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 63 10 23 96 EN 84 42 28 154 IS 54 31 30 115 ES 77 42 35 154 Column Total 278 125 116 519

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 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.

H0: Myers-Briggs type and area of study are 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 uniform binomial chi-square normal 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.)

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)

Chi-square Statistic:

O: Observed Frequency

	Arts and Science	Business	Allied Health	Total
IN	63	10	23	96
EN	84	42	28	154
IS	54	31	30	115
ES	77	42	35	154
Total	278	125	116	519

E : Expected Frequency

E:

	Arts and Science	Business	Allied Health	Total
IN	(96*278)/519	(96*125)/519	(96*116)/519	96
EN	(154*278)/519	(154*125)/519	(154*116)/519	154
IS	(115*278)/519	(115*125)/519	(115*116)/519	115
ES	(154*278)/519	(154*125)/519	(154*116)/519	154
Total	278	125	116	519

	Arts and Science	Business	Allied Health	Total
IN	51.4220	23.1214	21.4566	96
EN	82.4894	37.0906	34.4200	154
IS	61.5992	27.6975	25.7033	115
ES	82.4894	37.0906	34.4200	154
Total	278	125	116	519

O	E	(O-E)	(O-E)2	(O-E)2/E
63	51.4220	11.5780	134.0509	2.6069
84	82.4894	1.5106	2.2819	0.0277
54	61.5992	-7.5992	57.7483	0.9375
77	82.4894	-5.4894	30.1335	0.3653
10	23.1214	-13.1214	172.1708	7.4464
42	37.0906	4.9094	24.1026	0.6498
31	27.6975	3.3025	10.9065	0.3938
42	37.0906	4.9094	24.1026	0.6498
23	21.4566	1.5434	2.3819	0.1110
28	34.4200	-6.4200	41.2169	1.1975
30	25.7033	4.2967	18.4618	0.7183
35	34.4200	0.5800	0.3364	0.0098
			Total	15.1137

value of the chi-square statistic for the sample = 15.1137

Are all the expected frequencies greater than 5?

Yes

What sampling distribution will you use?

chi-square

Degrees of freedom =(Number of rows - 1) * (Number of columns - 1) = (4-1) * (3-1) = 3*2 = 6

Degrees of freedom = 6

(c)

P-value of the sample test statistic

p-value = 0.0194;

P-value is computed using the excel function ; CHISQ.DIST.RT(15.1137,6)

CHISQ.DIST.RT function

Returns the right-tailed probability of the chi-squared distribution.

The χ2 distribution is associated with a χ2 test. Use the χ2 test to compare observed and expected values. For example, a genetic experiment might hypothesize that the next generation of plants will exhibit a certain set of colors. By comparing the observed results with the expected ones, you can decide whether your original hypothesis is valid.

Syntax

CHISQ.DIST.RT(x,deg_freedom)

The CHISQ.DIST.RT function syntax has the following arguments:

• X     Required. The value at which you want to evaluate the distribution.

• Deg_freedom     Required. The number of degrees of freedom.