Question

A comparative study of job satisfaction was conducted in factories from different regions of the country. As part of the study, samples of workers from two factories were taken,
One in the north and the other in the south, and the employees were asked how satisfied they were with the work. Here are the results:

Region	Amount not Satisfied	Amount Satisfied	Amount very satisfied
Northern (X)	5	20	45
Southern (Y)	35	15	10

1) Does the location of the factory affect the satisfaction level? Test with significance level=3%.

2) Is the degree of satisfaction uniform? Does each level of satisfaction have a 1/3 chance of happening? Check if this is the case.

Answer 1

1)

Ho:  location of the factory does not affect the satisfaction level

Ha:  location of the factory affect the satisfaction level

Chi-Square Test of independence
	Observed Frequencies
	0
0	not	satistified	very satisfied				Total
X	5	20	45				70
Y	35	15	10				60
Total	40	35	55				130
	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	not	satistified	very satisfied				Total
X	40*70/130=21.538	35*70/130=18.846	55*70/130=29.615				70
Y	40*60/130=18.462	35*60/130=16.154	55*60/130=25.385				60
Total	40	35	55				130
	(fo-fe)^2/fe
X	12.6992	0.0706	7.992
Y	14.8157	0.0824	9.324

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   44.9840  
      
Level of Significance =   0.03  
Number of Rows =   2  
Number of Columns =   3  
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 3- 1 ) =   2  
      
p-Value =   0.0000   [Excel function: =CHISQ.DIST.RT(χ²,df) ]
Decision:    p-value < α , Reject Ho  

there is enough evidence that  location of the factory affect the satisfaction level

2)

observed frequencey, O	expected proportion	expected frequency,E	(O-E)		(O-E)²/E
40	0.333	43.33	-3.33		0.256
35	0.333	43.33	-8.33		1.603
55	0.333	43.33	11.67		3.141

chi square test statistic,X² = Σ(O-E)²/E =   5.000              
                  
level of significance, α=   0.03              
Degree of freedom=k-1=   3   -   1   =   2
                  
P value =   0.0821   [ excel function: =chisq.dist.rt(test-stat,df) ]          
Decision: P value >α , Do not reject Ho                  

there is no enough evidence to reject the claim that  degree of satisfaction is uniform