \text { Multiple } \\ \text { Locations } \end{array} & \text { Head } & \text { Neck } & \text { Thorax } & \begin{array}{l} \text { Abdomen/ } \\ \text { Lumbar/Spine } \end{array} \\ \hline \text { Proportion } & 0.57 & 0.31 & 0.03 & 0.06 & 0.03 \\ \hline \end{array} $$ The following data show the location of injury and number of fatalities for 2068 riders not wearing a helmet. $$ \begin{array}{lccccc} \hline \begin{array}{c} \text { Location } \\ \text { of injury } \end{array} & \begin{array}{l} \text { Multiple } \\ \text { Locations } \end{array} & \text { Head } & \text { Neck } & \text { Thorax } & \begin{array}{l} \text { Abdomen/ } \\ \text { Lumbar/Spine } \end{array} \\ \hline \text { Number } & 1036 & 864 & 38 & 83 & 47 \\ \hline \end{array} $$ (a) Does the distribution of fatal injuries for riders not wearing a helmet follow the distribution for all riders? Use the \(\alpha=0.05\) level of significance. (b) Compare the observed and expected counts for each category. What does this information tell you?

Religion in Congress Is the religious make-up of the United States Congress reflective of that in the general population? The following table shows the religious affiliation of the 535 members of the 114 th Congress along with the religious affiliation of a random sample of 1200 adult Americans. $$ \begin{array}{lcc} \text { Religion } & \begin{array}{c} \text { Number of } \\ \text { Members } \end{array} & \begin{array}{c} \text { Sample of } \\ \text { Residents } \end{array} \\ \hline \text { Protestant } & 306 & 616 \\ \hline \text { Catholic } & 164 & 287 \\ \hline \text { Mormon } & 16 & 20 \\ \hline \text { Orthodox Christian } & 5 & 7 \\ \hline \text { Jewish } & 28 & 20 \\ \hline \text { Buddhist/Muslim/Hindu/Other } & 6 & 57 \\ \hline \text { Unaffiliated/Don't Know/Refused } & 10 & 193 \\ \hline \end{array} $$ (a) Determine the probability distribution for the religious affiliation of the members of the 114 th Congress. (b) Assuming the distribution of the religious affiliation of the adult American population is the same as that of the Congress, determine the number of adult Americans we would expect for each religion from a random sample of 1200 individuals. (c) The data in the third column represent the declared religion of a random sample of 1200 adult Americans (based on data obtained from Pew Research). Do the sample data suggest that the American population has the same distribution of religious affiliation as the 114 th Congress? (d) Explain what the results of your analysis suggest.

Determine \((a)\) the \(\chi^{2}\) test statistic, \((b)\) the degrees of freedom, (c) the critical value using \(\alpha=0.05,\) and (d) test the hypothesis at the \(\alpha=0.05\) level of significance. \(H_{0}: p_{\mathrm{A}}=p_{\mathrm{B}}=p_{\mathrm{C}}=p_{\mathrm{D}}=\frac{1}{4}\) \(H_{1}\) : At least one of the proportions is different from the others. $$ \begin{array}{lcccc} \hline\text { Outcome } & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \\\ \hline \text { Observed } & 30 & 20 & 28 & 22 \\ \hline \text { Expected } & 25 & 25 & 25 & 25 \\ \hline \end{array} $$
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