1. 1 1 Slide 統計學 Spring 2004 授課教師：統計系余清祥 日期： 2004 年 3 月 23 日 第六週：配適度與獨立性檢定

2. 2 2 Slide Chapter 12 Tests of Goodness of Fit and Independence n Goodness of Fit Test: A Multinomial Population n Tests of Independence: Contingency Tables n Goodness of Fit Test: Poisson and Normal Distributions

3. 3 3 Slide Goodness of Fit Test: A Multinomial Population 1. Set up the null and alternative hypotheses. 2. Select a random sample and record the observed frequency, f i, for each of the k categories. 3. Assuming H 0 is true, compute the expected frequency, e i, in each category by multiplying the category probability by the sample size. continued

4. 4 4 Slide Goodness of Fit Test: A Multinomial Population 4. Compute the value of the test statistic. 5. Reject H 0 if (where  is the significance level and there are k - 1 degrees of freedom).

5. 5 5 Slide Example: Finger Lakes Homes (A) n Multinomial Distribution Goodness of Fit Test Finger Lakes Homes manufactures four models of prefabricated homes, a two-story colonial, a ranch, a split-level, and an A-frame. To help in production planning, management would like to determine if previous customer purchases indicate that there is a preference in the style selected.

6. 6 6 Slide Example: Finger Lakes Homes (A) n Multinomial Distribution Goodness of Fit Test The number of homes sold of each model for 100 sales over the past two years is shown below. Model Colonial Ranch Split-Level A-Frame # Sold 30 20 35 15

7. 7 7 Slide Example: Finger Lakes Homes (A) n Multinomial Distribution Goodness of Fit Test Notation Notation p C = popul. proportion that purchase a colonial p R = popul. proportion that purchase a ranch p S = popul. proportion that purchase a split-level p A = popul. proportion that purchase an A-frame Hypotheses Hypotheses H 0 : p C = p R = p S = p A =.25 H a : The population proportions are not p C =.25, p R =.25, p S =.25, and p A =.25 p C =.25, p R =.25, p S =.25, and p A =.25

8. 8 8 Slide Example: Finger Lakes Homes (A) n Multinomial Distribution Goodness of Fit Test Expected Frequencies Expected Frequencies e 1 =.25(100) = 25 e 2 =.25(100) = 25 e 1 =.25(100) = 25 e 2 =.25(100) = 25 e 3 =.25(100) = 25 e 4 =.25(100) = 25 e 3 =.25(100) = 25 e 4 =.25(100) = 25 Test Statistic Test Statistic = 1 + 1 + 4 + 4 = 1 + 1 + 4 + 4 = 10 = 10

9. 9 9 Slide n Multinomial Distribution Goodness of Fit Test Rejection Rule Rejection Rule With  =.05 and With  =.05 and k - 1 = 4 - 1 = 3 k - 1 = 4 - 1 = 3 degrees of freedom degrees of freedom 22 22 7.81 Do Not Reject H 0 Reject H 0 Example: Finger Lakes Homes (A)

10. 10 Slide Example: Finger Lakes Homes (A) n Multinomial Distribution Goodness of Fit Test Conclusion Conclusion  2 = 10 > 7.81, so we reject the assumption there is no home style preference, at the.05 level of significance. no home style preference, at the.05 level of significance.

11. 11 Slide Test of Independence: Contingency Tables 1. Set up the null and alternative hypotheses. 2. Select a random sample and record the observed frequency, f ij, for each cell of the contingency table. 3. Compute the expected frequency, e ij, for each cell.

12. 12 Slide Test of Independence: Contingency Tables 4. Compute the test statistic. 5. Reject H 0 if (where  is the significance level and with n rows and m columns there are ( n - 1)( m - 1) degrees of freedom).

13. 13 Slide Example: Finger Lakes Homes (B) n Contingency Table (Independence) Test Each home sold can be classified according to price and to style. Finger Lakes Homes’ manager would like to determine if the price of the home and the style of the home are independent variables. Each home sold can be classified according to price and to style. Finger Lakes Homes’ manager would like to determine if the price of the home and the style of the home are independent variables. The number of homes sold for each model and price for the past two years is shown below. For convenience, the price of the home is listed as either $65,000 or less or more than $65,000. Price Colonial Ranch Split-Level A-Frame Price Colonial Ranch Split-Level A-Frame < $65,000 18 6 19 12 < $65,000 18 6 19 12 > $65,000 12 14 16 3 > $65,000 12 14 16 3

14. 14 Slide n Contingency Table (Independence) Test Hypotheses Hypotheses H 0 : Price of the home is independent of the style of the home that is purchased H 0 : Price of the home is independent of the style of the home that is purchased H a : Price of the home is not independent of the H a : Price of the home is not independent of the style of the home that is purchased style of the home that is purchased Expected Frequencies Expected Frequencies Price Colonial Ranch Split-Level A-Frame Total Price Colonial Ranch Split-Level A-Frame Total < $99K 18 6 19 12 55 > $99K 12 14 16 3 45 Total 30 20 35 15 100 Total 30 20 35 15 100 Example: Finger Lakes Homes (B)

15. 15 Slide n Contingency Table (Independence) Test Test Statistic Test Statistic =.1364 + 2.2727 +... + 2.0833 = 9.1486 =.1364 + 2.2727 +... + 2.0833 = 9.1486 Rejection Rule Rejection Rule With  =.05 and (2 - 1)(4 - 1) = 3 d.f., With  =.05 and (2 - 1)(4 - 1) = 3 d.f., Reject H 0 if  2 > 7.81 Reject H 0 if  2 > 7.81 Conclusion Conclusion We reject H 0, the assumption that the price of the home is independent of the style of the home that is purchased. We reject H 0, the assumption that the price of the home is independent of the style of the home that is purchased. Example: Finger Lakes Homes (B)

16. 16 Slide Goodness of Fit Test: Poisson Distribution 1. Set up the null and alternative hypotheses. 2. Select a random sample and a. Record the observed frequency, f i, for each of the k values of the Poisson random variable. k values of the Poisson random variable. b. Compute the mean number of occurrences, . 3. Compute the expected frequency of occurrences, e i, for each value of the Poisson random variable. continued

17. 17 Slide Goodness of Fit Test: Poisson Distribution 4. Compute the value of the test statistic. 5. Reject H 0 if (where  is the significance level and there are k - 2 degrees of freedom).

18. 18 Slide Example: Troy Parking Garage n Poisson Distribution Goodness of Fit Test In studying the need for an additional entrance to a city parking garage, a consultant has recommended an approach that is applicable only in situations where the number of cars entering during a specified time period follows a Poisson distribution. In studying the need for an additional entrance to a city parking garage, a consultant has recommended an approach that is applicable only in situations where the number of cars entering during a specified time period follows a Poisson distribution.

19. 19 Slide Example: Troy Parking Garage n Poisson Distribution Goodness of Fit Test A random sample of 100 one-minute time intervals resulted in the customer arrivals listed below. A statistical test must be conducted to see if the assumption of a Poisson distribution is reasonable. A random sample of 100 one-minute time intervals resulted in the customer arrivals listed below. A statistical test must be conducted to see if the assumption of a Poisson distribution is reasonable. # Arrivals 0 1 2 3 4 5 6 7 8 9 10 11 12 # Arrivals 0 1 2 3 4 5 6 7 8 9 10 11 12 Frequency 0 1 4 10 14 20 12 12 9 8 6 3 1 Frequency 0 1 4 10 14 20 12 12 9 8 6 3 1

20. 20 Slide Example: Troy Parking Garage n Poisson Distribution Goodness of Fit Test Hypotheses Hypotheses H 0 : Number of cars entering the garage during H 0 : Number of cars entering the garage during a one-minute interval is Poisson distributed. a one-minute interval is Poisson distributed. H a : Number of cars entering the garage during a one-minute interval is not Poisson distributed H a : Number of cars entering the garage during a one-minute interval is not Poisson distributed

21. 21 Slide Example: Troy Parking Garage n Poisson Distribution Goodness of Fit Test Estimate of Poisson Probability Function Estimate of Poisson Probability Function  otal Arrivals = 0(0) + 1(1) + 2(4) +... + 12(1) = 600 Total Time Periods = 100 Total Time Periods = 100 Estimate of  = 600/100 = 6 Estimate of  = 600/100 = 6 Hence, Hence,

22. 22 Slide Example: Troy Parking Garage n Poisson Distribution Goodness of Fit Test Expected Frequencies Expected Frequencies x f ( x ) xf ( x ) x f ( x ) xf ( x ) 0.0025.25 7.138913.89 0.0025.25 7.138913.89 1.0149 1.49 8.104110.41 1.0149 1.49 8.104110.41 2.0446 4.46 9.06946.94 2.0446 4.46 9.06946.94 3.0892 8.9210.04174.17 3.0892 8.9210.04174.17 4.133913.3911.02272.27 4.133913.3911.02272.27 5.162016.2012.01551.55 5.162016.2012.01551.55 6.160616.06 Total1.0000100.00 6.160616.06 Total1.0000100.00

23. 23 Slide Example: Troy Parking Garage n Poisson Distribution Goodness of Fit Test Observed and Expected Frequencies Observed and Expected Frequencies i f i e i f i - e i i f i e i f i - e i 0 or 1 or 256.20-1.20 0 or 1 or 256.20-1.20 3108.921.08 41413.39.61 52016.203.80 61216.06-4.06 71213.89-1.89 8910.41-1.41 986.941.06 10 or more107.992.01 10 or more107.992.01

24. 24 Slide n Poisson Distribution Goodness of Fit Test Test Statistic Test Statistic Rejection Rule Rejection Rule With  =.05 and k - p - 1 = 9 - 1 - 1 = 7 d.f. (where k = number of categories and p = number of population parameters estimated), With  =.05 and k - p - 1 = 9 - 1 - 1 = 7 d.f. (where k = number of categories and p = number of population parameters estimated), Reject H 0 if  2 > 14.07 Reject H 0 if  2 > 14.07 Conclusion Conclusion We cannot reject H 0. There’s no reason to doubt the assumption of a Poisson distribution. We cannot reject H 0. There’s no reason to doubt the assumption of a Poisson distribution. Example: Troy Parking Garage

25. 25 Slide Goodness of Fit Test: Normal Distribution 1. Set up the null and alternative hypotheses. 2. Select a random sample and a. Compute the mean and standard deviation. b. Define intervals of values so that the expected frequency is at least 5 for each interval. frequency is at least 5 for each interval. c. For each interval record the observed frequencies 3. Compute the expected frequency, e i, for each interval. continued

26. 26 Slide Goodness of Fit Test: Normal Distribution 4. Compute the value of the test statistic. 5. Reject H 0 if (where  is the significance level and there are k - 3 degrees of freedom).

27. 27 Slide Example: Victor Computers n Normal Distribution Goodness of Fit Test Victor Computers manufactures and sells a Victor Computers manufactures and sells a general purpose microcomputer. As part of a study to evaluate sales personnel, management wants to determine if the annual sales volume (number of units sold by a salesperson) follows a normal probability distribution.

28. 28 Slide Example: Victor Computers n Normal Distribution Goodness of Fit Test A simple random sample of 30 of the salespeople was taken and their numbers of units sold are below. 33 43 44 45 52 52 56 58 63 64 64 65 66 68 70 72 73 73 74 75 83 84 85 86 91 92 94 98 102 105 (mean = 71, standard deviation = 18.54) (mean = 71, standard deviation = 18.54)

29. 29 Slide n Normal Distribution Goodness of Fit Test Hypotheses Hypotheses H 0 : The population of number of units sold has a normal distribution with mean 71 and standard deviation 18.54. H 0 : The population of number of units sold has a normal distribution with mean 71 and standard deviation 18.54. H a : The population of number of units sold does not have a normal distribution with mean 71 and standard deviation 18.54. H a : The population of number of units sold does not have a normal distribution with mean 71 and standard deviation 18.54. Example: Victor Computers

30. 30 Slide n Normal Distribution Goodness of Fit Test Interval Definition Interval Definition To satisfy the requirement of an expected To satisfy the requirement of an expected frequency of at least 5 in each interval we will divide the normal distribution into 30/5 = 6 equal probability intervals. Example: Victor Computers

31. 31 Slide Example: Victor Computers n Normal Distribution Goodness of Fit Test Interval Definition Interval Definition Areas = 1.00/6 =.1667 Areas = 1.00/6 =.1667 71 53.02 63.03 78.97 88.98 = 71 +.97(18.54)

32. 32 Slide n Normal Distribution Goodness of Fit Test Observed and Expected Frequencies Observed and Expected Frequencies i f i e i f i - e i i f i e i f i - e i Less than 53.02651 Less than 53.02651 53.02 to 63.0335-2 53.02 to 63.0335-2 63.03 to 71.00651 63.03 to 71.00651 71.00 to 78.97550 71.00 to 78.97550 78.97 to 88.9845-1 78.97 to 88.9845-1 More than 88.98651 More than 88.98651 Total3030 Total3030 Example: Victor Computers

33. 33 Slide n Normal Distribution Goodness of Fit Test Test Statistic Test Statistic Rejection Rule Rejection Rule With  =.05 and k - p - 1 = 6 - 2 - 1 = 3 d.f., With  =.05 and k - p - 1 = 6 - 2 - 1 = 3 d.f., Reject H 0 if  2 > 7.81 Reject H 0 if  2 > 7.81 Conclusion Conclusion We cannot reject H 0. There is little evidence to We cannot reject H 0. There is little evidence to support rejecting the assumption the population is normally distributed with  = 71 and  = 18.54. support rejecting the assumption the population is normally distributed with  = 71 and  = 18.54. Example: Victor Computers

34. 34 Slide End of Chapter 12