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統計學 Spring 2004

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. Chapter 12 Tests of Goodness of Fit and Independence. Goodness of Fit Test: A Multinomial Population Tests of Independence: Contingency Tables

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統計學 Spring 2004

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  1. 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. Chapter 12 Tests of Goodness of Fit and Independence • Goodness of Fit Test: A Multinomial Population • Tests of Independence: Contingency Tables • Goodness of Fit Test: Poisson and Normal Distributions

  3. 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, fi , for each of the k categories. 3. Assuming H0 is true, compute the expected frequency, ei , in each category by multiplying the category probability by the sample size. continued

  4. Goodness of Fit Test:A Multinomial Population 4. Compute the value of the test statistic. 5. Reject H0 if (where  is the significance level and there are k - 1 degrees of freedom).

  5. Example: Finger Lakes Homes (A) • 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. Example: Finger Lakes Homes (A) • 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. Example: Finger Lakes Homes (A) • Multinomial Distribution Goodness of Fit Test • Notation pC = popul. proportion that purchase a colonial pR = popul. proportion that purchase a ranch pS = popul. proportion that purchase a split-level pA = popul. proportion that purchase an A-frame • Hypotheses H0: pC = pR = pS = pA = .25 Ha: The population proportions are not pC = .25, pR = .25, pS = .25, and pA = .25

  8. Example: Finger Lakes Homes (A) • Multinomial Distribution Goodness of Fit Test • Expected Frequencies e1 = .25(100) = 25 e2 = .25(100) = 25 e3 = .25(100) = 25 e4 = .25(100) = 25 • Test Statistic = 1 + 1 + 4 + 4 = 10

  9. Example: Finger Lakes Homes (A) • Multinomial Distribution Goodness of Fit Test • Rejection Rule With  = .05 and k - 1 = 4 - 1 = 3 degrees of freedom Do Not Reject H0 Reject H0 2 7.81

  10. Example: Finger Lakes Homes (A) • Multinomial Distribution Goodness of Fit Test • Conclusion c2 = 10 > 7.81, so we reject the assumption there is no home style preference, at the .05 level of significance.

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

  12. Test of Independence: Contingency Tables 4. Compute the test statistic. 5. Reject H0 if (where  is the significance level and with n rows and m columns there are (n - 1)(m - 1) degrees of freedom).

  13. Example: Finger Lakes Homes (B) • 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. 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 < $65,000 18 6 19 12 > $65,000 12 14 16 3

  14. Example: Finger Lakes Homes (B) • Contingency Table (Independence) Test • Hypotheses H0: Price of the home is independent of the style of the home that is purchased Ha: Price of the home is not independent of the style of the home that is purchased • Expected Frequencies 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

  15. Example: Finger Lakes Homes (B) • Contingency Table (Independence) Test • Test Statistic = .1364 + 2.2727 + . . . + 2.0833 = 9.1486 • Rejection Rule With  = .05 and (2 - 1)(4 - 1) = 3 d.f., Reject H0 if 2 > 7.81 • Conclusion We reject H0, the assumption that the price of the home is independent of the style of the home that is purchased.

  16. 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, fi , for each of the k values of the Poisson random variable. b. Compute the mean number of occurrences, . 3. Compute the expected frequency of occurrences, ei , for each value of the Poisson random variable. continued

  17. Goodness of Fit Test: Poisson Distribution 4. Compute the value of the test statistic. 5. Reject H0 if (where  is the significance level and there are k - 2 degrees of freedom).

  18. Example: Troy Parking Garage • 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.

  19. Example: Troy Parking Garage • 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. # Arrivals0 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

  20. Example: Troy Parking Garage • Poisson Distribution Goodness of Fit Test • Hypotheses H0: Number of cars entering the garage during a one-minute interval is Poisson distributed. Ha: Number of cars entering the garage during a one-minute interval is not Poisson distributed

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

  22. Example: Troy Parking Garage • Poisson Distribution Goodness of Fit Test • Expected Frequencies xf (x ) xf (x ) xf (x ) xf (x ) 0 .0025 .25 7 .1389 13.89 1 .0149 1.49 8 .1041 10.41 2 .0446 4.46 9 .0694 6.94 3 .0892 8.92 10 .0417 4.17 4 .1339 13.39 11 .0227 2.27 5 .1620 16.20 12.01551.55 6 .1606 16.06 Total 1.0000 100.00

  23. Example: Troy Parking Garage • Poisson Distribution Goodness of Fit Test • Observed and Expected Frequencies ifieifi - ei 0 or 1 or 2 5 6.20 -1.20 3 10 8.92 1.08 4 14 13.39 .61 5 20 16.20 3.80 6 12 16.06 -4.06 7 12 13.89 -1.89 8 9 10.41 -1.41 9 8 6.94 1.06 10 or more 10 7.99 2.01

  24. Example: Troy Parking Garage • Poisson Distribution Goodness of Fit Test • Test Statistic • 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), Reject H0 if 2 > 14.07 • Conclusion We cannot reject H0. There’s no reason to doubt the assumption of a Poisson distribution.

  25. 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. c. For each interval record the observed frequencies 3. Compute the expected frequency, ei , for each interval. continued

  26. Goodness of Fit Test: Normal Distribution 4. Compute the value of the test statistic. 5. Reject H0 if (where  is the significance level and there are k - 3 degrees of freedom).

  27. Example: Victor Computers • Normal Distribution Goodness of Fit Test 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. Example: Victor Computers • 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)

  29. Example: Victor Computers • Normal Distribution Goodness of Fit Test • Hypotheses H0: The population of number of units sold has a normal distribution with mean 71 and standard deviation 18.54. Ha: The population of number of units sold does not have a normal distribution with mean 71 and standard deviation 18.54.

  30. Example: Victor Computers • Normal Distribution Goodness of Fit Test • Interval Definition 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.

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

  32. Example: Victor Computers • Normal Distribution Goodness of Fit Test • Observed and Expected Frequencies ifieifi - ei Less than 53.02 6 5 1 53.02 to 63.03 3 5 -2 63.03 to 71.00 6 5 1 71.00 to 78.97 5 5 0 78.97 to 88.98 4 5 -1 More than 88.98 6 5 1 Total 30 30

  33. Example: Victor Computers • Normal Distribution Goodness of Fit Test • Test Statistic • Rejection Rule With  = .05 and k - p - 1 = 6 - 2 - 1 = 3 d.f., Reject H0 if 2 > 7.81 • Conclusion We cannot reject H0. There is little evidence to support rejecting the assumption the population is normally distributed with  = 71 and  = 18.54.

  34. End of Chapter 12

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