1 Outline 1.Count data 2.Properties of the multinomial experiment 3.Testing the null hypothesis 4.Examples

2 Count data Sometimes, the data we have to analyze are produced by counting things. – How many people choose each of Brands A, B, and C of coffee?

3 Count data Usually, we count things in a sample in order to make an inference to a population. – E.g., are the proportions of people choosing each brand different from one another? – Or, are the proportions of people choosing each brand different from some hypothetical values in the population?

4 Count data To answer such questions, we need to know approximately how much difference between the various counts could be produced by sampling error. We determine that quantity using the ‘multinomial probability distribution,’ an extension of the binomial probability distribution.

5 Properties of the Multinomial Experiment 1.There are n identical trials 2.There are k possible outcomes on each trial 3.The probabilities of the outcomes are the same across trials 4.Trials are all independent of each other 5.The multinomial random variables are the k values n 1, n 2, …, n k.

6 Testing the null hypothesis We often want to test the null hypothesis that all the categories are equal in frequency. If we asked 60 people which of Brands A, B, and C they prefer, equal frequency would look like this: ABC

7 Testing the null hypothesis At other times, we might want to test a specific null hypothesis, such as that B and C are equally popular, but A is twice as popular as either: ABC In both cases, we call the values shown the “expected values.”

8 Testing the null hypothesis The null hypothesis can be tested using the statistic χ 2. χ 2 = Σ[n i – E(n i )] 2 E(n i ) χ 2 increases as the observed values, n i, get further from the expected values E(n i ).

9 Chi-square – example Suppose we want to know whether there is any population preference for brands of coffee among brands A, B, and C. We need to know two things: – How should choices among the brands be distributed in a sample if there is no preference (all are equally popular)? – How are choices distributed in our sample?

10 Chi-square – example We ask a sample of 90 people for their preference If there is no preference, each brand should be chosen by ⅓ of the people asked: ABC These are the “expected values” – – expected if the null hypothesis is true

11 Chi-square – example We ask a sample of 90 people for their preference The actual choices look like this: ABC These are the “observed values”

12 Expected vs. Observed Values ABC ABC Expected values – each value = ⅓ * 90 Observed values

13 Chi-square – example χ 2 = Σ[n i – E(n i )] 2 E(n i ) χ 2 = (15-30) 2 + (42-30) 2 + (33-30) = 12.6

14 Chi-square – the formal hypothesis test H O : P A = P B = P C = ⅓ H A : Something different – at least one P > ⅓ Test statistic: χ 2 = Σ[n i – E(n i )] 2 E(n i ) where d.f. = (k-1; k = number of categories)

15 Chi-square – the formal hypothesis test Rejection region: χ 2 obt > χ 2 crit = χ 2 (.05, 2) = (note: rejection region is always > χ 2 crit ) Decision: since χ 2 obt > χ 2 crit, reject HO. Brands are not equally popular

16 Chi-square – Example 1 At a recent meeting of the Coin Flippers Society, each member flipped three coins simultaneously and the number of tails occurring was recorded. Shown below are the numbers of members who had certain numbers of tails. Is there evidence that the coin flipping outcomes were different from what would be expected if all the coins used were fair? (α =.01) Number of TailsNumber of Members

17 Chi-square – Example 1 Shown below are the numbers of members who had certain numbers of tails. Number of tails = the categories people fall into Number of members = number of people in each category. Number of members is the dependent variable. Do you see why?

18 Chi-square – Example 1 To begin, we need to compute the expected values for each of the categories. That is, we need to figure out how many of our 500 members would fall into each category if all the coins used were fair. Wait a minute! How do we know there are 500 members?

19 Chi-square – Example 1 At a recent meeting of the Coin Flippers Society, each member flipped three coins simultaneously and the number of tails occurring was recorded. Shown below are the numbers of members who had certain numbers of tails. Is there evidence that the coin flipping outcomes were different from what would be expected if all the coins used were fair? (α =.01) Number of TailsNumber of Members Σ = 500

20 Chi-square – Example 1 How many possible outcomes are there for one trial? HHH HHT HTH THH HTT THT THH TTT There are 8 possible outcomes

21 Chi-square – Example 1 Of these eight possible outcomes, how many involve getting 0 tails? Just one – HHH. How many involve getting 1 tail? 3 – HHT, HTH, THH. How many involve getting 2 tails? 3 – HTT, THT, TTH. How many involve getting 3 tails? 1 - TTT

22 Chi-square – Example 1 H O : P 0 =.125, P 1 =.375, P 2 =.375, P 3 =.125 H A : At least one P is different from the value specified in H O. Test statistic: χ 2 = Σ[n i – E(n i )] 2 E(n i ) Rejection region: χ 2 obt > χ 2 crit = χ 2 (.01, 3) =

23 Chi-square – Example 1 Now we compute the expected values using (a) the probabilities in H O and (b) our sample n: P 0 * 500 =.125 * 500 = 62.5 P 1 * 500 =.375 * 500 = P 2 * 500 =.375 * 500 = P 3 * 500 =.125 * 500 = 62.5

24 Chi-square – Example 1 χ 2 = [65–62.5] 2 + [182–187.5] 2 + [194–187.5] 2 + [59–62.5] = Decision: Do not reject. There is no evidence that the coin flipping outcomes were different from what would be expected if all the coins used were fair.

25 Chi-square – Example 2 There is an “old wives’ tale” that babies don’t tend to be born randomly during the day but tend more to be born in the middle of the night, specifically between the hours of 1 AM and 5 AM. To investigate this, a researcher collects birth-time data from a large maternity hospital. The day was broken into 4 parts: Morning (5 AM to 1 PM), Mid-day (1 PM to 5 PM), Evening (5 PM to 1 AM), and Mid-night (1 AM to 5 AM). The number of births at these times for the last three months (January to March) are shown on the next slide.

26 Chi-square – Example 2 Morning110 Mid-day 50 Evening 100 Mid-night100 Does it appear that births are not randomly distributed throughout the day? (α =.01)

27 Chi-square – Example 2 The critical thing about a chi-square question is usually the expected values. In the previous example, we computed the expected values on the basis of probabilities of various outcomes for a fair coin. In this question, expected values for the number of births in each segment of the day will be based on one variable: how long in hours is each segment.

28 Chi-square – Example 2 Morning: 5 AM to 1 PM = 8 hours Mid-day: 1 PM to 5 PM = 4 hours Evening: 5 PM to 1 AM = 8 hours Mid-night: 1 AM to 5 AM = 4 hours These periods are not all equal in length!

29 Chi-square – Example 2 If time of day was irrelevant to when babies are born, we would expect every period of, say, 4 hours to produce the same number of babies. Since the Morning and Evening segments each contain two 4- hour periods and the Mid-day and Midnight segments each contain one 4-hour period, our expected values will be: MorningMid-dayEveningMidnight 1/3 1/6 1/3 1/6

30 Chi-square – Example 2 Our sample totals 360 babies. In 1/6 of a day (4 hours) we would expect 360/6 = 60 babies to be born, under the null hypothesis, giving these expected values for the four segments of the day: MorningMid-dayEveningMidnight

31 Chi-square – Example 2 H O : P morn = 1/3, P midday = 1/6, P even = 1/3, P midnight = 1/3 H A : At least one P different from value specified in H O. Test statistic: χ 2 = Σ[n i – E(n i )] 2 E(n i ) Rejection region: χ 2 obt > χ 2 crit = χ 2 (.05, 3) = 7.81

32 Chi-square – Example 2 χ 2 obt = [ ] 2 + … + [100-60] =

33 Chi-square – Example 2 χ 2 obt = Decision: Reject H O. Births are not randomly scattered throughout the day.