Question

The following table shows the Myers-Briggs personality preference and area of study for a random sample of 519 college students. In the table, IN refers to introvert, intuitive; EN refers to extrovert, intuitive; IS refers to introvert, sensing; and ES refers to extrovert, sensing.

Myers-Briggs Preference	Arts & Science	Business	Allied Health	Row Total
IN	68	17	11	96
EN	79	45	30	154
IS	54	37	24	115
ES	76	40	38	154
Column Total	277	139	103	519

Use a chi-square test to determine if Myers-Briggs preference type is independent of area of study at the 0.05 level of significance.

(A) What is the level of significance?

(B) State the null and alternate hypotheses.

H₀: Myers-Briggs type and area of study are not independent.
H₁: Myers-Briggs type and area of study are not independent.

H₀: Myers-Briggs type and area of study are not independent.
H₁: Myers-Briggs type and area of study are independent.  

H₀: Myers-Briggs type and area of study are independent.
H₁: Myers-Briggs type and area of study are not independent.

H₀: Myers-Briggs type and area of study are independent.
H₁: Myers-Briggs type and area of study are independent.

(C) 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.)

(D) Are all the expected frequencies greater than 5? (yes/no)

(E) What sampling distribution will you use? (uniform / Student's t / chi-square / binomial / normal)

(F) What are the degrees of freedom?

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

(H) Based on your answers in parts (a) to (g), 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.

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

At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs type and area of study are not independent.

At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs type and area of study are not independent.    

Answer 1

(A) level of significance: 0.05

(B) Null and alternate hypotheses:

H0: Myers-Briggs type and area of study are independent.

H1: Myers-Briggs type and area of study are not independent.

(C)

Observed Frequencies
	Arts and Science	Business	Allied Health	Total
IN	68	17	11	96
EN	79	45	30	154
IS	54	37	24	115
ES	76	40	38	154
Total	277	139	103	519
Expected Frequencies
	Arts and Science	Business	Allied Health	Total
IN	277 * 96 / 519 = 51.237	139 * 96 / 519 = 25.711	103 * 96 / 519 = 19.052	96
EN	277 * 154 / 519 = 82.1927	139 * 154 / 519 = 41.2447	103 * 154 / 519 = 30.5626	154
IS	277 * 115 / 519 = 61.3776	139 * 115 / 519 = 30.7996	103 * 115 / 519 = 22.8227	115
ES	277 * 154 / 519 = 82.1927	139 * 154 / 519 = 41.2447	103 * 154 / 519 = 30.5626	154
Total	277	139	103	519
	(fo-fe)²/fe
IN	(68 - 51.237)²/51.237 = 5.4843	(17 - 25.711)²/25.711 = 2.9513	(11 - 19.052)²/19.052 = 3.4031
EN	(79 - 82.1927)²/82.1927 = 0.124	(45 - 41.2447)²/41.2447 = 0.3419	(30 - 30.5626)²/30.5626 = 0.0104
IS	(54 - 61.3776)²/61.3776 = 0.8868	(37 - 30.7996)²/30.7996 = 1.2482	(24 - 22.8227)²/22.8227 = 0.0607
ES	(76 - 82.1927)²/82.1927 = 0.4666	(40 - 41.2447)²/41.2447 = 0.0376	(38 - 30.5626)²/30.5626 = 1.8099

Test statistic:  

χ² = ∑ ((fo-fe)²/fe) = 16.825

(D) Yes, all the expected frequencies are greater than 5.

(E) Sampling distribution: chi-square

(F) Degrees of freedom, df = (r-1)(c-1) = 6

(G) p-value = CHISQ.DIST.RT(16.8247, 6) = 0.0099

(H) Since the P-value ≤ α, we reject the null hypothesis.

(I) Interpret your conclusion:

At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs type and area of study are not independent.