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Inference for Relationships: Chi-Squared Test of Homogeneity

This chapter discusses the chi-squared test of homogeneity used to determine if different populations are the same across a specific characteristic. It provides an example of comparing conditional distributions to analyze the effect of background music on wine purchases.

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Inference for Relationships: Chi-Squared Test of Homogeneity

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  1. Chapter 11: Inference for Distributions of Categorical Data Section 11.2 Inference for Relationships The Practice of Statistics, 4th edition – For AP* STARNES, YATES, MOORE

  2. Chapter 11Inference for Distributions of Categorical Data • 11.1 Chi-Square Goodness-of-Fit Tests • 11.2 Inference for Relationships

  3. Chi-Squared Test of Homogeneity Are different populations the same across some characteristic?

  4. c2 test for homogeneity • Used with a single categorical variable from two (or more) independent samples • Used to see if the two (or more) populations are the same (homogeneous) • Several groups but STILL ONE VARIABLE

  5. Inference for Relationships • Example: Comparing Conditional Distributions 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.

  6. Inference for Relationships • Example: Comparing Conditional Distributions 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.

  7. Inference for Relationships • Expected Counts and the Chi-Square Statistic 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 familiar chi-square statistic and distributions.

  8. Assumptions & formula remain the same as for Goodness of Fit! • Samples are from a random sampling or randomized experiment • All expected counts are greater than 5 • Both the samples and observations are independent

  9. Hypotheses – written in words • To perform a test of homogeneity: • H0: There is no difference in the distribution of a categorical variable for several populations or treatments. • (H0: the proportions for the two (or more) distributions are the same) • Ha: There is a difference in the distribution of a categorical variable for several populations or treatments. • (Ha: At least one of the proportions for the distributions is different) • we compare the observed counts in a two-way table with the counts we would expect if H0were true. Be sure to write in context!

  10. The overall proportion of Italian wine bought during the study was 31/243 = 0.128. So the expected counts of Italian wine bought under each treatment are: The overall proportion of Other wine bought during the study was 113/243 = 0.465. So the expected counts of Other wine bought under each treatment are: The overall proportion of French wine bought during the study was 99/243 = 0.407. So the expected counts of French wine bought under each treatment are: Finding the expected counts is not that difficult, as the following example illustrates. Inference for Relationships • Expected Counts and the Chi-Square Statistic 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 H0is true. We can see from the two-way table that 99 of the 243 bottles of wine bought during the study were French wines. 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 = 0.407.

  11. Consider the expected count of French wine bought when no music was playing: Inference for Relationships • Finding Expected Counts 99 99 84 243 84 243 Finding Expected Counts 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: The expected count in any cell of a two-way table when H0 is true is • = 34.22 This suggests a general formula for the expected count in any cell of a two-way table:

  12. In order to calculate a chi-square statistic for the wine example, we must check to make sure the conditions are met: • All the expected counts in the music and wine study are at least 5. This satisfies the Large Sample Size condition. • The Random condition is met because the treatments were assigned at random. • We’re comparing three independent groups in a randomized experiment. But are individual observations (each wine bottle sold) independent? If a customer buys several bottles of wine at the same time, knowing that one bottle is French wine might give us additional information about the other bottles bought by this customer. In that case, the Independent condition would be violated. But if each customer buys only one bottle of wine, this condition is probably met. We can’t be sure, so we’ll proceed to inference with caution. Inference for Relationships • Calculating the Chi-Square Statistic Just as we did with the chi-square goodness-of-fit test, we compare the observed counts with the expected counts using the statistic This time, the sum is over all cells (not including the totals!) in the two-way table.

  13. Inference for Relationships • Calculating The Chi-Square Statistic The tables below show the observed and expected counts for the wine and music experiment. Calculate the chi-square statistic.

  14. Degrees of freedom Or cover up one row & one column & count the number of cells remaining!

  15. 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. • Example: Does Music Influence Purchases? We decided to proceed with caution because, although the Random and Large Sample Size conditions are met, we aren’t sure that individual observations (type of wine bought) are independent. Our calculated test statistic is χ2 = 18.28 with df = 4. The small P-value 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. Furthermore, the random assignment allows us to say that the difference is caused by the music that’s played.

  16. Inference for Relationships • The Chi-Square Test for Homogeneity Chi-Square Test for Homogeneity • Suppose the Random, Large Sample Size, and Independent conditions are • met. You can use the chi-square test for homogeneity to test • 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. • Start by finding the expected counts. Then calculate the chi-square statistic • where the sum is over all cells (not including totals) in the two-way table. If H0is true, the χ2statistic has approximately a chi-square distribution with degrees of freedom = (number of rows – 1) (number of columns - 1). The P-value is the area to the right of χ2 under the corresponding chi-square density curve.

  17. The chi-square test for homogeneity allows us to compare the distribution of a categorical variable for any number of populations or treatments. If the test allows us to reject the null hypothesis of no difference, we then want to do a follow-up analysis that examines the differences in detail. Start by examining which cells in the two-way table show large deviations between the observed and expected counts. Then look at the individual components to see which terms contribute most to the chi-square statistic. • Follow-up Analysis Looking at the output, we see that just two of the nine components that make up the chi-square statistic contribute about 14 (almost 77%) of the total χ2 = 18.28. We are led to a specific conclusion: sales of Italian wine are strongly affected by Italian and French music.

  18. Should Dentist Advertise? • It may seem hard to believe but until the 1970’s most professional organizations prohibited their members from advertising. In 1977, the U.S. Supreme Court ruled that prohibiting doctors and lawyers from advertising violated their free speech rights. • Why do you think professional organizations sought to prohibit their members from advertising? • The paper “Should Dentist Advertise?” (J. of Advertising Research (June 1982): 33 – 38) compared the attitudes of consumers and dentists toward the advertising of dental services. Separate samples of 101 consumers and 124 dentists were asked to respond to the following statement: “I favor the use of advertising by dentists to attract new patients.” • Possible responses were: strongly agree, agree, neutral, disagree, strongly disagree. • The authors were interested in determining whether the two groups—dentists and consumers—differed in their attitudes toward advertising.

  19. Should Dentist Advertise? • H0: • Ha: The true category proportions for all responses are the same for both populations of consumers and dentists. The true category proportions for all responses are not the same for both populations of consumers and dentists.

  20. Observed Data Response Strongly Strongly Group Agree Agree Neutral Disagree Disagree Consumers 34 49 9 4 5 Dentists 9 18 23 28 46 • How do we determine the expected cell count under the assumption of homogeneity?

  21. Observed Data 101 124 • The expected cell counts are estimated from the sample data (assuming that H0 is true) by using …

  22. Expected Values 101 124 19.30 • So the calculation for the first cell is …

  23. Observed Data 14.36 14.36 22.89 19.30 30.08 17.64 28.11 23.70 36.92 17.64 • Students on the right side of the classroom finish the first row and the left side find the expected values for the dentists.

  24. Conditions • So now we can consider the conditions of our analysis. • We will assume the data was randomly selected. • The sample was large enough because every cell in the contingency table had an expected frequency of at least 5. • We can assume each participants’ answer was independent from others.

  25. Test Statistic • Now we can calculate the c 2 test statistic:

  26. Sampling Distribution The two-way table for this situation has 2 rows and 5 columns, so the appropriate degrees of freedom is (2 – 1)(5 – 1) = 4. Since the likelihood of seeing such a large amount of difference between the observed frequencies and what we would expected to have seen if the two populations were homogeneous is so small (approx 0), there is strong evidence against the assumption that the proportions in the response categories are the same for the populations of consumers and dentists.

  27. Post-graduation activities of graduates from an upstate NY high school Have what kids do after graduation changed across three graduating classes?

  28. Could test whether two proportions are the same using a two-proportion z test…. but we have 3 groups. Chi-square goodness-of-fit tests against given proportions (theoretical models) …. but we want to know if choices have changed. So… we’ll use a chi-square test of homogeneity. Homogeneity means that things are the same so we have a built-in null hypothesis – the distribution does not change from group to group. This test looks for differences too large from what we might expect from random sample-to-sample variation.

  29. Ho: The post-high school choices made by classes of 1980, 1990, 2000 have the same distributions Ha: The post-high school choices made by classes of 1980, 1990, 2000 do not have the same distributions Conditions: * categorical data with counts * expected values are all at least 5 * years are independent of each other and randomly selected Degrees of freedom: (R – 1)(C – 1) = 3 * 2 = 6 Test statistic:

  30. = 72.77 P-value = P(x2 > 72.77) < 0.0001 The P-value is very small, so I reject the null hypothesis and conclude there is evidence that the choices made by high-school graduates have changed over the three classes examined.

  31. When we reject the null hypothesis, it’s a good idea to examine residuals. To standardize the residuals: What can this show us?

  32. Inference for Relationships • Example: Cell-Only Telephone Users Random digit dialing telephone surveys used to exclude cell phone numbers. If the opinions of people who have only cell phones differ from those of people who have landline service, the poll results may not represent the entire adult population. The Pew Research Center interviewed separate random samples of cell-only and landline telephone users who were less than 30 years old. Here’s what the Pew survey found about how these people describe their political party affiliation. State: We want to perform a test of H0: There is no difference in the distribution of party affiliation in the cell-only and landline populations. Ha: There is a difference in the distribution of party affiliation in the cell-only and landline populations. We will use α = 0.05.

  33. Inference for Relationships • Example: Cell-Only Telephone Users Plan: If the conditions are met, we should conduct a chi-square test for homogeneity. • Random The data came from separate random samples of 96 cell-only and 104 landline users. • Large Sample Size We followed the steps in the Technology Corner (page 705) to get the expected counts. The calculator screenshot confirms all expected counts ≥ 5. • Independent Researchers took independent samples of cell-only and landline phone users. Sampling without replacement was used, so there need to be at least 10(96) = 960 cell-only users under age 30 and at least 10(104) = 1040 landline users under age 30. This is safe to assume.

  34. Inference for Relationships • Example: Cell-Only Telephone Users Do: Since the conditions are satisfied, we can a perform chi-test for homogeneity. We begin by calculating the test statistic. P-Value: Using df = (3 – 1)(2 – 1) = 2, the P-value is 0.20. Conclude: Because the P-value, 0.20, is greater than α = 0.05, we fail to reject H0. There is not enough evidence to conclude that the distribution of party affiliation differs in the cell-only and landline user populations.

  35. Many studies involve comparing the proportion of successes for each of several populations or treatments. • The two-sample z test from Chapter 10 allows us to test the null hypothesis H0: p1= p2, where p1and p2are the actual proportions of successes for the two populations or treatments. • The chi-square test for homogeneity allows us to test H0: p1= p2= …= pk. This null hypothesis says that there is no difference in the proportions of successes for the k populations or treatments. The alternative hypothesis is Ha: at least two of the pi’sare different. Inference for Relationships • Comparing Several Proportions Caution: Many students incorrectly state Haas “all the proportions are different.” Think about it this way: the opposite of “all the proportions are equal” is “some of the proportions are not equal.”

  36. Inference for Relationships • Example: Cocaine Addiction is Hard to Break Cocaine addicts need cocaine to feel any pleasure, so perhaps giving them an antidepressant drug will help. A three-year study with 72 chronic cocaine users compared an antidepressant drug called desipramine with lithium (a standard drug to treat cocaine addiction) and a placebo. One-third of the subjects were randomly assigned to receive each treatment. Here are the results: State: We want to perform a test of H0: p1 = p2 = p3 there is no difference in the relapse rate for the three treatments. Ha: at least two of the pi’s there is a difference in the relapse rate for are different the three treatments. where pi= the actual proportion of chronic cocaine users like the ones in this experiment who would relapse under treatment i. We will use α = 0.01.

  37. Inference for Relationships • Example: Cocaine Addiction is Hard to Break Plan: If the conditions are met, we should conduct a chi-square test for homogeneity. • Random The subjects were randomly assigned to the treatment groups. • Large Sample Size We can calculate the expected counts from the two-way table assuming H0is true. All the expected counts are ≥5 so the condition is met. • Independent The random assignment helps create three independent groups. If the experiment is conducted properly, then knowing one subject’s relapse status should give us no information about another subject’s outcome. So individual observations are independent.

  38. Inference for Relationships • Example: Cocaine Addiction is Hard to Break Do: Since the conditions are satisfied, we can a perform chi-test for homogeneity. We begin by calculating the test statistic. P-Value: Using df = (3 – 1)(2 – 1) = 2, the calculator give a P-value of 0.0052. Conclude: Because the P-value, 0.0052, is less than α = 0.01, we reject H0. We have sufficient evidence to conclude that the true relapse rates for the three treatments are not all the same.

  39. c2 test for Independence • Used with categorical, bivariate data from ONEsample • Used to see if the two categorical variables are associated (dependent) or not associated (independent) • One sample but two variables

  40. Hypotheses – written in words H0: There is no association between the two variables (Or the variables are independent) Ha: There is an association between the two variables (Or the variables are not independent (dependent)) Be sure to write in context!

  41. Assumptions & formula remain the same! Expected counts & df are found the same way as test for homogeneity. Only change is the hypotheses!

  42. A study from the University of Texas Southwestern Medical Center examined whether the risk of hepatitis C was related to whether people had tattoos and to where they got their tattoos. Data differs from other kinds because they categorize subjects from a single group on two categorical variable rather than on only one.

  43. Is the chance of having hepatitis C independent of tattoo status? If hepatitis status is independent of tattoos, we expect the proportion of people testing positive for hepatitis to be the same for the three levels of tattoo status. Are the categorical variables tattoo status and hepatitis statistically independent? A chi-square test for independence

  44. Ho: Tattoo status and hepatitis status are independent Ha: Tattoo status and hepatitis status are not independent Conditions: * representative sample of people * expected counts are all at least 5 * status is independent Degrees of freedom: (R – 1)(C – 1) = 2 * 1 = 2 Test statistic:

  45. = 57.91 P-value = P(x2 > 57.91) = X2cdf ( 57.61, ∞, 2) ≈ 0

  46. The p-value is very small, so I reject the null hypothesis and conclude that hepatitis status is not independent of tattoo status. Because the expected cell frequency condition was violated, I need to check that the two cells with small expected counts did not influence this result too greatly.

  47. Whenever we reject the null hypothesis, it’s a good idea to examine the residuals. Since counts may be different for cells, we are better off standardizing the residuals. To standardize a cell’s residuals, divide by the square root of its expected value. The + and the – sign indicate whether we observed more cases than we expected, or fewer.

  48. Examining the residuals: largest component: Hepatitis C/Tattoo parlor – suggest that a principal source of infection may be tattoo parlors second largest component: Hepatitis C/no tattoo – those who have no tattoos are less likely to be infected with hepatitis C than we might expect if the two variables are independent

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