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 | 87 | 43 | 24 | 154 |
IS | 60 | 30 | 25 | 115 |
ES | 70 | 39 | 45 | 154 |
Column Total | 281 | 129 | 109 | 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
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?
YesNo
What sampling distribution will you use?
normaluniform chi-squarebinomialStudent'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.)
(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 > ?, we fail to reject the null hypothesis.Since the P-value > ?, we reject the null hypothesis. Since the P-value ? ?, we reject the null hypothesis.Since the P-value ? ?, 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.
The chi-square statistic is 16.2636. The p-value is
.012407. The result is significant at p < .05.
a)
H0: Myers-Briggs type and area of study are independent.
H1: Myers-Briggs type and area of study are not independent.
b)
The chi-square statistic is 16.2636.
Are all the expected frequencies greater than 5?
Yes
What sampling distribution will you use?
chi-square
What are the degrees of freedom?
= (3-1)(4-1) = 6
c)
p-value = 0.012407
d)
Since the P-value ? ?, 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.
Get Answers For Free
Most questions answered within 1 hours.