1. 1 1 Slide © 2005 Thomson/South-Western Chapter 12 Tests of Goodness of Fit and Independence n Goodness of Fit Test: A Multinomial Population Goodness of Fit Test: Poisson Goodness of Fit Test: Poisson and Normal Distributions and Normal Distributions Test of Independence Test of Independence

2. 2 2 Slide © 2005 Thomson/South-Western Hypothesis (Goodness of Fit) Test for Proportions of 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. 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 frequency, e i, in each category by multiplying the category probability by the sample size. category probability by the sample size.

3. 3 3 Slide © 2005 Thomson/South-Western Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population 4. Compute the value of the test statistic. Note: The test statistic has a chi-square distribution with k – 1 df provided that the expected frequencies are 5 or more for all categories. f i = observed frequency for category i e i = expected frequency for category i k = number of categories where:

4. 4 4 Slide © 2005 Thomson/South-Western Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population where  is the significance level and there are k - 1 degrees of freedom p -value approach: Critical value approach: Reject H 0 if p -value <  5. Rejection rule: Reject H 0 if

5. 5 5 Slide © 2005 Thomson/South-Western Multinomial Distribution Goodness of Fit Test n Example: Finger Lakes Homes (A) Finger Lakes Homes manufactures Finger Lakes Homes manufactures four models of prefabricated homes, four models of prefabricated homes, a two-story colonial, a log cabin, a a two-story colonial, a log cabin, a split-level, and an A-frame. To help split-level, and an A-frame. To help in production planning, management in production planning, management would like to determine if previous would like to determine if previous customer purchases indicate that there customer purchases indicate that there is a preference in the style selected. is a preference in the style selected.

6. 6 6 Slide © 2005 Thomson/South-Western Split- A- Split- A- Model Colonial Log Level Frame # Sold 30 20 35 15 The number of homes sold of each The number of homes sold of each model for 100 sales over the past two years is shown below. Multinomial Distribution Goodness of Fit Test n Example: Finger Lakes Homes (A)

7. 7 7 Slide © 2005 Thomson/South-Western n Hypotheses Multinomial Distribution Goodness of Fit Test where: p C = population proportion that purchase a colonial p C = population proportion that purchase a colonial p L = population proportion that purchase a log cabin p L = population proportion that purchase a log cabin p S = population proportion that purchase a split-level p S = population proportion that purchase a split-level p A = population proportion that purchase an A-frame p A = population proportion that purchase an A-frame H 0 : p C = p L = p S = p A =.25 H a : The population proportions are not p C =.25, p L =.25, p S =.25, and p A =.25 p C =.25, p L =.25, p S =.25, and p A =.25

8. 8 8 Slide © 2005 Thomson/South-Western n Rejection Rule 22 22 7.815 Do Not Reject H 0 Reject H 0 Multinomial Distribution Goodness of Fit Test With  =.05 and k - 1 = 4 - 1 = 3 k - 1 = 4 - 1 = 3 degrees of freedom degrees of freedom if p -value 7.815. Reject H 0 if p -value 7.815.

9. 9 9 Slide © 2005 Thomson/South-Western n Expected Frequencies n Test Statistic Multinomial Distribution Goodness of Fit Test 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 = 1 + 1 + 4 + 4 = 10

10. 10 Slide © 2005 Thomson/South-Western Multinomial Distribution Goodness of Fit Test n Conclusion Using the p -Value Approach The p -value < . We can reject the null hypothesis. The p -value < . We can reject the null hypothesis. Because  2 = 10 is between 9.348 and 11.345, the Because  2 = 10 is between 9.348 and 11.345, the area in the upper tail of the distribution is between area in the upper tail of the distribution is between.025 and.01..025 and.01. Area in Upper Tail.10.05.025.01.005  2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838

11. 11 Slide © 2005 Thomson/South-Western n Conclusion Using the Critical Value Approach Multinomial Distribution Goodness of Fit Test We reject, at the.05 level of significance, We reject, at the.05 level of significance, the assumption that there is no home style preference.  2 = 10 > 7.815

12. 12 Slide © 2005 Thomson/South-Western 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. frequency, f ij, for each cell of the contingency table. 3. Compute the expected frequency, e ij, for each cell.

13. 13 Slide © 2005 Thomson/South-Western Test of Independence: Contingency Tables 5. Determine the rejection rule. Reject H 0 if p -value <  or. 4. Compute the test statistic. where  is the significance level and, with n rows and m columns, there are ( n - 1)( m - 1) degrees of freedom.

14. 14 Slide © 2005 Thomson/South-Western Each home sold by Finger Lakes Each home sold by Finger Lakes Homes can be classified according to price and to style. Finger Lakes’ manager would like to determine if the price of the home and the style of the home are independent variables. Contingency Table (Independence) Test n Example: Finger Lakes Homes (B)

15. 15 Slide © 2005 Thomson/South-Western Price Colonial Log Split-Level A-Frame Price Colonial Log Split-Level A-Frame The number of homes sold for 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 $99,000 or less or more than $99,000. > $99,000 12 14 16 3 < $99,000 18 6 19 12 Contingency Table (Independence) Test n Example: Finger Lakes Homes (B)

16. 16 Slide © 2005 Thomson/South-Western n Hypotheses Contingency Table (Independence) Test H 0 : Price of the home is independent of the style of the home that is purchased style of the home that is purchased 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

17. 17 Slide © 2005 Thomson/South-Western n Expected Frequencies Contingency Table (Independence) Test Price Colonial Log Split-Level A-Frame Total Price Colonial Log Split-Level A-Frame Total < $99K > $99K Total Total 30 20 35 15 100 12 14 16 3 45 18 6 19 12 55

18. 18 Slide © 2005 Thomson/South-Western n Rejection Rule Contingency Table (Independence) Test With  =.05 and (2 - 1)(4 - 1) = 3 d.f., Reject H 0 if p -value 7.815 =.1364 + 2.2727 +... + 2.0833 = 9.149 n Test Statistic

19. 19 Slide © 2005 Thomson/South-Western n Conclusion Using the p -Value Approach The p -value < . We can reject the null hypothesis. The p -value < . We can reject the null hypothesis. Because  2 = 9.145 is between 7.815 and 9.348, the Because  2 = 9.145 is between 7.815 and 9.348, the area in the upper tail of the distribution is between area in the upper tail of the distribution is between.05 and.025..05 and.025. Area in Upper Tail.10.05.025.01.005  2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838 Contingency Table (Independence) Test

20. 20 Slide © 2005 Thomson/South-Western n Conclusion Using the Critical Value Approach Contingency Table (Independence) Test We reject, at the.05 level of significance, We reject, at the.05 level of significance, the assumption that the price of the home is independent of the style of home that is purchased.  2 = 9.145 > 7.815

21. 21 Slide © 2005 Thomson/South-Western Goodness of Fit Test: Poisson Distribution 1. Set up the null and alternative hypotheses. H 0 : Population has a Poisson probability distribution H 0 : Population has a Poisson probability distribution H a : Population does not have a Poisson distribution H a : Population does not have a Poisson distribution 3. Compute the expected frequency of occurrences e i for each value of the Poisson random variable. for each value of the Poisson random variable. 2. Select a random sample and a. Record the observed frequency f i for each value of a. Record the observed frequency f i for each value of the Poisson random variable. the Poisson random variable. b. Compute the mean number of occurrences . b. Compute the mean number of occurrences .

22. 22 Slide © 2005 Thomson/South-Western Goodness of Fit Test: Poisson Distribution 4. Compute the value of the test statistic. f i = observed frequency for category i e i = expected frequency for category i k = number of categories where:

23. 23 Slide © 2005 Thomson/South-Western where  is the significance level and there are k - 2 degrees of freedom there are k - 2 degrees of freedom p -value approach: Critical value approach: Reject H 0 if p -value <  5. Rejection rule: Reject H 0 if Goodness of Fit Test: Poisson Distribution

24. 24 Slide © 2005 Thomson/South-Western n Example: Troy Parking Garage In studying the need for an In studying the need for an additional entrance to a city parking garage, a consultant has recommended an analysis approach that is applicable only in situations where the number of cars entering during a specified time period follows a Poisson distribution. Goodness of Fit Test: Poisson Distribution

25. 25 Slide © 2005 Thomson/South-Western A random sample of 100 one- 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. Goodness of Fit Test: Poisson Distribution n Example: Troy Parking Garage # 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

26. 26 Slide © 2005 Thomson/South-Western n Hypotheses Goodness of Fit Test: Poisson Distribution H a : Number of cars entering the garage during a one-minute interval is not Poisson distributed one-minute interval is not Poisson distributed H 0 : Number of cars entering the garage during a one-minute interval is Poisson distributed a one-minute interval is Poisson distributed

27. 27 Slide © 2005 Thomson/South-Western n Estimate of Poisson Probability Function Goodness of Fit Test: Poisson Distribution  otal Arrivals = 0(0) + 1(1) + 2(4) +... + 12(1) = 600 Hence, Estimate of  = 600/100 = 6 Total Time Periods = 100

28. 28 Slide © 2005 Thomson/South-Western n Expected Frequencies Goodness of Fit Test: Poisson Distribution x f ( x ) nf ( x ) 0123456 13.77 13.77 10.33 10.33 6.88 6.88 4.13 4.13 2.25 2.25 2.01 2.01100.00.1377.1377.1033.1033.0688.0688.0413.0413.0225.0225.0201.02011.0000 7 8 9 10 10 11 11 12+ 12+Total.0025.0149.0446.0892.1339.1606.1606.25.25 1.49 1.49 4.46 4.46 8.92 8.9213.3916.0616.06 x f ( x ) nf ( x )

29. 29 Slide © 2005 Thomson/South-Western n Observed and Expected Frequencies Goodness of Fit Test: Poisson Distribution i f i e i f i - e i i f i e i f i - e i -1.20 1.08 1.08 0.61 0.61 3.94 3.94-4.06-1.77-1.33 1.12 1.12 1.61 1.61 6.20 6.20 8.92 8.9213.3916.0616.0613.7710.33 6.88 6.88 8.39 8.39 51014201212 9 810 0 or 1 or 2 3 4 5 6 7 8 9 10 or more

30. 30 Slide © 2005 Thomson/South-Western n Test Statistic Goodness of Fit Test: Poisson Distribution With  =.05 and k - p - 1 = 9 - 1 - 1 = 7 d.f. 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 p -value 14.067. n Rejection Rule

31. 31 Slide © 2005 Thomson/South-Western n Conclusion Using the p -Value Approach The p -value > . We cannot reject the null hypothesis. There is no reason to doubt the assumption of a Poisson distribution. The p -value > . We cannot reject the null hypothesis. There is no reason to doubt the assumption of a Poisson distribution. Because  2 = 3.268 is between 2.833 and 12.017 in the Chi-Square Distribution Table, the area in the upper tail Because  2 = 3.268 is between 2.833 and 12.017 in the Chi-Square Distribution Table, the area in the upper tail of the distribution is between.90 and.10. Area in Upper Tail.90.10.05.025.01  2 Value (df = 7) 2.833 12.017 14.067 16.013 18.475 Goodness of Fit Test: Poisson Distribution

32. 32 Slide © 2005 Thomson/South-Western Goodness of Fit Test: Normal Distribution 1. Set up the null and alternative hypotheses. 3. Compute the expected frequency, e i, for each interval. 2. Select a random sample and a. Compute the mean and standard deviation. a. Compute the mean and standard deviation. b. Define intervals of values so that the expected 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 c. For each interval record the observed frequencies

33. 33 Slide © 2005 Thomson/South-Western 4. Compute the value of the test statistic. Goodness of Fit Test: Normal Distribution 5. Reject H 0 if (where  is the significance level and there are k - 3 degrees of freedom). and there are k - 3 degrees of freedom).

34. 34 Slide © 2005 Thomson/South-Western Normal Distribution Goodness of Fit Test n Example: IQ Computers IQIQ IQ Computers (one better than HP?) IQ Computers (one better than HP?) manufactures and sells a general purpose microcomputer. As part of a study to evaluate sales personnel, management wants to determine, at a.05 significance level, if the annual sales volume (number of units sold by a salesperson) follows a normal probability distribution.

35. 35 Slide © 2005 Thomson/South-Western A simple random sample of 30 of the salespeople was taken and their numbers of units sold are below. Normal Distribution Goodness of Fit Test n Example: IQ Computers (mean = 71, standard deviation = 18.54) 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 IQIQ

36. 36 Slide © 2005 Thomson/South-Western n Hypotheses Normal Distribution Goodness of Fit Test H a : The population of number of units sold does not have a normal distribution with does not have a normal distribution with mean 71 and standard deviation 18.54. mean 71 and standard deviation 18.54. H 0 : The population of number of units sold has a normal distribution with mean 71 has a normal distribution with mean 71 and standard deviation 18.54. and standard deviation 18.54.

37. 37 Slide © 2005 Thomson/South-Western n Interval Definition Normal Distribution Goodness of Fit Test 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.

38. 38 Slide © 2005 Thomson/South-Western n Interval Definition Areas = 1.00/6 =.1667 Areas = 1.00/6 =.1667 71 53.02 71 .43(18.54) = 63.03 78.97 88.98 = 71 +.97(18.54) Normal Distribution Goodness of Fit Test

39. 39 Slide © 2005 Thomson/South-Western n Observed and Expected Frequencies Normal Distribution Goodness of Fit Test 1-2 1 0 1 5 5 5 5 5 530 6 3 6 5 4 630 Less than 53.02 53.02 to 63.03 53.02 to 63.03 63.03 to 71.00 63.03 to 71.00 71.00 to 78.97 71.00 to 78.97 78.97 to 88.98 78.97 to 88.98 More than 88.98 i f i e i f i - e i i f i e i f i - e i Total

40. 40 Slide © 2005 Thomson/South-Western n Test Statistic With  =.05 and k - p - 1 = 6 - 2 - 1 = 3 d.f. With  =.05 and k - p - 1 = 6 - 2 - 1 = 3 d.f. (where k = number of categories and p = number of population parameters estimated), Reject H 0 if p -value 7.815. n Rejection Rule Normal Distribution Goodness of Fit Test

41. 41 Slide © 2005 Thomson/South-Western Normal Distribution Goodness of Fit Test n Conclusion Using the p -Value Approach The p -value > . We cannot reject the null hypothesis. There is little evidence to support rejecting the assumption the population is normally distributed with  = 71 and  = 18.54. The p -value > . We cannot reject the null hypothesis. There is little evidence to support rejecting the assumption the population is normally distributed with  = 71 and  = 18.54. Because  2 = 1.600 is between.584 and 6.251 in the Chi-Square Distribution Table, the area in the upper tail Because  2 = 1.600 is between.584 and 6.251 in the Chi-Square Distribution Table, the area in the upper tail of the distribution is between.90 and.10. Area in Upper Tail.90.10.05.025.01  2 Value (df = 3).584 6.251 7.815 9.348 11.345

42. 42 Slide © 2005 Thomson/South-Western End of Chapter 12