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INDE 2333 ENGINEERING STATISTICS I GOODNESS OF FIT - PowerPoint PPT Presentation

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.

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ENGINEERING STATISTICS I

GOODNESS OF FIT

University of Houston

Dept. of Industrial Engineering

Houston, TX 77204-4812

(713) 743-4195

• Chi-square goodness of fit test

• 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?

• 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…

• Approach 1

• 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

• 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

• 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

• 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-SQUAREfor a particular number of degrees of freedom

f(X^2)

Right tail probability, alpha, typically 0.05

0

X^2

X^2 Critical value

• Cannot 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 1EQUAL 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

• Ho:

• Poisson distribution with mean of 4.6

• Ha:

• Not poisson distribution with a mean of 4.6

CHI-SQUAREfor 10-1 degrees of freedom

f(X^2)

Right tail probability, alpha = 0.01

0

X^2

16.919 Critical value

• 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 2EQUIPROBABLE APPROACH

• Were the scores from an INDE 2333 exam normally distributed?

• Sample statistics

• Mean=71.95

• Std=11.93

• N=43

• 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

• 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-SQUAREfor 5 of degrees of freedom

f(X^2)

0.05

0

X^2

11.070

• 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

• 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

• Frequency

• Data_array, bins_array

• Range operation

• CTRL-SHIFT-ENTER

• Norminv function

• Probability, mean, std

• Chiinv function

• Probability, df