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)	66	41	107
M.D.	70	92	162
Lawyer	60	77	137
Column Total	196	210	406

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

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: Myers-Briggs preference and profession are independent
H₁: Myers-Briggs preference and profession are independent.H₀: Myers-Briggs preference and profession are independent
H₁: Myers-Briggs preference and profession are not independent.    H₀: Myers-Briggs preference and profession are not independent
H₁: Myers-Briggs preference and profession are independent.H₀: Myers-Briggs preference and profession are not independent
H₁: Myers-Briggs preference and profession are not independent.

(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.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

normalStudent's t    chi-squareuniformbinomial

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic.

p-value > 0.1000.050 < p-value < 0.100    0.025 < p-value < 0.0500.010 < p-value < 0.0250.005 < p-value < 0.010p-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

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

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.

Answer 1

(a) What is the level of significance?

The level of significance is given as α = 0.05 or 5%.

State the null and alternate hypotheses.

H₀: Myers-Briggs preference and profession are independent
H₁: Myers-Briggs preference and profession are not independent.

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.)

Solution:

Here, we have to use chi square test for independence of two categorical variables.

H₀: Myers-Briggs preference and profession are independent
H₁: Myers-Briggs preference and profession are not independent.

We are given level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 3

Number of columns = c = 2

Degrees of freedom = df = (r – 1)*(c – 1) = 2*1 = 2

α = 0.05

Critical value = 5.991465

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	E	I	Total
Clergy	66	41	107
M.D.	70	92	162
Lawyer	60	77	137
Total	196	210	406

Expected Frequencies
	Column variable
Row variable	E	I	Total
Clergy	51.65517	55.34483	107
M.D.	78.2069	83.7931	162
Lawyer	66.13793	70.86207	137
Total	196	210	406

Are all the expected frequencies greater than 5?

Answer: Yes

Calculations
(O - E)
14.34483	-14.3448
-8.2069	8.206897
-6.13793	6.137931

(O - E)^2/E
3.98361	3.718036
0.861218	0.803803
0.569631	0.531655

What sampling distribution will you use?

Answer: Chi square

What are the degrees of freedom?

Answer: df = 2

Chi square = ∑[(O – E)^2/E] = 10.46795

Test statistic = 10.468

(c) Find or estimate the P-value of the sample test statistic.

P-value = 0.005332

(By using Chi square table or excel)

0.005 <p-value < 0.010

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

P-value < α = 0.05

So, we reject the null hypothesis

Answer: Since the P-value ≤ α, we reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs preference and profession are independent.