Mathematics, 11.06.2020 16:02 temily0303

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

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Answer from: adrianVocke5685

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!

Employees at a company produced refrigerators on three shifts. Each shift recorded their quality sta

Answer from: Quest

The answer should be 3%

Answer from: Quest

see below

step-by-step explanation:

7. there are 10 data points in the set. the middle point is between 5 and 6. the middle point in the top half is 8 and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term point 8 = 100 point 3 = 88

iqr = 100-88 = 12

final point 8 = 93 point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6. b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical . there is an outlier for the males that skews the data

5. the iqr is a better measure of spread for the movies and basketball games. since the basketball game has an outlier, we do not want to use standard deviation for measurement.

3. the exam median is much higher than the class median. the median line is farther to the right on the exams

2. there is no way to tell what the means are from the box and whisker plots.

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