12.10 (STUDENT CD-ROM TOPIC) CHI-SQUARE GOODNESS- OF-FIT TESTS

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1 CDR4_BERE601_11_SE_C1QXD 1//08 1:0 PM Page 1 110: (Student CD-ROM Topic) Chi-Square Goodness-of-Fit Tests CD (STUDENT CD-ROM TOPIC) CHI-SQUARE GOODNESS- OF-FIT TESTS In this section, χ goodness-of-fit tests are used to determine whether a set of data matches a specific probability distribution Goodness-of-fit tests compare the observed frequencies in a category to the frequencies that are theoretically expected if the data follow a specific probability distribution To perform a χ goodness-of-fit test, you follow several steps First, you determine the specific probability distribution to compare to the data Second, you estimate (from a sample) or hypothesize the value of each parameter (such as the mean) of the selected probability distribution Next, you determine the theoretical probability in each category using the selected probability distribution Finally, you use a χ -test statistic to test whether the selected distribution is a good fit to the data Chi-Square Goodness-of-Fit Test for a Poisson Distribution Recall from Section 54 that you used the Poisson distribution to find the probability of a specific number of arrivals per minute at a bank located in the central business district of a city Suppose that you recorded the actual arrivals per minute in 00 one-minute periods over the course of a week Table 1 summarizes the results TABLE 1 Distribution of Arrivals per Minute Arrivals To determine whether the number of arrivals per minute follows a Poisson distribution, the null and alternative hypotheses are: H 0 : The number of arrivals per minute follows a Poisson distribution H 1 : The number of arrivals per minute does not follow a Poisson distribution The Poisson distribution has one parameter, its mean λ, and you need to specify its value in the null and alternative hypotheses You can use either a λ value based on past knowledge, or use a λ value estimated from sample data In this example, you estimate λ using the data in Table 1 Using Equation (313) on page 114 and the computations in Table 13 X c mj fj j = 1 = n 580 = = 90 00

2 CDR4_BERE601_11_SE_C1QXD 1//08 1:0 PM Page CD1- CHAPTER TWELVE Chi-Square Tests and Nonparametric Tests TABLE 13 Computing the Sample Mean Number of Arrivals from the Distribution of Arrivals per Minute Arrivals f j m j f j To find the probabilities from the tables of the Poisson distribution (Table E7), you use 90 as an estimate of λ You then compute the expected frequency for each number of arrivals by multiplying the appropriate Poisson probability by the sample size n = 00 Table 14 summarizes these results TABLE 14 Observed and Expected Frequencies of the Arrivals per Minute Probability, P(X), for Observed Poisson Distribution with Expected Arrivals f o λ = 9 f e = n P(X) or more Observe in Table 14 that the expected frequency of 9 or more arrivals is less than 10 In order to have all categories contain a frequency of 10 or greater, you need to combine the category 9 or more with the category of 8 arrivals Equation (115) defines the chi-square test for determining whether the data follow a specific probability distribution Chi-Square Goodness-of-Fit Test fo fe χ STAT = ( ) f k e (115) where f o = observed frequency f e = expected frequency k = number of categories or classes remaining after combining classes p = number of parameters estimated The test statistic follows a chi-square distribution with k p 1 degrees of freedom χ STAT

3 CDR4_BERE601_11_SE_C1QXD 1//08 1:0 PM Page 3 110: (Student CD-ROM Topic) Chi-Square Goodness-of-Fit Tests CD1-3 Returning to the example concerning the arrivals at the bank, nine categories remain (0, 1,, 3, 4, 5, 6, 7, 8 or more) Because you estimated the mean of the Poisson distribution from the data, the number of degrees of freedom are k p 1 = = 7 Using the 005 level of significance, from Table E4, the critical value of χ with 7 degrees of freedom is The decision rule is STAT Reject H if χ > ; otherwise do not reject H 0 Table 15 shows the computation of the χ STAT 0 TABLE 15 Computing the Chi- Square Test Statistic for the Arrivals per Minute Arrivals f o f e (f o ) (f o ) (f o ) /f e or more χ STAT Because = 8954 < 14067, you do not reject H 0 There is insufficient evidence to conclude that the arrivals per minute do not fit a Poisson distribution Chi-Square Goodness-of-Fit Test for a Normal Distribution In Chapters 8 through 11, when constructing confidence intervals and testing hypotheses about numerical variables, you assumed that the underlying population was normally distributed Although the boxplot and the normal probability plot are useful for evaluating the validity of this assumption, when you have a large sample size, you can also use the χ goodness-of-fit test As an example of how you can use the χ goodness-of-fit test for a normal distribution, refer to the 006 returns achieved by the 868 funds summarized in Table 8 on page 38 To test whether these returns follow a normal distribution, the null and alternative hypotheses are: H 0 : The 006 returns follow a normal distribution H 1 : The 006 returns do not follow a normal distribution The normal distribution has two parameters, the mean μ and the standard deviation σ You must specify the values of these parameters in the null and alternative hypotheses You can use values based on past knowledge or use values estimated from sample data In this example, from the sample data: X = 1514 and S = 69 Table 8 uses class interval widths of 5 with class boundaries beginning at 10 Because the normal distribution is continuous, you need to determine the area in each class interval In addition, because a normally distributed variable theoretically ranges from to +, you must account for the area beyond the first and last class interval Thus, the area below 10 is the area below the Z value: X X Z = S = = From Table E, the area below Z = 358 is approximately

4 CDR4_BERE601_11_SE_C1QXD 1//08 1:0 PM Page 4 CD1-4 CHAPTER TWELVE Chi-Square Tests and Nonparametric Tests To find the area between 10 and 5, you compute the area below 5 as follows: X X Z = S = = From Table E, the area below Z = 78 is approximately 0007 Thus, the area between 10 and 5 is the difference in the area below 10 and the area below 5, which is = Continuing, to find the area between 5 and 00, you compute the area below 00 as follows: X X Z = S = = From Table E, the area below Z = 199 is approximately 0033 Thus, the area between 5 and 00 is the difference in the area below 5 and the area below 00, which is = 0006 In a similar manner, you can compute the area in each class interval Table 16 summarizes the complete set of computations needed to find the area and expected frequency in each class TABLE 16 Computing the Area and Expected Frequencies in Each Class Interval for the 006 Returns Classes (f o ) X X X Z Area Below Area in Class f e = n P(X) Below but less than but less than but less than but less than but less than but less than but less than but less than but less than but less than or more Note: The areas in the classes were calculated using Microsoft Excel P(X) Observe in Table 16 that the expected frequency in the category below 10, 35 but less than 40, and in the category 40 or more are each less than 10 Because all categories need to have a frequency of 10 or greater, you combine the category below 10 with the category 10 to 5, and the categories 35 but less than 40 and 40 or more with the category 30 but less than 35

5 CDR4_BERE601_11_SE_C1QXD 1//08 1:0 PM Page 5 110: (Student CD-ROM Topic) Chi-Square Goodness-of-Fit Tests CD1-5 Using Equation (115), you now compute the chi-square test statistic In this example, after combining classes, 9 classes remain Because you used the sample mean and sample standard deviation to estimate the population mean and population standard deviation, the number of degrees of freedom is equal to k p 1 = 9 1 = 6 Using a level of significance of 005, the critical value of chi-square with 6 degrees of freedom is 159 Table 17 summarizes the computations for the chi-square test Because χ STAT = 3185 > 159, reject H 0 There is sufficient evidence to conclude that the 006 returns do not fit a normal distribution TABLE 17 Computing the Chi- Square Test Statistic for the 006 Returns Classes f o f e (f o ) (f o ) (f o ) /f e Less than but less than but less than but less than but less than but less than but less than but less than or more PROBLEMS FOR SECTION 110 Applying the Concepts 1117 The manager of a computer network has collected data on the number of times that service has been interrupted on each day over the past 500 days The results are as follows: Interruptions Number of per Day Days Does the distribution of service interruptions follow a Poisson distribution? (Use the 001 level of significance) 1118 Referring to the data in Problem 1117, at the 001 level of significance, does the distribution of service interruptions follow a Poisson distribution with a population mean of 15 interruptions per day? 1119 The manager of a commercial mortgage department of a large bank has collected data during the past two years concerning the number of commercial mortgages approved per week The results from these two years (104 weeks) indicate the following: Number of Commercial Mortgages Approved Does the distribution of commercial mortgages approved per week follow a Poisson distribution? (Use the 001 level of significance) 110 A random sample of 500 car batteries revealed the following distribution of battery life (in years)

6 CDR4_BERE601_11_SE_C1QXD 1//08 1:0 PM Page 6 CD1-6 CHAPTER TWELVE Chi-Square Tests and Nonparametric Tests Life (in Years) 0 under under 94 under under under under For these data, X = 80 and S = 097 At the 005 level of significance, does battery life follow a normal distribution? 111 A random sample of 500 long distance telephone calls revealed the following distribution of call length (in minutes) Length (in Minutes) 0 under under under under under under a Compute the mean and standard deviation of this frequency distribution b At the 005 level of significance, does call length follow a normal distribution?

Statistics for Managers Using Microsoft Excel

Statistics for Managers Using Microsoft Excel 7 th Edition Chapter 1 Chi-Square Tests and Nonparametric Tests Statistics for Managers Using Microsoft Excel 7e Copyright 014 Pearson Education, Inc. Chap