The chi-square goodness-of-fit test is a statistics used to compare and decide whether two or more populations, variables or characteristics are the same. It does not matter what the distributions of the populations are so long as the relative frequencies are known for each population or the population and some standard population frequencies.

1. Know the procedure used to evaluate Goodness-of-Fit chi-square tests.

Problem: The number of defects for a new thermometer is classified by the following defects types with their expected defects percent obtained from historical statistics from an older model thermometer:
 

Defects (category)	Number of defect Observed, n	Expected % defect
N. None	1,188(n1)	87
A. Class A	91(n2)	9
B. Class B	47(n3)	3
C. Class C	10(n4)	1
Since number of cells within a category is 4, k = 4-1=3	Sample size, n = 1336

Step 1. Make a problems statement: (becomes the hypothesis statement, H_o ).

Is the new thermometer the same quality (based on defect profile) as the old? or

Are the distribution of defects from the new thermometer the same as the distribution of defects from the old?.

Is the new thermometer (based on the characteristics of defect proportions) the same as the standard (old thermometer)?

Hypothesis:

So H_o (null hypothesis): Old = New or Old is same as New or Distribution of New is same as Old.

H_a: H_o is not true. (alternate hypothesis)

Step 2. Choose , the significance level of the test.

If you want the be 95 % certain that the test is true, then  = 0.05 =(100-95)/100

Step 3. Look up  from chi-square table:
 

df
3	0.07	0.12	0.22	0.35	0.58	6.25	7.81	9.35	11.34	12.84

With 4 types of defects (categories) d.f. = 4-1 = 3, so 

Step 4. (4) Determine or compute , The Expected Frequencies:
 

Defects Type, k =3	pi	Number of defect Observed, O, ni	E=np  Expected Defects, E	O-E	(O-E)2
N. None	0.87	1,188(n1)	1,162.32	25.68	659.4624	0.5674
A. Class A	0.09	91(n2)	120.24	-29.24	854.9776	7.1106
B. Class B	0.03	47(n3)	40.08	6.92	47.8864	1.1948
C. Class C	0.01	10(n4)	13.36	-3.36	11.2896	0.845
Total	1	n = 1336	1,336			= 9.7178

Step 5. Compute 

= 9.72 from computational table above.

Step 6 Perform test chi-square test: , 

Since , i.e. 9.72 > 7.81, Then we reject the null hypothesis ( that ) is true (there is a difference between defects levels of the new and the old).

Make Conclusion or inference:

The old thermometer is different than the old (pertaining to the characteristics of defect distribution)

So H_o (null hypothesis) is rejected; there seems to be a difference between old and new.

Workshop Problem 11a (goodness-of-fit):

There are 4 types of response to a questionnaire about students' study habits, If the types of responses are equally likely, and the actual number of response per each type is given in the table below, do we have evidence to conclude that response types are not all equal?
 

Response Types	1	2	3	4
Number of Responses	30	40	33	47

(a) What is the chi-square statistics?

(b) What is your inference about based on chi-square anaylsis?

(c) Why did you select this particular chi-square anaylsis?

(d) Is there any other inference you can make about the data from the conclusion of the chi-square anaylsis?