O represents the observed frequency of an outcome.

Transcription

1 The Multinomial Experiment Unlike a binomial experiment which only has two possible outcomes (e.g. heads or tails), a multinomial experiment: Stat 104: Quantitative Methods for Economists Class 6: Chi-Square Tests Consists of a fixed number, n, of trials. Each trial can have one of k outcomes, called cells. Each probability p i remains constant. Our usual notion of probabilities holds, namely: p 1 + p + + p k = 1, and Each trial is independent of the other trials. 1 4 Two Techniques The first is a goodness-of-fit test applied to data produced by a multinomial experiment, a generalization of a binomial experiment and is used to describe one population of data. The second uses data arranged in a contingency table to determine whether two classifications of a population of nominal data are statistically independent; this test can also be interpreted as a comparison of two or more populations. Chi-squared Goodness-of-Fit Test We test whether there is sufficient evidence to reject a specified set of values for p i. To illustrate, our null hypothesis is: H 0 : p 1 = a 1, p = a,, p k = a k where a 1, a,, a k are the values we want to test. In both cases, we use the chi-squared ( ) distribution. Our research hypothesis is: H a : At least one p i is not equal to its specified value 5 Goodness of Fit Goodness-of-Fit Notation A goodness-of-fit test is used to test the hypothesis that an observed frequency distribution fits (or conforms to) some claimed distribution. It can be used to test for normality, or if some observed data follow some other distribution. We will use it simply in the multinomial setting. O represents the observed frequency of an outcome. E represents the expected frequency of an outcome. k n represents the number of different categories or outcomes. represents the total number of trials

2 Two companies, A and B, have recently conducted aggressive advertising campaigns to maintain and possibly increase their respective shares of the market for fabric softener. These two companies enjoy a dominant position in the market. Before the advertising campaigns began, the market share of company A was 45%, whereas company B had 40% of the market. Other competitors accounted for the remaining 15%. Has their market share changed after the campaign? Test Statistic If the null hypothesis is true, we would expect the number of customers selecting brand A, brand B, and other to be 00 times the proportions specified under the null hypothesis. That is, e 1 = 00(.45) = 90 e = 00(.40) = 80 e 3 = 00(.15) = 30 In general, the expected frequency for each cell is given by e i = np i This expression is derived from the formula for the expected value of a binomial random variable, which we covered a few weeks ago A marketing analyst solicited the preferences of a random sample of 00 customers of fabric softener. Of the 00 customers, 10 indicated a preference for company A's product, 8 preferred company B's fabric softener, and the remaining 16 preferred the products of one of the competitors. Can the analyst infer at the 5% significance level that customer preferences have changed from their levels before the advertising campaigns were launched? If the expected frequencies and the observed frequencies are quite different, we would conclude that the null hypothesis is false, and we would reject it. However, if the expected and observed frequencies are similar, we would not reject the null hypothesis. The test statistic measures the similarity of the expected and observed frequencies We compare market share before and after an advertising campaign to see if there is a difference (i.e. if the advertising was effective in improving market share). We hypothesize values for the parameters equal to the before-market share. That is, H 0 : p 1 =.45, p =.40, p 3 =.15 The alternative hypothesis is a denial of the null. That is, H a : At least one p i is not equal to its specified value Chi-squared Goodness-of-Fit Test The Chi-squared goodness of fit test statistic is given by: observed frequency ( oi ei ) χ = e What is the smallest value possible for this statistic? What do small values imply? What do large values imply? expected frequency Note: this statistic is approximately Chi-squared with k 1 degrees of freedom provided the sample size is large. The decision rule is reject if: i 9 1

3 The Chi-Square Distribution The Chi-Square distribution is a probability distribution like the Normal, Binomial, or t. It is a continuous probability distribution, whose shape depends on its degrees of freedom. In order to calculate our test statistic, we lay-out the data in a tabular fashion for easier calculation by hand: Company Observed Frequency Expected Frequency Delta Summation Component o i e i (o i ei) (o i e i ) /e i A χ function for different degrees of freedom B Others Total Check that these are equal ( oi e ) χ = e i i Critical Chi-Square Critical values for chi-square are found on tables, sorted by degrees of freedom and probability levels. Be sure to use the 0.05 confidence level. If your calculated chi-square value is greater than the critical value from the table, you reject the null hypothesis. If your chi-square value is less than the critical value, you fail to reject the null hypothesis. Our rejection region is: Since our test statistic is 8.18 which is greater than our critical value for Chi-squared, we reject H 0 in favor of H a, that is, There is sufficient evidence to infer that the proportions have changed since the advertising campaigns were implemented Chi-Square Table Using the Computer Stata does not have this procedure built in (it s a macro one can download from the web). There are many online calculators, such as the one at

4 For our Data Another Vista Health Guard, a Pennsylvania health clinic with 5 offices, is open seven days a week. The new operations manager was recently hired from a similar position at a smaller chain of clinics in Florida. She is naturally concerned that the level of staffing physicians, nurses, and other support personnel be balanced with patient demand. Currently, the staffing level is balanced Monday through Friday, with reduced staffing on Saturday and Sunday. Her predecessor explained that patient demand is fairly level throughout the week and about 5% less on weekends, but the new manager suspects that the staff members want to have weekends free. Although she was willing to operate with this schedule for a while, she has decided to study patient demand to see whether the assumed demand pattern still applies. 19 Goodness-of-Fit Requirements 1. The data have been randomly selected.. The sample data consist of frequency counts for each of the different categories. 3. For each category, the expected frequency is at least 5. (The expected frequency for a category is the frequency that would occur if the data actually have the distribution that is being claimed. There is no requirement that the observed frequency for each category must be at least 5.) The operations manager requested a random sample of 0 days for each day of the week that showed the number of patients on each of the sample days. For the 140 days observed, the total count was 56,000 patients Relationships Among the χ Test Statistic, P-Value, and Goodness-of-Fit Recall that the previous operations manager at Vista Health Guard based his staffing on the premise that from Monday to Friday the patient count remained essentially the same and on Saturdays and Sundays it went down 5%. If this is so, how many of the 56,000 patients would we expect on Monday? How many on Tuesday, and so forth? To figure out this demand, let x represent the weekday demand. Then 5x+.75x+.75x=56000, and solving for x we find x=8615. Then weekend demand is 0.75x = 6461 per weekend day

5 As you can see below, the actual data and the expected data do not match perfectly. However, is the difference enough to warrant a change in staffing patterns? Is there a relationship between undergraduate major and MBA major? Suppose the undergraduate degrees are BA, BEng, BBA, and several others. There are three possible majors for the MBA students, accounting, finance, and marketing. Can the statistician conclude that the undergraduate degree affects the choice of major? 5 8 Computer Output The data are stored in two columns. The first column consist of integers 1,, 3, and 4 representing the undergraduate degree where 1 = BA = BEng 3 = BBA 4 = other The second column lists the MBA major where Conclusion? 1= Accounting = Finance 3 = Marketing 6 9 Chi-squared Test of a Contingency Table Tabulate in Stata The Chi-squared test of a contingency table is used to: determine whether there is enough evidence to infer that two nominal variables are related, and to infer that differences exist among two or more populations of nominal variables. In order to use these techniques, we need to classify the data according to two different criteria

6 The problem objective is to determine whether two variables (undergraduate degree and MBA major) are related. Both variables are nominal. Thus, the technique to use is the chisquared test of a contingency table. The alternative hypotheses specifies what we test. That is, H a : The two variables are dependent If the null hypothesis is true (Remember we always start with this assumption.) and the two nominal variables are independent, then, for example P(BA and Accounting) = [P(BA)] [P(Accounting)] Since we don t know the values of P(BA) or P(Accounting) We need to use the data to estimate the probabilities. The null hypothesis is a denial of the alternative hypothesis. H 0 : The two variables are independent Test Statistic The test statistic is the same as the one used to test proportions in the goodness-of-fit-test. That is, the test statistic is ( oi ei ) χ = e i Note however, that there is a major difference between the two applications. In this one the null does not specify the proportions p i, from which we compute the expected values e i, which we need to calculate the χ test statistic. That is, we cannot use e = np i Estimating the Probabilities There are 15 students of which 61 who have chosen accounting as their MBA major. Thus, we estimate the probability of accounting as Similarly P(Accounting) P(BA) 60 = = because we don t know the p i (they are not specified by the null hypothesis). It is necessary to estimate the p i from the data We start with what is called the contingency table of counts. Undergrad Degree MBA Major Accounting Finance Marketing Total BA BEng BBA If the null hypothesis is true P(BA and Accounting) = (60/15)(61/15) Now that we have the probability we can calculate the expected value. That is, E(BA and Accounting) = 15(60/15)(61/15) = (60)(61)/15 = 4.08 We can do the same for the other 11 cells. Other Total

7 We can now compare observed with expected frequencies Undergrad Degree MBA Major Accounting Finance Marketing BA BEng BBA Is there a difference between handedness patterns in men and women? A good set of data to help you answer this question comes from the government s 5-year Health and Nutrition Survey (HANES) of , which recorded the gender and handedness of a random sample of 37 individuals from across the country. Other and calculate our test statistic: Using Stata Stata Output The p-value is.03. There is enough evidence to infer that the MBA major and the undergraduate degree are related. We can also interpret the results of this test in two other ways. 1.There is enough evidence to infer that there are differences in MBA major between the four undergraduate categories. The p-value is so we Reject Ho. It appears that women are more likely to be right-handed and men are more likely to be left-handed or ambidextrous.. There is enough evidence to infer that there are differences in undergraduate degree between the majors

8 -The Titanic Titanic Survival Data 43 Things you should know Chi-Square Goodness of Fit Test Chi-Square Test for Independence Readings from the book Not in book 44 8