Question

In order to receive credit for the homework, you must show all steps for doing a chi square test of independence. These steps include: null and alternative hypothesis); critical value; calculations (show work); reject or accept H; results statement using no relationship or a relationship.

Answer 1

Ho: there is no relationship between quality and shift

H1: there is a relationship between quality and shift

Chi-Square Test of independence
	Observed Frequencies
	Y
0	perfect	satisfactory	defective				Total
shift 1	106	124	1				231
shift 2	67	85	2				154
Total	173	209	3				385
	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	perfect	satisfactory	defective				Total
shift 1	173*231/385=103.8	209*231/385=125.4	3*231/385=1.8				231
shift 2	173*154/385=69.2	209*154/385=83.6	3*154/385=1.2				154
Total	173	209	3				385
	(fo-fe)^2/fe
shift 1	0.047	0.016	0.356
shift 2	0.070	0.023	0.533

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   1.0445
  
Level of Significance =   0.05
Number of Rows =   2
Number of Columns =   3
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 3- 1 ) =   2

Critical Value =   5.991   [ Excel function: =CHISQ.INV.RT(α,DF) ]  
Decision: test statistic,X² < critical value , So, Do not reject the null hypothesis          

conclusion: there is no enough evidence to conclude that there is a relationship between quality and shift