Question

Employees at a company produced refrigerators on three shifts. Each shift recorded their quality stats below. A unit was considered defective if it at least one part was assembled wrong or was missing. Management believes that quality depends on the the shift it was produced. Test the claim that shifts are independent of quality using chi-square at alpha = 0.05. SHOW YOUR WORK

Answer 1

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

$X^2= sum\frac{(O_{ij}-E_{ij})^2}{E_{ij}} ~X_{(r-1)(c-1)}$

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:

$E_{ij}= \frac{O_{i.}*O_{.j}}{n}$

$O_{i.}$ Represents the marginal value of the i-row

$O_{.j}$ Represents the marginal value of the j-column

$E_{11}= \frac{O_{1.}*O_{.1}}{n}= \frac{21*40}{120}= 7$

$E_{12}= \frac{O_{1.}*O_{.2}}{n}= \frac{21*40}{120}= 7$

$E_{13}= \frac{O_{1.}*O_{.3}}{n}= \frac{21*40}{120}= 7$

$E_{21}= \frac{O_{2.}*O_{.1}}{n}= \frac{99*40}{120}= 33$

$E_{22}= \frac{O_{2.}*O_{.2}}{n}= \frac{99*40}{120}= 33$

$E_{23}= \frac{O_{2.}*O_{.3}}{n}= \frac{99*40}{120}= 33$

$X^2_{H_0}= \frac{(7-7)^2}{7} + \frac{(5-7)^2}{7} + \frac{(9-7)^2}{7} + \frac{(33-33)^2}{33} + \frac{(35-33)^2}{33} + \frac{(31-33)^2}{33} = 1.385= 1.34$

Using the critical value approach, the rejection region for this test is one-tailed to the right, the critical value is:

$X^2_{(c-1)(r-1);1-\alpha }= X^2_{2; 0.95}= 5.991$

Decision rule:

If $X^2_{H_0}$ ≥ 5.991, reject the null hypothesis.

If $X^2_{H_0}$ < 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!