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Chapter 11: Problem 25

To make a test of independence or homogeneity, what should be the minimum expected frequency for each cell? What are the alternatives if this condition is not satisfied?

Short Answer

Expert verified
Each cell in a contingency table should have a minimum expected frequency of 5 for the Chi-Square test of independence or homogeneity to be valid. If this condition is not met, alternative methods such as combining categories, using exact methods, or resorting to a different test can be considered.

Step by step solution

01

Minimum Expected Frequency

In statistics, it is generally expected that each cell in a contingency table should have a minimum expected frequency of 5 for the Chi-Square test of independence or homogeneity to be valid. This requirement ensures that the sampling distribution of the test statistic approximates a chi-square distribution. If this condition is not met, the use of the Chi-square test could lead to inaccurate results.
02

Alternatives When The Condition Is Not Met

If the minimum expected frequency condition is not met, there are several alternative approaches to continue with the statistical analysis. These include: 1. Combining adjacent categories: Categories with small expected frequencies can be merged with other categories in the contingency table. 2. Using Exact methods: Exact methods such as Fisher's exact test could be used regardless of the sample size, particularly for 2x2 contingency tables. 3. Using a different test: Another statistical test that does not have the minimum expected frequency requirement, like the likelihood ratio test, may be used instead of the Chi-square test.

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Most popular questions from this chapter

Two random samples, one of 95 blue-collar workers and a second of 50 white- collar workers, were taken from a large company. These workers were asked about their views on a certain company issue. The following table gives the results of the survey. \begin{tabular}{lccc} \hline & \multicolumn{3}{c} { Opinion } \\ \cline { 2 - 4 } & Favor & Oppose & Uncertain \\ \hline Blue-collar workers & 44 & 39 & 12 \\ White-collar workers & 21 & 26 & 3 \\ \hline \end{tabular} Using the \(2.5 \%\) significance level, test the null hypothesis that the distributions of opinions are homogeneous for the two groups of workers.

The following are the prices (in dollars) of the same brand of camcorder found at eight stores in Los Angeles. \(\begin{array}{llllllll}568 & 628 & 602 & 642 & 550 & 688 & 615 & 604\end{array}\) A. Using the formula from Chapter 3 , find the sample variance, \(s^{2}\), for these data. b. Make the \(95 \%\) confidence intervals for the population variance and standard deviation. Assume that the prices of this camcorder at all stores in Los Angeles follow a normal distribution. c. Test at the \(5 \%\) significance level whether the population variance is different from 750 square dollars.

You have collected data on a variable, and you want to determine if a normal distribution is a reasonable model for these data. The following table shows how many of the values fall within certain ranges of \(z\) values for these data. \begin{tabular}{lr} \hline Category & Count \\ \hline\(z\) score below \(-2\) & 48 \\ \(z\) score from \(-2\) to less than \(-1.5\) & 67 \\ \(z\) score from \(-1.5\) to less than \(-1\) & 146 \\ \(z\) score from \(-1\) to less than \(-0.5\) & 248 \\ \(z\) score from \(-0.5\) to less than 0 & 187 \\ \(z\) score from 0 to less than \(0.5\) & 125 \\ \(z\) score from \(0.5\) to less than 1 & 88 \\ \(z\) score from 1 to less than \(1.5\) & 47 \\ \(z\) score from \(1.5\) to less than 2 & 25 \\ \(z\) score of 2 or above & 19 \\ \hline Total & 1000 \\ \hline \end{tabular} Perform a hypothesis test to determine if a normal distribution is an appropriate model for these data. Use a significance level of \(5 \%\).

A student who needs to pass an elementary statistics course wonders whether it will make a difference if she takes the course with instructor A rather than instructor B. Observing the final grades given by each instructor in a recent elementary statistics course, she finds that Instructor A gave 48 passing grades in a class of 52 students and Instructor \(\mathrm{B}\) gave 44 passing grades in a class of 54 students. Assume that these classes and grades make simple random samples of all classes and grades of these instructors. a. Compute the value of the standard normal test statistic \(z\) of Section \(10.5 .3\) for the data and use it to find the \(p\) -value when testing for the difference between the proportions of passing grades given by these instructors. b. Construct a \(2 \times 2\) contingency table for these data. Compute the value of the \(\chi^{2}\) test statistic for the test of independence and use it to find the \(p\) -value. c. How do the test statistics in parts a and b compare? How do the \(p\) -values for the tests in parts a and \(\mathrm{b}\) compare? Do you think this is a coincidence, or do you think this will always happen?

Suppose that you have a two-way table with the following row and column totals. \begin{tabular}{l|cccc|c} \hline & & \multicolumn{3}{c|} { Variable 1 } & \\ \cline { 3 - 5 } \multicolumn{2}{l|} {} & A & B & C & Total \\ \hline & \(\mathbf{X}\) & & & & 120 \\ Variable 2 & \(\mathbf{Y}\) & & & & 205 \\ & \(\mathbf{Z}\) & & & & 175 \\ \hline & Total & 165 & 140 & 195 & 500 \\ \hline \end{tabular} The observed values in the cells must be counts, which are nonnegative integers. Calculate the expected counts for the cells under the assumption that the two variables are independent. Based on your calculations, explain why it is impossible for the test statistic to have a value of zero.
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