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 8 24 96 EN 77 38 39 154 IS 60 37 18 115 ES 76 45 33 154 Column Total 277 128 114 519 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 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 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? uniform binomial chi-square normal Student's t 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

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 of independence:

Expected	Ei=row total*column total/grand total	arts	Business	Allied	Total
	IN	51.237	23.676	21.087	96
	EN	82.193	37.981	33.827	154
	IS	61.378	28.362	25.260	115
	ES	82.193	37.981	33.827	154
	total	277	128	114	519
chi square    χ2	=(Oi-Ei)2/Ei	arts	Business	Allied	Total
	IN	3.179	10.379	0.402	13.9612
	EN	0.328	0.000	0.791	1.1193
	IS	0.031	2.631	2.087	4.7482
	ES	0.467	1.297	0.020	1.7840
	total	4.0048	14.3073	3.3006	21.613
test statistic X2 =		21.613

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)

p-value < 0.005

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.