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Statistics Tests of Goodness of Fit and Independence 1/71 Contents STATISTICS in PRACTICE Goodness of Fit Test: A Multinomial Population Test of Independence Goodness of Fit Test: Poisson and Normal Distributions 2/71 STATISTICS in PRACTICE United Way of Greater Rochester is a nonprofit organization dedicated to improving the quality of life for people. Because of enormous volunteer involvement, United Way of Greater Rochester is able to hold its operating costs at just eight cents of every dollar raised. 3/71 STATISTICS in PRACTICE One of statistical tests was to determine whether perceptions of administrative expenses were independent of occupation. In this chapter, you will learn how a statistical test of independence. 4/71 Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population Consider the case that each element of a population is assigned to one and only one of several classes or categories. Such a population is a multinomial population(多項母體). We want to test if the population follows a multinominal distribution with specified probabilities for each of the k categories. 5/71 Hypothesis Test for Proportions of a Multinomial Population-- Procedures 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. 6/71 Hypothesis Test for Proportions of a Multinomial Population-- Procedures 4. Compute the value of the test statistic. ( f i ei ) 2 k 2 i 1 ei 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. 7/71 Hypothesis Test for Proportions of a Multinomial Population-- Procedures 5. Rejection rule: p-value approach: Reject H0 if p-value < Critical value approach: Reject H0 if 2 2 where is the significance level and there are k - 1 degrees of freedom 8/71 Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population Example: The market share study being conducted by Scott Marketing Research. Over the past year market shares stabilized at 30% for company A, 50% for company B, and 20% for company C. Recently company C developed a ―new and improved‖ product to replace its current entry in the market. Company C retained Scott Marketing Research to determine whether the new product will alter market shares. 9/71 Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population Scott Marketing Research conduct a sample survey and compute the proportion preferring each company’s product. A hypothesis test will then be conducted to see whether the new product caused a change in market shares. The null and alternative hypotheses are 10/71 Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population Perform a goodness of fit test that will determine whether the sample of 200 customer purchase preferences is consistent with the null hypothesis. Data 11/71 Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population Expected: the expected frequency for each category is found by multiplying the sample size of 200 by the hypothesized proportion for the category. 12/71 Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population Computation of the Chi-square Test Statistic for The Scott Marketing Research Market Share Study 13/71 Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population p-Value Approach The test statistic χ2 = 7.34 and 5.991 < 7.34 < 7.378. Thus, the corresponding upper tail area or p-value must be between .05 and .025. with p-value < .05, we reject H0. Minitab or Excel can be used to show χ 2 = 7.34 provides a p-value = .0255. 14/71 Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population Critical Value Approach With α = .05 and 2 degrees of freedom, the critical value for the test statistic is = 5.991. The upper tail rejection rule becomes Reject H0 if χ 2 > 5.991 With 7.34 > 5.991, we reject H0. Comparisons of the observed and expected frequencies for the other two companies indicate that company C’s gain in market share will hurt company A more than company B. 15/71 Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population Note: The chi-square test is an approximate test and the test result may not be valid when the expected value for a category is less than 5. If one or more categories have expected values less than 5, you can combine them with adjacent categories to achieve the minimum required expected value. 16/71 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. 17/71 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 18/71 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 19/71 Multinomial Distribution Goodness of Fit Test Rejection Rule Reject H0 if p-value < .05 or χ2 > 7.815. With = .05 and k-1=4-1=3 degrees of freedom Do Not Reject H0 Reject H0 χ2 7.815 20/71 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 ( 30 25) 2 ( 20 25) 2 ( 35 25) 2 (15 25) 2 2 25 25 25 25 =1+1+4+4 = 10 21/71 Multinomial Distribution Goodness of Fit Test Conclusion Using the p-Value Approach Area in Upper Tail .10 .05 .025 .01 .005 2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838 Because χ2 = 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 < . We can reject the null hypothesis. 22/71 Multinomial Distribution Goodness of Fit Test Conclusion Using the Critical Value Approach χ2 = 10 > 7.815 We reject, at the .05 level of significance, the assumption that there is no home style preference. 23/71 Test of Independence: Contingency Tables(列聯表） Another important application of the chi- square distribution involves using sample data to test for the independence of two variables. Contingency table is a table that lists all possible combinations of the two variables. 24/71 Test of Independence: Contingency Tables Example: A test of independence addresses the question of whether the beer preference (light, regular, or dark) is independent of the gender of the beer drinker (male, female). If the independence assumption is valid, we argue that the fraction of drinking light beer, regular beer and dark beer must be applicable to both male and female beer drinker. 25/71 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. (Row i Total)(Column j Total) eij Sample Size 26/71 Test of Independence: Contingency Tables 4. Compute the test statistic. ( f ij eij ) 2 2 i j eij 5. Determine the rejection rule. Reject H0 if p -value < or . 2 2 where is the significance level and, with n rows and m columns, there are (n - 1)(m - 1) degrees of freedom. 27/71 Test of Independence: Contingency Tables Example: A test of independence addresses the question of whether the beer preference (light, regular, or dark) is independent of the gender of the beer drinker (male, female). The hypotheses for this test of independence are: H0: Beer preference is independent of the gender of the beer drinker Ha: Beer preference is not independent of the 28/71 gender of the beer drinker Test of Independence: Contingency Tables Sample Results 1/3 7/15 1/5 1 Expected Frequencies 80*1/3=26.67 29/71 Test of Independence: Contingency Tables 30/71 Test of Independence: Contingency Tables Computation of the Chi-square Test Statistic p-value = .0468. At the .05 level of significance, p-value < α= .05. We reject the null hypothesis 31/71 Test of Independence: Contingency Tables 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. 32/71 Test of Independence: Contingency Tables 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 33/71 Test of Independence: Contingency Tables 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 34/71 Test of Independence: Contingency Tables Expected Frequencies Price Colonial Log Split-Level A-Frame Total < $99K 18 6 19 12 55 > $99K 12 14 16 3 45 Total 30 20 35 15 100 35/71 Contingency Table (Independence) Test Rejection Rule .2 =7.815 With α = .05 and (2 - 1)(4 - 1) = 3 d.f., 05 Reject H0 if p-value < .05 or 2 > 7.815 Test Statistic (18 16.5) 2 (6 11) 2 (3 6.75) 2 2 16.5 11 6.75 = .1364 + 2.2727 + . . . + 2.0833 = 9.149 36/71 Contingency Table (Independence) Test Conclusion Using the p-Value Approach Area in Upper Tail .10 .05 .025 .01 .005 2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838 Because χ2 = 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 < . We can reject the null hypothesis. 37/71 Contingency Table (Independence) Test Conclusion Using the Critical Value Approach χ2 = 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. 38/71 Goodness of Fit Test: Poisson Distribution In general, the goodness of fit test can be used with any hypothesized probability distribution. Here we will demonstrate the cases of Poisson distribution and Normal distribution. 39/71 Goodness of Fit Test: Poisson Distribution 1. Set up the null and alternative hypotheses. H0: Population has a Poisson probability distribution Ha: 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 . 40/71 Goodness of Fit Test: Poisson Distribution 3. Compute the expected frequency of occurrences ei for each value of the Poisson random variable. 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 41/71 Goodness of Fit Test: Poisson Distribution 5. Rejection rule: p-value approach: Reject H0 if p-value < Critical value approach: Reject H0 if 2 2 where is the significance level and there are k - 2 degrees of freedom 42/71 Goodness of Fit Test: Poisson Distribution—Example Consider the arrival of customers at Dubek’s Food Market in Tallahassee, Florida. A statistical test conducted to see whether an assumption of a Poisson distribution for arrivals is reasonable. 43/71 Goodness of Fit Test: Poisson Distribution—Example Hypotheses are H0: The number of customers entering the store during 5-minute intervals has a Poisson probability distribution Ha: The number of customers entering the store during 5-minute intervals does not have a Poisson distribution 44/71 Goodness of Fit Test: Poisson Distribution—Example The Poisson probability function, x e f ( x) x! where --μ represents the expected number of customers arriving per 5-minute period, --x is the random variable indicating the number of customers arriving during a 5-minute period, and --f (x) is the probability that x customers will arrive in a 5-minute interval. 45/71 Goodness of Fit Test: Poisson Distribution—Example Sample Data μ known, an estimate of μ is 640/128 = 5 customers 46/71 Goodness of Fit Test: Poisson Distribution—Example Expected frequency 47/71 Goodness of Fit Test: Poisson Distribution—Example Computation of the Chi-square Test Statistic p-value = .1403. With p-value > α = .05, we cannot reject H0. 48/71 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. 49/71 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. # 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 50/71 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 51/71 Goodness of Fit Test: Poisson Distribution Estimate of Poisson Probability Function otal Arrivals = 0(0) + 1(1) + 2(4) + . . . + 12(1) = 600 Estimate of = 600/100 = 6 Total Time Periods = 100 Hence, x 6 6 e f ( x) x! 52/71 Goodness of Fit Test: Poisson Distribution Expected Frequencies x f (x ) nf (x ) x f (x ) nf (x ) 0 .0025 .25 7 .1377 13.77 1 .0149 1.49 8 .1033 10.33 2 .0446 4.46 9 .0688 6.88 3 .0892 8.92 10 .0413 4.13 4 .1339 13.39 11 .0225 2.25 5 .1606 16.06 12+ .0201 2.01 6 .1606 16.06 Total 1.0000 100.00 53/71 Goodness of Fit Test: Poisson Distribution Observed and Expected Frequencies i fi ei f i - ei 0 or 1 or 2 5 6.20 -1.20 3 10 8.92 1.08 4 14 13.39 0.61 5 20 16.06 3.94 6 12 16.06 -4.06 7 12 13.77 -1.77 8 9 10.33 -1.33 9 8 6.88 1.12 10 or more 10 8.39 1.61 54/71 Goodness of Fit Test: Poisson Distribution Rejection Rule .2 14.067 With = .05 and k - p - 1 = 9 - 1 - 1 = 7 d.f. 05 (where k = number of categories and p = number of population parameters estimated), Reject H0 if p-value < .05 or 2 > 14.067. Test Statistic (1.20) 2 (1.08) 2 (1.61) 2 2 3.268 6.20 8.92 8.39 55/71 Goodness of Fit Test: Poisson Distribution Conclusion Using the p-Value Approach Area in Upper Tail .90 .10 .05 .025 .01 2 Value (df = 7) 2.833 12.017 14.067 16.013 18.475 Because χ2 = 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 > . We cannot reject the null hypothesis. There is no reason to doubt the assumption of a Poisson distribution. 56/71 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. 57/71 Goodness of Fit Test: Normal Distribution 4. Compute the value of the test statistic. 5. Reject H0 if 2 (where is the significance 2 level and there are k - 3 degrees of freedom). 58/71 Goodness of Fit Test: Normal Distribution--Example Chemline Employee Aptitude Scores To test the null hypothesis that the population of test scores has a normal distribution. Sample Data 59/71 Goodness of Fit Test: Normal Distribution—Example Estimates of μ and are x= 68.42, s = 10.41. Hypotheses H0: The population of test scores has a normal distribution with mean 68.42 and standard deviation 10.41. Ha: The population of test scores does not have a normal distribution with mean 68.42 and standard deviation 10.41 60/71 Goodness of Fit Test: Normal Distribution—Example Expected Frequency Normal Distribution with 10 Equal- Probability Intervals (The Chemiline Example) Note: 55.10 = 68.42 - 1.28(10.41) 61/71 Goodness of Fit Test: Normal Distribution—Example 62/71 Goodness of Fit Test: Normal Distribution—Example Computation of the Chi-square Test Statistic p-value = .4084 >α = .10. Cannot reject H0. 63/71 Normal Distribution Goodness of Fit Test IQ 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 64/71 distribution. Normal Distribution Goodness of Fit Test IQ Example: IQ Computers 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) 65/71 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. 66/71 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. 67/71 Normal Distribution Goodness of Fit Test Interval Definition Areas = 1.00/6 = .1667 53.02 71 88.98 = 71 + .97(18.54) 71 .43(18.54) = 63.03 78.97 68/71 Normal Distribution Goodness of Fit Test Observed and Expected Frequencies i fi ei f i - 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 69/71 Normal Distribution Goodness of Fit Test Rejection Rule 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 H0 if p-value < .05 or 2 > 7.815. Test Statistic (1)2 (2)2 (1) 2 (0) 2 (1) 2 (1) 2 2 1.600 5 5 5 5 5 5 70/71 Normal Distribution Goodness of Fit Test Conclusion Using the p-Value Approach Area in Upper Tail .90 .10 .05 .025 .01 2 Value (df = 3) .584 6.251 7.815 9.348 11.345 Because χ2 = 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 > . 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. 71/71