Comparing Multiple Proportions, Test of Independence and Goodness of Fit

Comparing Multiple Proportions, Test of Independence and Goodness of Fit

Content Testing the Equality of Population Proportions for Three or More Populations Test of Independence Goodness of Fit Test 2

Comparing Multiple Proportions, Test of Independence and Goodness of Fit In this chapter we introduce three additional hypothesis-testing procedures. The test statistic and the distribution used are based on the chi-square ( 2 ) distribution. In all cases, the data are categorical. 3

Testing the Equality of Population Proportions for Three or More Populations Using the notation p 1 = population proportion for population 1 p 2 = population proportion for population 2 and p k = population proportion for population k The hypotheses for the equality of population proportions for k 3 populations are as follows: H 0 : p 1 = p 2 =... = p k H a : Not all population proportions are equal 4

Testing the Equality of Population Proportions for Three or More Populations If H 0 cannot be rejected, we cannot detect a difference among the k population proportions. If H 0 can be rejected, we can conclude that not all k population proportions are equal. Further analyses can be done to conclude which population proportions are significantly different from others. 5

Testing the Equality of Population Proportions for Three or More Populations Example: Finger Lakes Homes Finger Lakes Homes manufactures three models of prefabricated homes, a two-story colonial, a log cabin, and an A-frame. To help in product-line planning, management would like to compare the customer satisfaction with the three home styles. p 1 = proportion likely to repurchase a Colonial for the population of Colonial owners p 2 = proportion likely to repurchase a Log Cabin for the population of Log Cabin owners p 3 = proportion likely to repurchase an A-Frame for the population of A-Frame owners 6

Testing the Equality of Population Proportions for Three or More Populations We begin by taking a sample of owners from each of the three populations. Each sample contains categorical data indicating whether the respondents are likely or not likely to repurchase the home. 7

Testing the Equality of Population Proportions for Three or More Populations Observed Frequencies (sample results) Home Owner Colonial Log A-Frame Total Likely to Yes 100 81 83 264 Repurchase No 35 20 41 96 Total 135 101 124 360 8

Testing the Equality of Population Proportions for Three or More Populations Next, we determine the expected frequencies under the assumption H 0 is correct. Expected Frequencies Under the Assumption H 0 is True (Row i Total)(Column j Total) eij Total Sample Size If a significant difference exists between the observed and expected frequencies, H 0 can be rejected. 9

Testing the Equality of Population Proportions for Three or More Populations Expected Frequencies (computed) Home Owner Colonial Log A-Frame Total Likely to Yes 99.00 74.07 90.93 264 Repurchase No 36.00 26.93 33.07 96 Total 135 101 124 360 10

Testing the Equality of Population Proportions for Three or More Populations Next, compute the value of the chi-square test statistic. ( f ) 2 ij eij e i j ij where: f ij = observed frequency for the cell in row i and column j e ij = expected frequency for the cell in row i and column j under the assumption H 0 is true 2 Note: The test statistic has a chi-square distribution with k 1 degrees of freedom, provided the expected frequency is 5 or more for each cell. 11

Testing the Equality of Population Proportions for Three or More Populations Computation of the Chi-Square Test Statistic. Obs. Exp. Sqd. Sqd. Diff. / Likely to Home Freq. Freq. Diff. Diff. Exp. Freq. Repurch. Owner f ij e ij (f ij - e ij ) (f ij - e ij ) 2 (f ij - e ij ) 2 /e ij Yes Colonial 97 97.50-0.50 0.2500 0.0026 Yes Log Cab. 83 72.94 10.06 101.1142 1.3862 Yes A-Frame 80 89.56-9.56 91.3086 1.0196 No Colonial 38 37.50 0.50 0.2500 0.0067 No Log Cab. 18 28.06-10.06 101.1142 3.6041 No A-Frame 44 34.44 9.56 91.3086 2.6509 Total 360 360 2 = 8.6700 12

Testing the Equality of Population Proportions for Three or More Populations Rejection Rule p-value approach: Reject H 0 if p-value < a Critical value approach: Reject H 0 if 2 2 where is the significance level and there are k - 1 degrees of freedom 13

Testing the Equality of Population Proportions for Three or More Populations Rejection Rule (using a =.05) Reject H 0 if p-value.05 or 2 5.991 With =.05 and k - 1 = 3-1 = 2 degrees of freedom Do Not Reject H 0 Reject H 0 5.991 2 14

Testing the Equality of Population Proportions for Three or More Populations Conclusion Using the p-value Approach Area in Upper Tail.10.05.025.01.005 2 Value (df = 2) 4.605 5.991 7.378 9.210 10.597 Because 2 = 8.670 is between 9.210 and 7.378, the area in the upper tail of the distribution is between.01 and.025. The p-value. We can reject the null hypothesis. Actual p-value is.0131 15

Testing the Equality of Population Proportions for Three or More Populations We have concluded that the population proportions for the three populations of home owners are not equal. To identify where the differences between population proportions exist, we will rely on a multiple comparisons procedure. 16

Multiple Comparisons Procedure We begin by computing the three sample proportions. Colonial: p 100 1.741 135 Log Cabin: p 81 2.802 101 A-Frame: p 83 3.669 124 We will use a multiple comparison procedure known as the Marascuillo procedure. 17

Multiple Comparisons Procedure Marascuillo Procedure We compute the absolute value of the pairwise difference between sample proportions. Colonial and Log Cabin: Colonial and A-Frame: Log Cabin and A-Frame: p 1 p 2.741.802.061 p 1 p 3.741.669.072 p 2 p 3.802.669.133 18

Multiple Comparisons Procedure Critical Values for the Marascuillo Pairwise Comparison For each pairwise comparison compute a critical value as follows: CV ij 2 p (1 ) i(1 p) pj p i j ; k 1 n n i j For =.05 and k = 3: 2 = 5.991 19

Multiple Comparisons Procedure Pairwise Comparison Tests Pairwise Comparison Significant if p p CV ij p p > CV ij i j i j Colonial vs. Log Cabin.061.0923 Not Significant Colonial vs. A-Frame.072.0971 Not Significant Log Cabin vs. A-Frame.133.1034 Significant 20

Test of Independence 1. Set up the null and alternative hypotheses. H 0 : The column variable is independent of the row variable H a : The column variable is not independent of the row variable 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. (Row i Total)(Column j Total) eij Sample Size 21

Test of Independence 4. Compute the test statistic. ( f ) 2 ij eij e i j ij 5. Determine the rejection rule. 2 2 Reject H 0 if p -value or. where is the significance level and, with n rows and m columns, there are (n -1)(m - 1) degrees of freedom. 2 22

Test of Independence 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. 23

Test of Independence 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 24

Test of Independence 1. Hypotheses 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 style of the home that is purchased 2. Test of Independence 2 Test 3. =.05 2 2 4. Reject H 0 if 25

Test of Independence With =.05 and (2-1)(4-1) = 3 d.f., 2.05 7.815 Reject H 0 if p-value.05 or 2 7.815 5. Calculate Test Statistic 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 26

Test of Independence Test Statistic 2 (18 16.5) (6 11) (3 6.75)... 16.5 11 6.75.1364 2.2727 2.0833 9.149 2 2 2 27

Test of Independence 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 2 = 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. Actual p-value is.0274 28

Test of Independence Conclusion Using the Critical Value Approach 2 = 9.145 7.815 6. Conclusion 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. 29

Goodness of Fit Test: Multinomial Probability Distribution 1. State the null and alternative hypotheses. H 0 : The population follows a multinomial distribution with specified probabilities for each of the k categories H a : The population does not follow a multinomial distribution with specified probabilities for each of the k categories 30

Goodness of Fit Test: Multinomial Probability Distribution 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. 31

Goodness of Fit Test: Multinomial Probability Distribution 4. Compute the value of the test statistic. 2 k ( fi ei) e i 1 where: f i = observed frequency for category i e i = 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. i 2 32

Goodness of Fit Test: Multinomial Probability Distribution 5. Rejection rule: p-value approach: Reject H 0 if p-value < Critical value approach: Reject H 0 if 2 2 where is the significance level and there are k -1 degrees of freedom 33

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. 34

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 35

Multinomial Distribution Goodness of Fit Test 1. Hypotheses H 0 : p C = p L = p S = p A =.25 H a : The population proportions are not p where: C =.25, p L =.25, p S =.25, and p A =.25 p C = population proportion that purchase a colonial p L = population proportion that purchase a log cabin p S = population proportion that purchase a split-level p A = population proportion that purchase an A-frame 2. Goodness of Fit Test 2 Test 3. =.05 4. Reject H 0 if 2 2 36

Multinomial Distribution Goodness of Fit Test Rejection Rule Reject H0 if p-value.05 or 2 7.815. With =.05 and k - 1 = 4-1 = 3 degrees of freedom Do Not Reject H 0 Reject H 0 7.815 2 37

Multinomial Distribution Goodness of Fit Test 5. Calculate Test Statistic Expected Frequencies Test Statistic e 1 =.25(100) = 25 e 2 =.25(100) = 25 e 3 =.25(100) = 25 e 4 =.25(100) = 25 2 (30 25) (20 25) (35 25) (15 25) 25 25 25 25 =1 1 4 4 10 2 2 2 2 38

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 2 = 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. 39

Multinomial Distribution Goodness of Fit Test Conclusion Using the Critical Value Approach 2 = 10 > 7.815 6. Conclusion We reject, at the.05 level of significance, the assumption that there is no home style preference. 40

Goodness of Fit Test: Normal Distribution 1. State the null and alternative hypotheses. H 0 : The population has a normal distribution H a : 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 3. Compute the expected frequency, e i, for each interval. (Multiply the sample size by the probability of a normal random variable being in the interval.) 41

Goodness of Fit Test: Normal Distribution 4. Compute the value of the test statistic. 2 k i 1 ( f e ) i e i i 2 2 2 5. Reject H0 if (where is the significance level and there are k 2 1 degrees of freedom). 42

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. 43

Goodness of Fit Test: Normal Distribution Example: IQ Computers A simple random sample of 30 of the salespeople was taken and their numbers of units sold are listed 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) 44

Goodness of Fit Test: Normal Distribution 1. Hypotheses 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. 2. Goodness of Fit Test 2 Test 3. =.05 4. Reject H 0 if 2 2 45

Goodness of Fit Test: Normal Distribution 5. 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. 46

Goodness of Fit Test: Normal Distribution Interval Definition Areas = 1.00/6 =.1667 53.02 71 71.43(18.54) = 63.03 78.97 88.98 = 71 +.97(18.54) 47

Goodness of Fit Test: Normal Distribution Observed and Expected Frequencies i f i e i f i - e i 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 48

Goodness of Fit Test: Normal Distribution Rejection Rule With =.05 and k - p - 1 = 6-2 - 1 = 3 d.f. (where k = number of categories and p = number 2 of population parameters estimated),.05 7.815 Test Statistic Reject H 0 if p-value.05 or 2 7.815. 2 2 2 2 2 2 2 (1) ( 2) (1) (0) ( 1) (1) 5 5 5 5 5 5 1.600 49

Goodness of Fit Test: Normal Distribution 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 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. 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. Actual p-value is.6594 50

Goodness of Fit Test: Normal Distribution Conclusion Using the Critical Value Approach 2 = 1.6 < 7.815 6. Conclusion We cannot reject, at the.05 level of significance, the assumption that the population is normally distributed with = 71 and = 18.54. 51

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. b. Compute the mean number of occurrences,. 3. Compute the expected frequency of occurrences, e i, for each value of the Poisson random variable. 52

Goodness of Fit Test: Poisson Distribution 4. Compute the value of the test statistic. 5. Reject H 0 if 2 k i 1 2 2 f e 2 i e i i (where is the significance level and there are k 1 1 degrees of freedom). 53

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. 54

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. # 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 55

Example: Troy Parking Garage Poisson Distribution Goodness of Fit Test 1. Hypotheses H 0 : Number of cars entering the garage during a oneminute interval is Poisson distributed. H a : Number of cars entering the garage during a oneminute interval is not Poisson distributed 2. Goodness of Fit Test 2 Test 3. =.05 4. Reject H 0 if 2 2 56

Example: Troy Parking Garage Poisson Distribution Goodness of Fit Test 5. Calculation 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, f( x) 6 x 6 e x! 57

Example: Troy Parking Garage Poisson Distribution Goodness of Fit Test Expected Frequencies x f (x ) nf (x ) x f (x ) nf (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 12.0155 1.55 6.1606 16.06 Total 1.0000 100.00 58

Example: Troy Parking Garage Poisson Distribution Goodness of Fit Test Observed and Expected Frequencies i f i e i f i - e i 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 59

Example: Troy Parking Garage Poisson Distribution Goodness of Fit Test Test Statistic 2 2 2 2 ( 1.20) (1.08) (2.01)... 3.42 6.20 8.92 7.99 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), 2.05 14.07 Reject H 0 if 2 > 14.07 6. Conclusion We cannot reject H 0. There s no reason to doubt the assumption of a Poisson distribution. 60