Question

The 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)	65	42	107
M.D.	63	99	162
Lawyer	58	79	137
Column Total	186	220	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?

B. State the null and alternate hypotheses. (from the following)

H₀: Myers-Briggs preference and profession are independent
H₁: Myers-Briggs preference and profession are independent.

H₀: Myers-Briggs preference and profession are not independent
H₁: Myers-Briggs preference and profession are not independent.    

H₀: Myers-Briggs preference and profession are independent
H₁: Myers-Briggs preference and profession are not independent.

H₀: Myers-Briggs preference and profession are not independent
H₁: Myers-Briggs preference and profession are independent.

C. 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.)

D. Are all the expected frequencies greater than 5?

E. What sampling distribution will you use?

F. What are the degrees of freedom?

G. Find or estimate the P-value of the sample test statistic.

H. Based on your answers in parts (a) to (g), will you reject or fail to reject the null hypothesis of independence?

Answer 1

Tabulated Statistics: C1, Worksheet columns

Rows: C1   Columns: Worksheet columns

	E	I	All
All	65	42	107
	49.02	57.98
Md	63	99	162
	74.22	87.78
L	58	79	137
	62.76	74.24
All	186	220	406

Cell Contents
      Count
      Expected count

Chi-Square Test

	Chi-Square	DF	P-Value
Pearson	13.410	2	0.001

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A)

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B) H0: Myers-Briggs preference and profession are independent
     H1: Myers-Briggs preference and profession are not independent.

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Where E is expected value of each cell

E = (column sum* Row sum)/ Total

Example - E (for first cell) = 107*186/406 =49.02

similarly for others

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D)

	E	I	All
All	65	42	107
	49.02	57.98
Md	63	99	162
	74.22	87.78
L	58	79	137
	62.76	74.24
All	186	220	406

Yes

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F) df= (no. of rows-1)*(no. of column -1) = (3-1)(2-1) = 2

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G) P-value = 0.001

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H) Reject Ho

As P-value is less than level of significance