Example. χ 2 = Continued on the next page. All cells

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1 Section 11.1 Chi Square Statistic k Categories 1 st 2 nd 3 rd k th Total Observed Frequencies O 1 O 2 O 3 O k n Expected Frequencies E 1 E 2 E 3 E k n O 1 + O 2 + O O k = n E 1 + E 2 + E E k = n Another word for category is cell. The above table consists of k categories. The first row contains the observed (actual) frequencies of k categories. The second row consists of the expected (theoretical) frequencies of k categories. We want to determine if the observed and expected frequencies agree with each other. This is accomplished with hypothesis tests that employ the χ 2 statistic. Example χ 2 = All cells ( 0 E) 2 Suppose that you want to test the fairness of a six sided die. You roll it 60 times. How many times do you expect the faces 1, 2, 3, 4, 5, and 6 to appear? Since the theoretical probability of rolling any one number of a die is 1/6, we would expect any one number to appear (1/6)(60) = 10 times out of 60 rolls. That is E 1 = 10, E 2 = 10, E 3 = 10, E 4 = 10, E 5 = 10 and E 6 = 10. Suppose that you want to test the fairness of a coin. You toss it times. How many times do you expect heads and tails to appear? Since the theoretical probability of tossing a head or a tail is ½, we would expect a head to appear (1/2)() = 50 and a tail to appear (1/2)() = 50. Hence, E 1 = 50 and E 2 = 50. E 1

2 Section 11.2 Inferences Concerning Multinomial Experiments Multinomial Experiment An experiment with the following characteristics: 1. It consists of n repeated (identical) independent trials. 2. The outcome of each trial fits into exactly on of k possible cells. 3. There is a probability associated with each particular cell, and these individual probabilities remain constant during the experiment. (It must be true that p 1 + p p k = 1.) 4. The experiment will result in a set of k observed frequencies, O 1, O 2, O k, where each O i is the number of times a trial outcome falls into that particular cell. (It must be the case that O 1 + O O k = n.) Elementary Statistics, Eighth Edition, Robert Johnson Patricia Kuby, pp ( 0 E) 2 The hypothesis tests that we conduct use the χ 2 =, where the All cells E degrees of freedom, df = k 1, where k is the number of categories. It is also very important to remember that whenever we conduct a multinomial experiment, it will be a right tail test. The hypothesis statements for multinomial tests follow. H o : p 1 = p 0,1, p 2 = p 0,2,, p k = p 0,k Vs. H a : At least one inequality exists. Examples A certain type of flower seed will produce magenta, chartreuse, and ochre flowers in the ratio 6 : 3 : 1 (one flower per seed). A total of seeds are planted and all germinate, yielding the following results. Magenta Chartreuse Ochre a) If the null hypothesis is true, what is the expected number of magenta, chartreuse, and ochre flowers? seeds are planted. The expected frequency of colors is 6 : 3 : = 10. The expected number of magenta is (6/10)() = 60. 2

3 The expected number of chartreuse is (3/10)() = 30. The expected number of ochre is (1/10)() = 10. A table of expected vs. observed follows. Magenta Chartreuse Ochre Observed Expected b) How many degrees of freedom are associated with χ 2? Degrees of freedom, df = k 1. Since there are three categories, df = 3 1 = 2. c) Complete the hypothesis test using α =.10. State the hypotheses, and identify the claim. H o : p 1 =.6, p 2 =.3, p 3 =.1 (claim) vs. H a : At least one inequality exists. Let s first obtain χ 2. ( 0 E) ( 52 60) ( 36 30) ( 12 10) χ 2 = = + + = All cells E Now let s get the p value. Remember that this is a right tail test, so that will affect the way we obtain our p value. p value = χ 2 cdf(2.667, E99, 2) =.264 The p value is described in the following diagram. p value =.264 χ 2 = Decision: Fail to reject H o. Conclusion: There is enough evidence to support the claim that the hypothesized frequencies of flower colors are correct. The traditional method of hypothesis testing follows on the next page. 3

4 Find the critical value for this test using the INVCHI program. χ 2 (2,.10) = Now draw the picture. α =.10 χ 2 = χ 2 (2,.10) = Clearly, χ 2 is outside of the rejection region. Hence, our decision to fail to reject H o is supported. To obtain χ 2 using your calculator, enter the O cells in L1 and the E cells in L2. Go to L3, hit the UP arrow to highlight L3, then hit ENTER. Now type in the following: (L1 L2) 2 /L2 ENTER. Hit Stat, go to Calc, hit ENTER once. Then L3 ENTER. Your calculator output should look like this. x = s x x = = = σ = x x n = 3 χ 2 = x =

5 Section 11.3 Inferences Concerning Contingency Tables A contingency table is an arrangement of data into a two way classification. The data are sorted into cells, and the number of data in each cell is reported. Elementary Statistics, Eighth Edition, Robert Johnson Patricia Kuby, pp Contingency tables are useful for tests of independence, and tests of homogeneity. We will be using the χ 2 distribution to conduct these tests. Test of Independence Example A survey of randomly selected travelers who visited the service station restrooms of a large U.S. petroleum distributor showed the following results. Quality of Restroom Facilities (Observed) Gender of Respondent Above Average Average Below Average Totals Female Male Totals Using α =.05, does the sample present sufficient evidence to reject the hypothesis, Quality of responses is independent of the gender of the respondent? State the hypotheses. H o : The quality of responses is independent of the gender of the respondent. Vs. H a : The quality of responses is dependent of the gender of the respondent. Obtain the test statistic, χ 2. To do this, we must first obtain a table of expected frequencies. Recall that if two events are independent, then the probability of their intersection is equal to the product of their individual probabilities. Assuming that gender and response are independent, we obtain the EXPECTED FREQUENCY contingency table using the column and row totals of the OBSERVED FREQUENCY contingency table as follows. Row Total Column Total Row Total Column Total Expected Frequency = n = n n n 5

6 Quality of Restroom Facilities (Expected) Gender of Respondent Above Average Average Below Average Female = 8.85 = 29.5 = Male = 6.15 = 20.5 = χ 2 Recall that = All cells ( 0 E) ( ) ( ) ( ) E = ( ) ( ) ( ) = For a two way contingency table, the degrees of freedom are equal to df = (number of rows 1)(number of columns 1) =(r 1 )(c 1) = 1(2) = 2. For a two way contingency table we will always conduct a right tail test. Let s find the p value. p value = χ 2 cdf(9.825, E99, 2) = The p value is described in the following diagram. p value = Decision: Reject H o. χ 2 =

7 We perform the same hypothesis test, but in the traditional manner. Find the critical value, χ 2 (df,.05) = χ 2 (2,.05) = α =.05 χ 2 (2,.05) = χ 2 = Clearly, χ 2 is in the rejection region. Hence, we reject H o. Test of Homogeneity A test of homogeneity is used when the experimenter controls one of the two variables so that the row (or column) totals are predetermined. Elementary Statistics, Eighth Edition, Robert Johnson Patricia Kuby, pp The hypotheses for a test of homogeneity follow. H o : The distribution of proportions in row (or column) 1 is the same as in row (or column) 2 is the same as in row (or column) k. Vs. H a : The distribution of proportions in the rows (or columns) is not the same. These are the only differences between a test of independence and a test of homogeneity. KNOW HOW TO DISTINGUISH BETWEEN THE TWO TESTS! 7

8 Example A study of the Harvard Business School conducted in 1998 (Fortune, Tales of the Trailblazers, 10/12/98) concentrated on the career paths and life styles of the women who were graduates of the program. The focus was on the class of 1973 and the class of The class of 1973 was the first to include a solid number of women. But did things change ten years later? Consider the following table. Harvard Business School Class Men Women Total Total At the.01 level of significance, did the distribution of men and women who completed the program significantly change between 1973 and 1983? Clearly this is a test of homogeneity. Which totals were predetermined? The total number of people graduating from the classes is the answer to this question. State the hypotheses. H o : The proportions of men and women graduating from the Harvard School of Business are the same for the years 1973 and Vs. H a : The proportions of men and women graduating from the Harvard School of Business are not the same for the years 1973 and Obtain the expected frequency table. Harvard Business Men School Class = = Women = =

9 2 Recall that χ = All cells ( 0 E) ( ) ( ) E = ( ) ( ) + + = Degrees of freedom: df = (number of rows 1)(number of columns 1) = (r 1)(c 1) = (2 1)(2 1) = 1. Remembering that we perform a right tail test, p value = χ 2 cdf( , E99, 1) = Decision: Reject H o. The proportions are not the same in the years 1973 and It appears that women are making some headway! A picture of the p value follows. p value = χ 2 =

10 Let s conduct this test the traditional way. Find the critical value, χ 2 (1,.01). χ 2 (1,.01) = Draw the picture of the rejection region, and determine where χ 2 resides. α =.01 χ 2 (1,.01) = χ 2 = The critical is WAY INSIDE the rejection region, so this supports our p value decision to reject H o. 10