Question

The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions. E refers to extroverted and I refers to introverted.

	Personality Type
Occupation	E	I	Row Total
Clergy (all denominations)	67	40	107
M.D.	72	90	162
Lawyer	60	77	137
Column Total	199	207	406

Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.05 level of significance.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)

(c) What sampling distribution will you use?

uniformnormal  

chi-square

Student's t

binomial

(d) What are the degrees of freedom?


(e) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)


(f) Will you reject or fail to reject the null hypothesis of independence?

Since the P-value > ?, we fail to reject the null hypothesis.

Since the P-value > ?, we reject the null hypothesis.

Since the P-value ? ?, we reject the null hypothesis.

Since the P-value ? ?, we fail to reject the null hypothesis.

Answer 1

We have to test the hypothesis that

Occupations and personality preferences are indepndent or not .

i.e. Null Hypothesis : occupations and personality preferences are independent.

Alternative Hypothesis : Occupations and personality preferences are dependent.

Expected Frequency Table.

	Personality Type		Raw Total
Occupation	E	I
Clergy	52.446	54.554	107.000
M.D.	79.404	82.596	162.000
Lawyer	67.150	69.850	137.000
Column Total	199.000	207.000	406.000

b) Chi-square test statistic for testing the hypothesis is

Where Oi = Observed frequency ,Ei = expected frequency.

r= number of rows and c = number of columns.

Oi	Ei	(Oi-Ei)2/Ei
67	52.446	4.0389
72	79.404	0.6904
60	67.150	0.7614
40	54.554	3.8828
90	82.596	0.6637
77	69.850	0.7319
Total		10.7691

The value of chi-square statistic is

c) Since chi-square statistic follows chi-square distribution with (r-1)*(c-1) degrees of freedom.

We use chi-square distribution as a sampling distribution.

d) Degrees of freedom is (r-1)*(c-1)

(r-1)*(c-1) = 2 *1 = 2

Degrees of freedom= 2

e)

By using excel function CHIDIST( x, deg_freedom)

= CHIDIST(10.7691,2) = 0.004587

p-value = 0.004587

f) alpha: level of significance = 0.05

Since p-value is small as compared to level of significance alpha. The test is significant we reject the null hypothesis.

Since the p-value < = alpha we reject the null hypothesis.

Conclusion : Occupations and personality preferences are dependent.