Answer:
Step-by-step explanation:
Hello!
So in the refrigerator factory there are three shifts. Each shift records their quality based on the quantity of defective and working parts assembled.
Using a Chi-Square test of independence you have to test the claim that quality and shifts are independent.
The hypotheses are:
H₀: The variables are independent.
H₁: The variables are not independent.
α: 0.05
[tex]X^2= sum\frac{(O_{ij}-E_{ij})^2}{E_{ij}} ~X_{(r-1)(c-1)}[/tex]
r= total number of rows
c= total number of columns
i= 1, 2 (categories in rows)
j=1, 2, 3 (categories in columns)
To calculate the statistic you have to calculate the expected frequencies for each category:
[tex]E_{ij}= \frac{O_{i.}*O_{.j}}{n}[/tex]
[tex]O_{i.}[/tex] Represents the marginal value of the i-row
[tex]O_{.j}[/tex] Represents the marginal value of the j-column
[tex]E_{11}= \frac{O_{1.}*O_{.1}}{n}= \frac{21*40}{120}= 7[/tex]
[tex]E_{12}= \frac{O_{1.}*O_{.2}}{n}= \frac{21*40}{120}= 7[/tex]
[tex]E_{13}= \frac{O_{1.}*O_{.3}}{n}= \frac{21*40}{120}= 7[/tex]
[tex]E_{21}= \frac{O_{2.}*O_{.1}}{n}= \frac{99*40}{120}= 33[/tex]
[tex]E_{22}= \frac{O_{2.}*O_{.2}}{n}= \frac{99*40}{120}= 33[/tex]
[tex]E_{23}= \frac{O_{2.}*O_{.3}}{n}= \frac{99*40}{120}= 33[/tex]
[tex]X^2_{H_0}= \frac{(7-7)^2}{7} + \frac{(5-7)^2}{7} + \frac{(9-7)^2}{7} + \frac{(33-33)^2}{33} + \frac{(35-33)^2}{33} + \frac{(31-33)^2}{33} = 1.385= 1.34[/tex]
Using the critical value approach, the rejection region for this test is one-tailed to the right, the critical value is:
[tex]X^2_{(c-1)(r-1);1-\alpha }= X^2_{2; 0.95}= 5.991[/tex]
Decision rule:
If [tex]X^2_{H_0}[/tex] ≥ 5.991, reject the null hypothesis.
If [tex]X^2_{H_0}[/tex] < 5.991, do not reject the null hypothesis.
The value of the statistic is less than the critical value, the decision is to not reject the null hypothesis.
At 5% significance level, you can conclude that the shift the pieces were assembled and the quality of said pieces are independent.
I hope this helps!
