Chi Square Goodness of Fit Test Help
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Example (from Statistics and Data Analysis, by Ajit Tamhane and Dorothy Dunlop, Prentice-Hall, 2000)
A 1959 study measured the number of passengers in a car in urban traffic at one intersection. The data are given below. Do these data represent a Poisson distribution? There are 1,011 observations.
![](https://www.spcforexcel.com/wp-content/uploads/Chi-Fit-Data1.jpg)
To use this technique, you must determine the expected frequency or number. To do this, first determine the average of the observed frequencies. This is defined as the following:
? =[ (678*0)+(227*1)+(56*2)+(28*3)+(8*4)+(14*5)]/(678+227+56+28+8+14) = 0.519
You can use Excel’s Poisson function to find the expected values. For each number of passengers, use POISSON(x, 0.519, False) to find the expected value where x is the number of passengers. For example, for x = 0, the expected value is 602. Once this is complete, you can apply the Chi-Square Goodness of Fit test.
- 1. Enter the results into an Excel worksheet as shown below. The data can be downloaded at this link.
![](https://www.spcforexcel.com/wp-content/uploads/Chi-Fit-Data2.jpg)
- 2. Select all the data in the table above including the headings.
- 3. Select “Misc. Tools” from the “Statistical Tools” panel on the SPC for Excel ribbon.
- 4. Select the “Chi Square Goodness of Fit” option and then OK.
![](https://www.spcforexcel.com/wp-content/uploads/Chi-Fit-Input.jpg)
- Enter Data Range with Labels: enter the range containing the data and the labels; default is the range selected on the worksheet.
- Alpha: this is the confidence level; 1-alpha is the confidence interval. Default is 0.05 for 95% confidence.
- Select OK to generate the results.
- Select Cancel to end the program.
Chi Square Goodness of Fit Test Output
The output from the Chi Square Goodness of Fit Test is shown below.
![](https://www.spcforexcel.com/wp-content/uploads/Chi-Fit-Output.jpg)
The null and alternate hypothesis are printed at the top of a new worksheet. The data are printed and the contribution to chi-square determine. The value of alpha is printed and the calculated and critical values for chi-square are calculated. The p value is determined. If it is less than 0.05, it is in red. The conclusion is then generated based on the values of p and alpha. In this example, the null hypothesis is rejected. There is evidence that the data does not follow a Poisson distribution.