In: Statistics and Probability
he following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions. E refers to extroverted and I refers to introverted.
Personality Type | |||
Occupation | E | I | Row Total |
Clergy (all denominations) | 67 | 40 | 107 |
M.D. | 66 | 96 | 162 |
Lawyer | 56 | 81 | 137 |
Column Total | 189 | 217 | 406 |
Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.05 level of significance.
(a) What is the level of significance?
______________________
State the null and alternate hypotheses.
a. H0: Myers-Briggs preference and
profession are independent
b. H1: Myers-Briggs preference and profession
are independent.H0: Myers-Briggs preference and
profession are not independent
c. H1: Myers-Briggs preference and profession
are not independent. H0:
Myers-Briggs preference and profession are not independent
d. H1: Myers-Briggs preference and profession
are independent.H0: Myers-Briggs preference and
profession are independent
e. H1: Myers-Briggs preference and profession
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?
a. Yes
b. No
What sampling distribution will you use?
a. binomial
b. Student's t
c. normal
d. chi-square
e. uniform
What are the degrees of freedom?
_________________
(c) Find or estimate the P-value of the sample test
statistic.
a. p-value > 0.100
b. 0.050 < p-value < 0.100
c. 0.025 < p-value < 0.050
d. 0.010 < p-value < 0.025
e. 0.005 < p-value < 0.010
f. 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?
a. Since the P-value > α, we fail to reject the null hypothesis.
b. Since the P-value > α, we reject the null hypothesis.
c. Since the P-value ≤ α, we reject the null hypothesis.
d. 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 insufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.
a) level of significance =0.05
H0: Myers-Briggs preference and profession
are independent
e. H1: Myers-Briggs preference and profession
are not independent.
b)
\applying chi square test:
Expected | Ei=row total*column total/grand total | E | I | Total |
Clergy | 49.810 | 57.190 | 107 | |
MD | 75.414 | 86.586 | 162 | |
Lawyer | 63.776 | 73.224 | 137 | |
total | 189 | 217 | 406 | |
chi square χ2 | =(Oi-Ei)2/Ei | E | I | Total |
Clergy | 5.9322 | 5.1667 | 11.099 | |
MD | 1.1751 | 1.0235 | 2.199 | |
Lawyer | 0.9481 | 0.8257 | 1.774 | |
total | 8.055 | 7.016 | 15.071 |
value of the chi-square statistic =15.071
Are all the expected frequencies greater than 5? :YEs
What sampling distribution will you use? Chi square
degrees of freedom=2
c) f. p-value < 0.005
d)
c. 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 preference and the profession are not independent.