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	59	10	27	96
EN	85	41	28	154
IS	61	34	20	115
ES	75	43	36	154
Column Total	280	128	111	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?
0.05

State the 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.

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 independent.

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

Yes

No     

What sampling distribution will you use?

Student's t

normal     

chi-square

uniform

binomial

What are the degrees of freedom?
_____

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

p-value > 0.100

0.050 < p-value < 0.100     

0.025 < p-value < 0.050

0.010 < p-value < 0.025

0.005 < p-value < 0.010

p-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 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

	c1	c2	c3	c4	sum
r1	59	85	61	75	280
r2	10	41	34	43	128
r3	27	28	20	36	111
sum	96	154	115	154	519
expected
Eij	1	2	3	4
1	51.7919	83.0829	62.0424	83.0829
2	23.6763	37.9807	28.3622	37.9807
3	20.5318	32.9364	24.5954	32.9364
Oi	59	85	61	75	10	41	34	43	27	28.0000	20.0000	36.0000
Ei	51.7919	83.0829	62.0424	83.0829	23.6763	37.9807	28.3622	37.9807	20.5318	32.9364	24.5954	32.9364
(Oi-Ei)^2/Ei	1.0032	0.0442	0.0175	0.7864	7.8999	0.2400	1.1207	0.6633	2.0377	0.7399	0.8586	0.2850
sum	15.6963
alpha	0.05
critical value	12.5916
p-value	0.0155

Formulas

a)
option a)
Ho : independent
Ha: not independent

b)
TS = 15.696

yes, all frequencies > 5

chi-square

df = (3-1)(4-1) = 6

c)
p-value = 0.0155
hence
0.010 < p-value < 0.025

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

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

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