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	11	22	96
EN	82	47	25	154
IS	60	40	15	115
ES	74	38	42	154
Column Total	279	136	104	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 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 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 tnormal    chi-squarebinomialuniform

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) What is the level of significance?
ans: 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.

(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.)

Assuming that Myers-Briggs type and area of study are independent, we can find the expected frequency of any cell using

where RT= is the row total for the row containing the cell

CT= is the column total for the column containing the cell

N=519 is the total number of observations

We calculate the expected values as given below

Myers-Briggs	Arts & Science	Business	Allied Health	Row Total
Preference
IN	(96*279)/519 = 51.6069	(96*136)/519 = 25.1561	(96*104)/519 = 19.237	96
EN	(154*279)/519 = 82.7861	(154*136)/519 = 40.3545	(154*104)/519 = 30.8593	154
IS	(115*279)/519 = 61.8208	(115*136)/519 = 30.1349	(115*104)/519 = 23.0443	115
ES	(154*279)/519 = 82.7861	(154*136)/519 = 40.3545	(154*104)/519 = 30.8593	154
Column Total	279	136	104	519

Are all the expected frequencies greater than 5?

ans: Yes

The chi-square statistics is calculated using

ans: the value of the chi-square statistic for the sample is 24.275

What sampling distribution will you use?

ans: chi-square

What are the degrees of freedom?
The degrees of freedom is (number of rows -1) *( number of columns -1) = (4-1)*(3-1) = 6

ans: the degrees of freedom are 6

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

The test statistics is 24.275. The p-value is

Using the chi-square table, against the row corresponding to df=6, we can see that for 0.005 area in right tail, we get a critical value =18.5476. That is . that means must be less than 0.005

ans: 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?

We will reject the null hypothesis if the p-value is less than the significance level.

Here, the p-value is less than 0.005 and hence is less than .

ans: Since the P-value ≤ α, we reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

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