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	64	17	15	96
EN	84	44	26	154
IS	53	30	32	115
ES	70	44	40	154
Column Total	271	135	113	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?


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.

(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?

Student's tbinomial    uniformchi-squarenormal

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 > 0.1000.050 < 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 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

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

b)

applying chi square test:

Expected	Ei=row total*column total/grand total	Arts	Business	Allied	Total
	IN	50.127	24.971	20.902	96
	EN	80.412	40.058	33.530	154
	IS	60.048	29.913	25.039	115
	ES	80.412	40.058	33.530	154
	total	271	135	113	519
chi square    χ2	=(Oi-Ei)2/Ei	Arts	Business	Allied	Total
	IN	3.8393	2.5445	1.6664	8.0502
	EN	0.1601	0.3880	1.6910	2.2390
	IS	0.8273	0.0003	1.9355	2.7630
	ES	1.3483	0.3880	1.2485	2.9847
	total	6.1750	3.3207	6.5414	16.037

value of the chi-square statistic =16.037

Are all the expected frequencies greater than 5? :Yes

What sampling distribution will you use? chi-square

degrees of freedom =(row-1)*(column-1)=6

c)

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 independen