Question

The following table shows the Myers-Briggs personality preferences for a random sample of 309 people in the listed professions.

Occupation	Extroverted	Introverted	Row Total
Clergy (all denominations)	44	34	78
M.D.	50	72	122
Lawyer	62	47	109
Column Total	156	153	309

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Suppose that we use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.05 level of significance. Depending on the P-value, will you reject or fail to reject the null hypothesis of independence?

Group of answer choices

Since the P-value is less than α, we fail to reject the null hypothesis that the Myers-Briggs personality preference and profession are independent. At 0.05 level of significance, we conclude that the Myers-Briggs personality preference and profession are not independent.

Since the P-value is greater than α, we reject the null hypothesis that the Myers-Briggs personality preference and profession are not independent. At 0.05 level of significance, we conclude that the Myers-Briggs personality preference and profession are independent.

Since the P-value is greater than α, we fail to reject the null hypothesis that the Myers-Briggs personality preference and profession are not independent. At 0.05 level of significance, we conclude that the Myers-Briggs personality preference and profession are independent.

Since the P-value is less than α, we reject the null hypothesis that the Myers-Briggs personality preference and profession are independent. At 0.05 level of significance, we conclude that the Myers-Briggs personality preference and profession are not independent.

Since the P-value is greater than α, we fail to reject the null hypothesis that the Myers-Briggs personality preference and profession are independent. At 0.05 level of significance, we conclude that the Myers-Briggs personality preference and profession are not independent.

Answer 1

Chi-Square Test of independence
	Observed Frequencies
	0
0	E	I					Total
C	44	34					78
M	50	72					122
L	62	47					109
Total	156	153					309
	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	E	I					Total
C	156*78/309=39.379	153*78/309=38.621					78
M	156*122/309=61.592	153*122/309=60.408					122
L	156*109/309=55.029	153*109/309=53.971					109
Total	156	153					309
	(fo-fe)^2/fe
C	0.5423	0.5530
M	2.1818	2.2245
L	0.8830	0.9004

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   7.285  
      
Level of Significance =   0.05  
Number of Rows =   3  
Number of Columns =   2  
Degrees of Freedom=(#row - 1)(#column -1) = (3- 1 ) * ( 2- 1 ) =   2  
      
p-Value =   0.0261862   [Excel function: =CHISQ.DIST.RT(χ²,df) ]
Decision:    p-value < α , Reject Ho  

Since the P-value is less than α, we reject the null hypothesis that the Myers-Briggs personality preference and profession are independent. At 0.05 level of significance, we conclude that the Myers-Briggs personality preference and profession are not independent.