ENGINEERING STATISTICS I INDE 2333 ENGINEERING STATISTICS I GOODNESS OF FIT University of Houston Dept. of Industrial Engineering Houston, TX 77204-4812 (713) 743-4195

AGENDA Chi-square goodness of fit test

GOODNESS OF FIT TESTS Used to determine if a sample could have come from a distribution with the specified parameters Commonly used to determine if data is normally distributed Many tests such as the ones that we have been using require normally distributed data. If data is not normally distributed, non-parametric tests must be used (next subject in the course) Also used for input distributions in system modeling Customers or jobs arrive exponentially distributed? Service times follow what distribution? Failures occur according to what distribution?

GOODNESS OF FIT TESTS Based on a comparison of observations between Observed data Theoretical data The comparison utilizes a set of intervals or cells Each cell has a lower and upper boundary values The determination of the boundaries are a function of Theoretical distribution Number of observations in the sample 2 different approaches…

TWO DIFFERENT APPROACHES Used in the book Equal interval approach No cell grouping can have less than 5 expected observations Approach 2 Used in other books Equiprobable approach Maximum number of cells not to exceed 100 such that the expected number of observations is at least 5 = Int ( obs/5 ) Expected number of obs in each cell = obs / cells More statistically robust

HYPOTHESES TEST PROCEDURE Identify Ho and Ha Determine level of significance (generally 0.05 or 0.01) Determine “critical value” criterion from level of significance Calculate “test statistic” Make decision Fail to reject Ho Reject Ho

HYPOTHESES Ho The sample could have come from a distribution with the specified parameters Ha The sample could not have come from a distribution with the specified parameters

CRITICAL VALUE Chi-square distribution chart One sided test Alpha typically 0.05 Degrees of freedom # of cells - # of parameters used from sample -1 The -1 is always used due to the known sample size n Note, if the parameters are specified not sampled then they do not reduce the number of degrees of freedom in the above equation

CHI-SQUARE for a particular number of degrees of freedom f(X^2) Right tail probability, alpha, typically 0.05 X^2 X^2 Critical value

TEST STATISTIC

DECISION Cannot reject Reject Test statistic is less than the critical value Sample could have come from a distribution with the specified parameters Reject Test statistic is greater than the critical value Sample could not have come from a distribution with the specified parameters

EXAMPLE 1 EQUAL INTERVAL APPROACH 400 5 minute intervals were observed for air traffic control messages At alpha=0.01, is the distribution of the number of messages able to be considered as having a poisson distribution with a mean of 4.6? Approach Lamba parameter of 4.6 is given Use the poisson table probability table for 4.6 Multiply the probability by 400 to obtain the expected observations Compare the actual observations to the expected observations

HYPOTHESES Ho: Ha: Poisson distribution with mean of 4.6 Not poisson distribution with a mean of 4.6

Messages Observed Probability Expected 3 Combine 0.010 4.0 1 15 for 18 0.046 18.4 For 22.4 2 47 0.107 42.8 76 0.163 65.2 4 68 0.187 74.8 5 74 0.173 69.2 6 46 0.132 52.8 7 39 0.087 34.8 8 0.050 20.0 9 0.025 10.0 10 0.012 4.8 11 these 4 0.005 2.0 These 4 12 cells for a 0.002 0.8 Cells for a 13 total of 8 0.001 0.4

CHI-SQUARE for 10-1 degrees of freedom f(X^2) Right tail probability, alpha = 0.01 X^2 16.919 Critical value

TEST STATISTIC

DECISION Test statistic of 6.749 is less than the critical value of 16.919 Cannot reject Ho of distribution being poisson with a mean of 4.6 There is evidence to support the claim that the data came from a poisson distribution with a mean of 4.6 at an alpha level of 0.01

EXAMPLE 2 EQUIPROBABLE APPROACH Were the scores from an INDE 2333 exam normally distributed? Sample statistics Mean=71.95 Std=11.93 N=43

HYPOTHESES Ho The sample could have come from a normally distributed population with a mean of 71.95 and a std of 11.93 Ha The sample could not have come from a normally distributed population with a mean of 71.95 and a std of 11.93

CRITICAL VALUE Chi-square distribution chart One sided test 0.05 Degrees of freedom The sample size is 43 Want the maximum number of cells not to exceed 100 with a minimum expected number of observation of 5 43/5=8.6 cells With 8 cells, the expected number of observations is 5.375 Degrees of freedom is number of cells – number of parameters used from sample-1 Degrees of freedom=8-2-1=5

CHI-SQUARE for 5 of degrees of freedom f(X^2) 0.05 X^2 11.070

TEST STATISTIC

CELL BOUNDARIES To calculate observed values in each cell, we must determine the actual x cell boundaries from the 8 equiprobable cells For normal distributions Look up z value corresponding to probability Boundaries =mean+std * Z

CALCULATING OBSERVATIONS Cell Lower% Upper% Lower Z Upper lowerx upperx obs exp 1 0.125 -inf 58.227 5 5.375 2 0.250 63.905 6 3 0.375 68.151 4 0.500 71.953 0.625 75.756 0.750 80.002 7 0.875 85.680 8 1.000 +inf 100.00

CALCULATING TEST STATISTIC Cell obs exp ((O-E)^2)/E 1 5 5.375 0.026 2 6 0.072 3 4 1.049 7 8 1.282 Total 2.534

DECISION 2.581 < 11.070 Cannot reject the Ho Evidence to support the claim that the test scores are normally distributed with a mean of 71.95 and std of 11.93

IN EXCEL Frequency Range operation Norminv function Chiinv function Data_array, bins_array Range operation CTRL-SHIFT-ENTER Norminv function Probability, mean, std Chiinv function Probability, df