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Give an example of a situation where it would be appropriate to use a chi- square test of homogeneity. Describe the populations that would be sampled and the variable that would be recorded.

Short Answer

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The Chi-square test of homogeneity can be used in a situation where a company wants to understand the preference for different soda flavors in different cities. The populations sampled would be the residents of the three cities, and the variable recorded would be the preferred soda flavor.

Step by step solution

01

Understanding Chi-Square Test of Homogeneity

A chi-square test of homogeneity is applied when there's a need to determine whether different populations have the same distribution for one variable. It's a non-parametric statistical test. Non-parametric means it makes no assumptions about population parameters and the distribution's shape. Hence, it could be used in various situations, including market research, clinical trials, voter preferences, among others.
02

Example Scenario

Let's consider an example scenario in market research. Suppose a company sells soda drink in three different flavors: cola, lemon, and orange, and is planning to launch a new flavor into three different cities (city A, city B, city C). Before investing further in the production of the new flavor, the company must first understand which existing flavor is most preferred in each city. The company will conduct taste tests in different cities and record the flavor each participant prefers.
03

Describing the Populations and Variable

In this situation, the populations sampled would be the residents of the three different cities (city A, city B, city C). The variable recorded would be the preferred soda flavor (cola, lemon, or orange). Once the data is collected, a chi-square test of homogeneity can be performed to analyze if there is a significant difference among the populations in their preference for soda flavors.

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Explain the difference between situations that would lead to a chi-square test for homogeneity and those that would lead to a chi-square test for independence.

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