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	61	8	27	96
EN	83	38	33	154
IS	52	31	32	115
ES	79	44	31	154
Column Total	275	121	123	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 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?

YesNo    

What sampling distribution will you use?

Student's tbinomial    chi-squarenormaluniform

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

Ans:

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)

	Observed(fo)
Myers-Briggs	Arts & Science	Business	Allied Health	Row Total
Preference
IN	61	8	27	96
EN	83	38	33	154
IS	52	31	32	115
ES	79	44	31	154
Column Total	275	121	123	519
	Expected(fe)
Myers-Briggs	Arts & Science	Business	Allied Health	Row Total
Preference
IN	50.867	22.382	22.751	96
EN	81.599	35.904	36.497	154
IS	60.934	26.811	27.254	115
ES	81.599	35.904	36.497	154
Column Total	275	121	123	519
	Chi square=(fo-fe)^2/fe
Myers-Briggs	Arts & Science	Business	Allied Health	Row Total
Preference
IN	2.019	9.241	0.793	12.053
EN	0.024	0.122	0.335	0.482
IS	1.310	0.654	0.826	2.791
ES	0.083	1.826	0.828	2.736
Column Total	3.435	11.844	2.783	18.062

Test statistic:

Chi square=18.062

yes

Chi square

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

c)p-value=CHIDIST(18.062,6)=0.0061

0.005 <p-value < 0.01

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.