Respuesta :
The missing figures in the question is shown in bold format.
Also the table is better constructed for clearer understanding when answering the question.
A Deloitte employment survey asked a sample of human resource executives how their company planned to change its workforce over the next 12 months. A categorical response variable showed three options: The company plans to hire and add to the number of employees, the company plans no change in the number of employees, or the company plans to lay off and reduce the number of employees. Another categorical variable indicated if the company was private or public. Sample data for 180 companies are summarized as follows.
Company
Employment Plan Private Public
Add Employees 39 32
No Change 21 36
Lay Off Employees 12 44
a. Conduct a test of independence to determine if the employment plan for the next 12 months is independent of the type of company. At a level of 0.05 significance. Compute the value of the test statistic (to 2 decimals).
Answer:
Explanation:
From the table in the question; we can see the changes in employees adding, shedding, or not changing their staffing.
Company
Plan Private Public
Add 39 32
Number Change 21 36
Lay Off 12 44
The hypothesis are:
[tex]\mathbf{ H_o : Column \ independent \ of \ row}\\ \\ \mathbf{ H_a : Column \ is \ dependent \ of \ row}[/tex]
Using the following relation of variables given to determine expected frequencies ; we have :
[tex]\mathbf{e_f = \dfrac{(row _i)(column_j)}{Total \ sample}}[/tex]
From the above table ; the first row show the total entries of 72
The first column shows the total of 72
[tex]\mathbf{e_f = \dfrac{(39+32)(72)}{180}} \\ \\ \mathbf{e_f = 28.80}}[/tex]
The expected value for the first row, first column is 28.80
Repeating the same process for others;
For the first row ; second column we have :
[tex]\mathbf{e_f = \dfrac{(39+32)(112)}{180}} \\ \\ \mathbf{e_f = 44.80}}[/tex]
For the second row ; first column we have :
[tex]\mathbf{e_f = \dfrac{(21+36)(72)}{180}} \\ \\ \mathbf{e_f = 22.80}}[/tex]
For the second row ; second column we have :
[tex]\mathbf{e_f = \dfrac{(21+36)(112)}{180}} \\ \\ \mathbf{e_f = 35.47}}[/tex]
For the third row ; first column we have :
[tex]\mathbf{e_f = \dfrac{(12+44)(72)}{180}} \\ \\ \mathbf{e_f = 22.40}}[/tex]
For the third row ; second column we have :
[tex]\mathbf{e_f = \dfrac{(12+44)(112)}{180}} \\ \\ \mathbf{e_f = 34.84}}[/tex]
Company
Plan Private Public Total
Add 28.80 44.80 73.60
Number Change 22.80 35.47 58.27
Lay Off 22.40 34.84 57.24
Converting the table to chi- squared using the relation.
[tex]\mathbf{x^2 = \sum_i ( \dfrac{f_y-e_f}{e_f})^2}[/tex]
where;
[tex]f_y[/tex] = observed frequency from the original table
From the original above table ;
for the first row (1)
the observed frequency is = 39
the expected frequency is = 28.80
[tex]\mathbf{x^2 = \sum_i ( \dfrac{39-28.80}{28.80})^2} \\ \\ \mathbf{x^2 =3.6125}[/tex]
for the first row (2)
the observed frequency is = 32
the expected frequency is = 44.80
[tex]\mathbf{x^2 = \sum_i ( \dfrac{32-44.80}{44.80})^2} \\ \\ \mathbf{x^2 =3.6571}[/tex]
for the second row (1)
the observed frequency is = 21
the expected frequency is = 22.80
[tex]\mathbf{x^2 = \sum_i ( \dfrac{21-22.80}{22.80})^2} \\ \\ \mathbf{x^2 =0.1421}[/tex]
for the second row (2)
the observed frequency is = 36
the expected frequency is = 35.47
[tex]\mathbf{x^2 = \sum_i ( \dfrac{36-35.47}{35.47})^2} \\ \\ \mathbf{x^2 =0.0079}[/tex]
for the third row (1)
the observed frequency is = 12
the expected frequency is = 22.40
[tex]\mathbf{x^2 = \sum_i ( \dfrac{12-22.40}{22.40})^2} \\ \\ \mathbf{x^2 =4.8286}[/tex]
for the third row (2)
the observed frequency is = 44
the expected frequency is = 34.84
[tex]\mathbf{x^2 = \sum_i ( \dfrac{44-34.84}{34.84})^2} \\ \\ \mathbf{x^2 =2.4083}[/tex]
Company
Plan Private Public Total
Add 3.6125 3.6571 7.2696
Number Change 0.1421 0.0079 0.15
Lay Off 4.8286 2.4083 7.2369
Total [tex]x^2 =[/tex] 14.657
Hence, the total chi-square = 14.657;
To find the value for p; we need to determine the degree of freedom
df = (2-1)(3-1)
that result to a degree of freedom of 2
From the chi square chart at the chi-square is 14.657 and degree of freedom is 2 ; the p value is between 0.1 and 0.005. Since this makes p-value less than 0.05.
We rejected [tex]\mathbf{ H_o}[/tex]
Thus; the variables are dependent. We can conclude that the employment plan and the company are significantly related.
