Loading in 5 sec....

INDE 2333 ENGINEERING STATISTICS I GOODNESS OF FITPowerPoint Presentation

INDE 2333 ENGINEERING STATISTICS I GOODNESS OF FIT

- By
**suki** - Follow User

- 191 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' INDE 2333 ENGINEERING STATISTICS I GOODNESS OF FIT' - suki

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

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

- 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

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-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

DECISION

- 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

HYPOTHESES

- 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

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 2EQUIPROBABLE 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

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

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
- Data_array, bins_array

- Range operation
- CTRL-SHIFT-ENTER

- Norminv function
- Probability, mean, std

- Chiinv function
- Probability, df

Download Presentation

Connecting to Server..