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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
The minimum expected frequency for each cell in a chi-square test of independence or homogeneity should be at least 5 to maintain validity. When this condition is not met, alternatives include employing a Fisher's exact test or merging categories to increase cell frequencies.

Step by step solution

01

Understanding the concept of expected frequency

In Chi-square test for independence or homogeneity, expected frequency is given by the formulation \( (Row \ total * Column \ total) / Grand \ total \). This test assumes that the variables are independent.
02

Minimum Expected Frequency

For a chi-square test of independence or homogeneity to be valid, it is often noted that the minimum expected frequency for each cell should be at least 5. This is a general rule of thumb, ensuring that the approximation to the Chi-square distribution is sufficiently accurate.
03

Alternatives for violation of expected frequency

If the expected frequency condition is not satisfied, it means the Chi-square test might lead to unreliable results. In such cases, alternative methods can be employed. These include using a different test like Fisher's Exact Test or implementing a data transformation method such as collapsing categories together to increase cell frequencies.

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Most popular questions from this chapter

Sandpaper is rated by the coarseness of the grit on the paper. Sandpaper that is more coarse will remove material faster. Jobs such as the final sanding of bare wood prior to painting or sanding in between coats of paint require sandpaper that is much finer. A manufacturer of sandpaper rated 220, which is used for the final preparation of bare wood, wants to make sure that the variance of the diameter of the particles in their 220 sandpaper does not exceed \(2.0\) micrometers. Fifty-one randomly selected particles are measured. The variance of the particle diameters is \(2.13\) micrometers. Assume that the distribution of particle diameter is approximately normal. a. Construct the \(95 \%\) confidence intervals for the population variance and standard deviation. b. Test at a \(2.5 \%\) significance level whether the variance of the particle diameters of all particles in 220-rated sandpaper is greater than \(2.0\) micrometers.

Construct the \(98 \%\) confidence intervals for the population variance and standard deviation for the following data, assuming that the respective populations are (approximately) normally distributed. $$ \text { a. } n=21, s^{2}=9.2 \quad \text { b. } n=17, s^{2}=1.7 $$

The October 2011 ISACA Shopping on the Job Survey asked employees, "During the holiday season (November and December), how much total time do you think an average employee at your enterprise spends shopping online using a work- supplied computer or smartphone?" Among those who responded, \(3 \%\) said 0 hours, \(24 \%\) said 1 to 2 hours, \(22 \%\) said 3 to 5 hours, and \(51 \%\) said 6 or more hours (www.isaca. org/SiteCollectionDocuments/2011-ISACA-Shopping- on-the-Job-Survey-US.pdf). Suppose that another poll conducted recently asked the same question of 215 randomly selected business executives, which produced the frequencies listed in the following table. $$ \begin{array}{l|cccc} \hline \text { Response/category } & 0 \text { hours } & 1-2 \text { hours } & 3-5 \text { hours } & 6 \text { or more hours } \\ \hline \text { Frequency } & 2 & 41 & 55 & 117 \\ \hline \end{array} $$ Test at a \(2.5 \%\) significance level whether the distribution of responses for the executive survey differs from that of October 2011 survey of employees.

Find the value of \(\chi^{2}\) for 4 degrees of freedom and A. \(.005\) area in the right tail of the chi-square distribution curve b. \(.05\) area in the left tail of the chi-square distribution curve

The manufacturer of a certain brand of lightbulbs claims that the variance of the lives of these bulbs is 4200 square hours. A consumer agency took a random sample of 25 such bulbs and tested them. The variance of the lives of these bulbs was found to be 5200 square hours. Assume that the lives of all such bulbs are (approximately) normally distributed. a. Make the \(99 \%\) confidence intervals for the variance and standard deviation of the lives of all such bulbs. b. Test at a \(5 \%\) significance level whether the variance of such bulbs is different from 4200 square hours.
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