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Slides by John Loucks St. Edward’s University

Slides by John Loucks St. Edward’s University. Chapter 12 Tests of Goodness of Fit and Independence. Goodness of Fit Test: A Multinomial Population. Test of Independence. Goodness of Fit Test: Poisson and Normal Distributions.

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Slides by John Loucks St. Edward’s University

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  1. Slides by John Loucks St. Edward’s University

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

  3. Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial Population 1. State the null and alternative hypotheses. H0: The population follows a multinomial distribution with specified probabilities for each of the k categories Ha: The population does not follow a multinomial distribution with specified probabilities for each of the k categories

  4. Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial Population 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.

  5. Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial Population 4. Compute the value of the test statistic. where: fi = observed frequency for category i ei = expected frequency for category i k = number of categories 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.

  6. Reject H0 if Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial Population 5. Rejection rule: Reject H0 if p-value <a p-value approach: Critical value approach: where  is the significance level and there are k - 1 degrees of freedom

  7. Multinomial Distribution Goodness of Fit Test • Example: Finger Lakes Homes (A) Finger Lakes Homes manufactures four models of prefabricated homes, a two-story colonial, a log cabin, 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.

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

  9. Multinomial Distribution Goodness of Fit Test • Hypotheses H0: pC = pL = pS = pA = .25 Ha: The population proportions are not pC = .25, pL = .25, pS = .25, and pA = .25 where: pC = population proportion that purchase a colonial pL = population proportion that purchase a log cabin pS = population proportion that purchase a split-level pA = population proportion that purchase an A-frame

  10. Multinomial Distribution Goodness of Fit Test • Rejection Rule Reject H0 if p-value < .05 or c2 > 7.815. With  = .05 and k - 1 = 4 - 1 = 3 degrees of freedom Do Not Reject H0 Reject H0 2 7.815

  11. Multinomial Distribution Goodness of Fit Test • Expected Frequencies • Test Statistic • e1 = .25(100) = 25 e2 = .25(100) = 25 e3 = .25(100) = 25 e4 = .25(100) = 25 = 1 + 1 + 4 + 4 = 10

  12. Multinomial Distribution Goodness of Fit Test • Conclusion Using the p-Value Approach Area in Upper Tail .10 .05 .025 .01 .005 c2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838 Because c2= 10 is between 9.348 and 11.345, the area in the upper tail of the distribution is between .025 and .01. The p-value <a . We can reject the null hypothesis. Actual p-value is .0186

  13. Multinomial Distribution Goodness of Fit Test • Conclusion Using the Critical Value Approach c2 = 10 > 7.815 We reject, at the .05 level of significance, the assumption that there is no home style preference.

  14. Test of Independence: Contingency Tables 1. Set up the null and alternative hypotheses. H0: The column variable is independent of the row variable Ha: The column variable is not independent of the row variable 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.

  15. Reject H0 if p -value <a or . Test of Independence: Contingency Tables 4. Compute the test statistic. 5. Determine the rejection rule. where  is the significance level and, with n rows and m columns, there are (n - 1)(m - 1) degrees of freedom.

  16. Contingency Table (Independence) Test • Example: Finger Lakes Homes (B) 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.

  17. Contingency Table (Independence) Test • Example: Finger Lakes Homes (B) 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. Price Colonial Log Split-Level A-Frame < $99,000 18 6 19 12 > $99,000 12 14 16 3

  18. 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

  19. Contingency Table (Independence) Test • Expected Frequencies Price Colonial Log Split-Level A-Frame Total < $99K > $99K Total 18 6 19 12 55 12 14 16 3 45 30 20 35 15 100

  20. With  = .05 and (2 - 1)(4 - 1) = 3 d.f., Contingency Table (Independence) Test • Rejection Rule Reject H0 if p-value < .05 or 2> 7.815 • Test Statistic = .1364 + 2.2727 + . . . + 2.0833 = 9.149

  21. Contingency Table (Independence) Test • Conclusion Using the p-Value Approach Area in Upper Tail .10 .05 .025 .01 .005 c2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838 Because c2= 9.145 is between 7.815 and 9.348, the area in the upper tail of the distribution is between .05 and .025. The p-value <a . We can reject the null hypothesis. Actual p-value is .0274

  22. Contingency Table (Independence) Test • Conclusion Using the Critical Value Approach c2 = 9.145 > 7.815 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.

  23. Goodness of Fit Test: Poisson Distribution 1. State the null and alternative hypotheses. H0: The population has a Poisson distribution Ha: The population does not have a Poisson distribution 2. Select a random sample and a. Record the observed frequency fi for each value of the Poisson random variable. b. Compute the mean number of occurrences .

  24. Goodness of Fit Test: Poisson Distribution 3. Compute the expected frequency of occurrences ei for each value of the Poisson random variable. • Multiply the sample size by the Poisson probability • of occurrence for each value of the Poisson random • variable. • If there are fewer than five expected occurrences • for some values, combine adjacent values and • reduce the number of categories as necessary.

  25. Goodness of Fit Test: Poisson Distribution 4. Compute the value of the test statistic. where: fi = observed frequency for category i ei = expected frequency for category i k = number of categories

  26. Reject H0 if Goodness of Fit Test: Poisson Distribution 5. Rejection rule: Reject H0 if p-value <a p-value approach: Critical value approach: where  is the significance level and there are k - 2 degrees of freedom

  27. Goodness of Fit Test: Poisson Distribution • Example: Troy Parking Garage 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.

  28. Goodness of Fit Test: Poisson Distribution 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. • Example: Troy Parking Garage # 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

  29. Goodness of Fit Test: Poisson Distribution • 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

  30. Goodness of Fit Test: Poisson Distribution • 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,

  31. Goodness of Fit Test: Poisson Distribution • Expected Frequencies x f (x ) nf (x ) xf (x ) nf (x ) 0 1 2 3 4 5 6 7 8 9 10 11 12+ Total .0025 .0149 .0446 .0892 .1339 .1606 .1606 .25 1.49 4.46 8.92 13.39 16.06 16.06 .1377 .1033 .0688 .0413 .0225 .0201 1.0000 13.77 10.33 6.88 4.13 2.25 2.01 100.00

  32. Goodness of Fit Test: Poisson Distribution • Observed and Expected Frequencies ifieifi - ei -1.20 1.08 0.61 3.94 -4.06 -1.77 -1.33 1.12 1.61 5 10 14 20 12 12 9 8 10 0 or 1 or 2 3 4 5 6 7 8 9 10 or more 6.20 8.92 13.39 16.06 16.06 13.77 10.33 6.88 8.39

  33. With  = .05 and k - p - 1 = 9 - 1 - 1 = 7 d.f. (where k = number of categories and p = number of population parameters estimated), Goodness of Fit Test: Poisson Distribution • Test Statistic • Rejection Rule Reject H0 if p-value < .05 or 2> 14.067.

  34. Goodness of Fit Test: Poisson Distribution • Conclusion Using the p-Value Approach Area in Upper Tail .90 .10 .05 .025 .01 c2 Value (df = 7) 2.833 12.017 14.067 16.013 18.475 Because c2= 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. The p-value > a . We cannot reject the null hypothesis. There is no reason to doubt the assumption of a Poisson distribution. Actual p-value is .8591

  35. Goodness of Fit Test: Normal Distribution 1. State the null and alternative hypotheses. H0: The population has a normal distribution Ha: The population does not have a normal distribution 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 • Compute the expected frequency, ei , for each interval. • (Multiply the sample size by the probability of a • normal random variable being in the interval.

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

  37. Goodness of Fit Test: Normal Distribution • Example: IQ Computers 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.

  38. Goodness of Fit Test: Normal Distribution A simple random sample of 30 of the salespeople was taken and their numbers of units sold are listed below. • Example: IQ Computers 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)

  39. Goodness of Fit Test: Normal Distribution • 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.

  40. Goodness of Fit Test: Normal Distribution • 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.

  41. Goodness of Fit Test: Normal Distribution • Interval Definition Areas = 1.00/6 = .1667 53.02 71 88.98 = 71 + .97(18.54) 71 - .43(18.54) = 63.03 78.97

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

  43. With  = .05 and k - p - 1 = 6 - 2 - 1 = 3 d.f. (where k = number of categories and p = number of population parameters estimated), Goodness of Fit Test: Normal Distribution • Rejection Rule Reject H0 if p-value < .05 or 2> 7.815. • Test Statistic

  44. Goodness of Fit Test: Normal Distribution • Conclusion Using the p-Value Approach Area in Upper Tail .90 .10 .05 .025 .01 c2 Value (df = 3) .584 6.251 7.815 9.348 11.345 Because c2= 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. The p-value > a . 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. Actual p-value is .6594

  45. End of Chapter 12

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