Question

2. Below is a contingency table consisting of seat belt use and amount of smoking per day. Use a .05 significance level to test the claim that the amount of smoking is independent of seat belt use.

	Number of Cigarettes Smoked per Day
	0	1 – 14	15 – 34	35 and over
Wear Seat Belts	175	20	42	6
Don’t Wear Seat Belts	149	17	41	9

Test the claim using the critical value method (Check your answer with the p-value on the calculator).

Answer 1

Chi-Square Test of independence
	Observed Frequencies
	0
0	0	1-14	15-34	35 over			Total
wear seat bels	175	20	42	6			243
do not wear	149	17	41	9			216
Total	324	37	83	15			459
	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	0	1-14	15-34	35 over			Total
wear seat bels	324*243/459=171.529	37*243/459=19.588	83*243/459=43.941	15*243/459=7.941			243
do not wear	324*216/459=152.471	37*216/459=17.412	83*216/459=39.059	15*216/459=7.059			216
Total	324	37	83	15			459
	(fo-fe)^2/fe
wear seat bels	0.070	0.009	0.086	0.475
do not wear	0.079	0.010	0.096	0.534

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   1.358          
              
Level of Significance =   0.05          
Number of Rows =   2          
Number of Columns =   4          
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 4- 1 ) =   3          
              

Critical Value =   7.815
Decision: test statistic,X² < critical value , So, Do not reject the null hypothesis  

.....

p-Value =   0.7154
Decision:    p value > α , do not reject Ho                

same resluts with p value

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