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**CHAPTER 23: Two Categorical Variables: The Chi-Square Test**

Basic Practice of Statistics - 3rd Edition CHAPTER 23: Two Categorical Variables: The Chi-Square Test The Basic Practice of Statistics 6th Edition Moore / Notz / Fligner Lecture PowerPoint Slides Chapter 5

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**Chapter 23 Concepts Two-Way Tables The Problem of Multiple Comparisons**

Expected Counts in Two-Way Tables The Chi-Square Test Statistic Cell Counts Required for the Chi-Square Test Uses of the Chi-Square Tests The Chi-Square Distributions

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**Chapter 23 Objectives Construct and interpret two-way tables**

Describe the problem of multiple comparisons Calculate expected counts in two-way tables Describe the chi-square test statistic Describe the cell counts required for the chi- square test Describe uses of the chi-square test Describe the chi-square distributions Perform a chi-square goodness of fit test

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Two-Way Tables The two-sample z procedures of Chapter 21 allow us to compare the proportions of successes in two populations or for two treatments. What if we want to compare more than two samples or groups? More generally, what if we want to compare the distributions of a single categorical variable across several populations or treatments? We need a new statistical test. The new test starts by presenting the data in a two-way table. Two-way tables of counts have more general uses than comparing distributions of a single categorical variable. They can be used to describe relationships between any two categorical variables.

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Two-Way Tables Market researchers suspect that background music may affect the mood and buying behavior of customers. One study in a supermarket compared three randomly assigned treatments: no music, French accordion music, and Italian string music. Under each condition, the researchers recorded the numbers of bottles of French, Italian, and other wine purchased. Here is a table that summarizes the data: PROBLEM: (a) Calculate the conditional distribution (in proportions) of the type of wine sold for each treatment. (b) Make an appropriate graph for comparing the conditional distributions in part (a). (c) Are the distributions of wine purchases under the three music treatments similar or different? Give appropriate evidence from parts (a) and (b) to support your answer.

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Two-Way Tables The type of wine that customers buy seems to differ considerably across the three music treatments. Sales of Italian wine are very low (1.3%) when French music is playing but are higher when Italian music (22.6%) or no music (13.1%) is playing. French wine appears popular in this market, selling well under all music conditions but notably better when French music is playing. For all three music treatments, the percent of Other wine purchases was similar.

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**The Problem of Multiple Comparisons**

To perform a test of H0: There is no difference in the distribution of a categorical variable for several populations or treatments. Ha: There is a difference in the distribution of a categorical variable for several populations or treatments. we compare the observed counts in a two-way table with the counts we would expect if H0 were true. The problem of how to do many comparisons at once with an overall measure of confidence in all our conclusions is common in statistics. This is the problem of multiple comparisons. Statistical methods for dealing with multiple comparisons usually have two parts: 1. An overall test to see if there is good evidence of any differences among the parameters that we want to compare. 2. A detailed follow-up analysis to decide which of the parameters differ and to estimate how large the differences are. The overall test uses the chi-square statistic and distributions.

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**Expected Counts in Two-Way Tables**

Finding the expected counts is not that difficult, as the following example illustrates. The overall proportion of French wine bought during the study was 99/243 = So the expected counts of French wine bought under each treatment are: The null hypothesis in the wine and music experiment is that there’s no difference in the distribution of wine purchases in the store when no music, French accordion music, or Italian string music is played. To find the expected counts, we start by assuming that H0 is true. We can see from the two-way table that 99 of the 243 bottles of wine bought during the study were French wines. The overall proportion of Italian wine bought during the study was 31/243 = So the expected counts of Italian wine bought under each treatment are: If the specific type of music that’s playing has no effect on wine purchases, the proportion of French wine sold under each music condition should be 99/243 = The overall proportion of Other wine bought during the study was 113/243 = So the expected counts of Other wine bought under each treatment are:

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**Expected Counts in Two-Way Tables**

Consider the expected count of French wine bought when no music was playing: 99 99 84 84 243 243 The values in the calculation are the row total for French wine, the column total for no music, and the table total. We can rewrite the original calculation as: = 34.22 • The expected count in any cell of a two-way table when H0 is true is: Expected Counts

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**The Chi-Square Statistic**

To see if the data give convincing evidence against the null hypothesis, we compare the observed counts from our sample with the expected counts assuming H0 is true. The test statistic that makes the comparison is the chi-square statistic. The chi-square statistic is a measure of how far the observed counts are from the expected counts. The formula for the statistic is:

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**Cell Counts Required for the Chi-Square Test**

The chi-square test is an approximate method that becomes more accurate as the counts in the cells of the table get larger. We must therefore check that the counts are large enough to allow us to trust the P-value. Fortunately, the chi-square approximation is accurate for quite modest counts. Cell Counts Required for the Chi-Square Test You can safely use the chi-square test with critical values from the chi-square distribution when no more than 20% of the expected counts are less than 5 and all individual expected counts are 1 or greater. In particular, all four expected counts in a 2 2 table should be 5 or greater.

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**Chi-Square Calculation**

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The Chi-Square Test The chi-square test is an overall test for detecting relationships between two categorical variables. If the test is significant, it is important to look at the data to learn the nature of the relationship. We have three ways to look at the data: Compare selected percents: which cells occur in quite different percents of all cells? Compare observed and expected cell counts: which cells have more or less observations than we would expect if H0 were true? Look at the terms of the chi-square statistic: which cells contribute the most to the value of χ2?

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**Uses of the Chi-Square Test**

One of the most useful properties of the chi-square test is that it tests the null hypothesis “the row and column variables are not related to each other” whenever this hypothesis makes sense for a two-way table. Uses of the Chi-Square Test Use the chi-square test to test the null hypothesis H0: there is no relationship between two categorical variables when you have a two-way table from one of these situations: Independent SRSs from two or more populations, with each individual classified according to one categorical variable. A single SRS, with each individual classified according to both of two categorical variables.

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**The Chi-Square Distributions**

Software usually finds P-values for us. The P-value for a chi-square test comes from comparing the value of the chi-square statistic with critical values for a chi-square distribution. The chi-square distributions are a family of distributions that take only positive values and are skewed to the right. A particular chi-square distribution is specified by giving its degrees of freedom. The chi-square test for a two-way table with r rows and c columns uses critical values from the chi-square distribution with (r – 1)(c – 1) degrees of freedom. The P-value is the area under the density curve of this chi-square distribution to the right of the value of the test statistic. The Chi-Square Distributions

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**Chi-Square Calculation**

H0: There is no difference in the distributions of wine purchases at this store when no music, French accordion music, or Italian string music is played. Ha: There is a difference in the distributions of wine purchases at this store when no music, French accordion music, or Italian string music is played. Our calculated test statistic is χ2 = To find the P-value using a chi-square table look in the df = (3-1)(3-1) = 4. P df .0025 .001 4 16.42 18.47 The small P-value (between and ) gives us convincing evidence to reject H0 and conclude that there is a difference in the distributions of wine purchases at this store when no music, French accordion music, or Italian string music is played.

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**The Chi-Square Test for Goodness of Fit**

Mars, Inc. makes milk chocolate candies. Here’s what the company’s Consumer Affairs Department says about the color distribution of its M&M’S milk chocolate candies: On average, the new mix of colors of M&M’S milk chocolate candies will contain 13 percent of each of browns and reds, 14 percent yellows, 16 percent greens, 20 percent oranges, and 24 percent blues. The one-way table below summarizes the data from a sample bag of M&M’S milk chocolate candies. In general, one-way tables display the distribution of a categorical variable for the individuals in a sample. Color Blue Orange Green Yellow Red Brown Total Count 9 8 12 15 10 6 60

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**The Chi-Square Test for Goodness of Fit**

Since the company claims that 24% of all M&M’S milk chocolate candies are blue, we might believe that something fishy is going on. We could use the z test for a proportion to test the hypotheses: H0: p = 0.24 Ha: p ≠ 0.24 where p is the true population proportion of blue M&M’S. We could then perform additional significance tests for each of the remaining colors. However, performing a one-sample z test for each proportion would be pretty inefficient and would lead to the problem of multiple comparisons. More important, performing one-sample z tests for each color wouldn’t tell us how likely it is to get a random sample of 60 candies with a color distribution that differs as much from the one claimed by the company as this bag does (taking all the colors into consideration at one time). For that, we need a new kind of significance test, called a chi-square test for goodness-of-fit.

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**The Chi-Square Test for Goodness of Fit**

We can write the hypotheses in symbols as: H0: pblue = 0.24, porange = 0.20, pgreen = 0.16, pyellow = 0.14, pred = 0.13, pbrown = 0.13, Ha: At least one of the pi’s is incorrect where pcolor = the true population proportion of M&M’S milk chocolate candies of that color. The idea of the chi-square test for goodness-of-fit is this: we compare the observed counts from our sample with the counts that would be expected if H0 is true. The more the observed counts differ from the expected counts, the more evidence we have against the null hypothesis. In general, the expected counts can be obtained by multiplying the proportion of the population distribution in each category by the sample size.

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**The Chi-Square Test for Goodness of Fit**

Assuming that the color distribution stated by Mars, Inc. is true, 24% of all M&M’S milk chocolate candies produced are blue. For random samples of 60 candies, the average number of blue M&M’S should be (0.24)(60) = This is our expected count of blue M&M’S. Using this same method, we can find the expected counts for the other color categories: Orange: (0.20)(60) = 12.00 Green: (0.16)(60) = 9.60 Yellow: (0.14)(60) = 8.40 Red: (0.13)(60) = 7.80 Brown: (0.13)(60) = 7.80

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**The Chi-Square Test for Goodness of Fit**

To calculate the chi-square statistic, use the same formula as you did earlier in the chapter.

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**The Chi-Square Test for Goodness of Fit**

A categorical variable has k possible outcomes, with probabilities p1, p2, p3, … , pk. That is, pi is the probability of the ith outcome. We have n independent observations from this categorical variable. To test the null hypothesis that the probabilities have specified values H0: p1 = ___, p2 = ___, …, pk = ___. find the expected count for each category assuming that H0 is true. Then calculate the chi-square statistic: The Chi-Square Test for Goodness-of-Fit

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**The Chi-Square Test for Goodness of Fit**

P df .15 .10 .05 4 6.74 7.78 9.49 5 8.12 9.24 11.07 6 9.45 10.64 12.59 Since our P-value is between 0.05 and 0.10, it is greater than α = Therefore, we fail to reject H0. We don’t have sufficient evidence to conclude that the company’s claimed color distribution is incorrect.

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**Chapter 23 Objectives Review**

Construct and interpret two-way tables Describe the problem of multiple comparisons Calculate expected counts in two-way tables Describe the chi-square test statistic Describe the cell counts required for the chi- square test Describe uses of the chi-square test Describe the chi-square distributions Perform a chi-square goodness of fit test

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