The operations manager of a company that manufactures tires wants to determine whether there are any differences in the quality of workmanship among the three daily shifts. She randomly selects 499 tires and carefully inspects them. Each tire is either classified as perfect, satisfactory, or defective, and the shift that produced it is also recorded. The two categorical variables of interest are: shift and condition of the tire produced. The data can be summarized by the accompanying two-way table.   Perfect Satisfactory Defective Total Shift 1 106 124 2 232 Shift 2 66 85 3 154 Shift 3 36 74 3 113 Total 208 283 8 499 A) This sample has degree of freedom: ��=  If shift and quality are independent, then B) With any expected count accurate to 2 decimal places, the expected count in row 1 column 1 is �1,1=  C) With any component accurate to 2 decimal places, the component value in row 1 column 1 of the table is component1,1=

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The operations manager of a company that manufactures tires wants to determine whether there are any differences in the quality of workmanship among the three daily shifts. She randomly selects 499 tires and carefully inspects them. Each tire is either classified as perfectsatisfactory, or defective, and the shift that produced it is also recorded. The two categorical variables of interest are: shift and condition of the tire produced. The data can be summarized by the accompanying two-way table.

  Perfect Satisfactory Defective Total
Shift 1 106 124 2 232
Shift 2 66 85 3 154
Shift 3 36 74 3 113
Total 208 283 8 499


A) This sample has degree of freedom:

��= 

If shift and quality are independent, then

B) With any expected count accurate to 2 decimal places, the expected count in row 1 column 1 is

�1,1= 

C) With any component accurate to 2 decimal places, the component value in row 1 column 1 of the table is

component1,1= 

D) �2 value of this sample is

�2=  . Round to 2 decimal places.

E) At significance level 0.02, is the sample showing strong evidence that shift and quality are dependent? Use 4 decimal places for p-value.

  • Yes
  • No


F) The p-value for this sample was:

p-value = 
### Quality of Workmanship Among Daily Shifts of Tire Manufacturing

#### Overview
The operations manager of a tire manufacturing company is investigating whether there are differences in the quality of workmanship among three daily shifts. A total of 499 randomly selected tires were inspected, and each tire was classified as either perfect, satisfactory, or defective. The shift that produced each tire was also recorded. The categorical variables of interest are the shift and the condition of the tire. The summarized data is presented in the following two-way table:

#### Two-Way Table
|           | Perfect | Satisfactory | Defective | Total |
|-----------|---------|--------------|-----------|-------|
| **Shift 1** | 106     | 124          | 2         | 232   |
| **Shift 2** | 66      | 85           | 3         | 154   |
| **Shift 3** | 36      | 74           | 3         | 113   |
| **Total**   | 208     | 283          | 8         | 499   |

#### Analysis

**A) Degree of Freedom Calculation**

The degrees of freedom for this sample can be calculated using the formula for the chi-square test of independence:

\[ DF = (r - 1) \times (c - 1) \]

where \( r \) is the number of rows and \( c \) is the number of columns in the two-way table.

\[ DF = \_\_\_\_ \]

**B) Calculation of Expected Count**

If shift and quality are independent, the expected count in row 1, column 1 (\( E_{1,1} \)) is calculated using the formula:

\[ E_{1,1} = \frac{(\text{Row Total} \times \text{Column Total})}{\text{Grand Total}} \]

With any expected count accurate to 2 decimal places,

\[ E_{1,1} = \_\_\_\_ \]

**C) Calculation of Component Value**

The component value in row 1, column 1 of the table can be computed as follows:

\[ \text{component}_{1,1} = \frac{(O_{1,1} - E_{1,1})^2}{E_{1,1}} \]

where \( O_{1,1} \) is the observed count and \( E_{1,1
Transcribed Image Text:### Quality of Workmanship Among Daily Shifts of Tire Manufacturing #### Overview The operations manager of a tire manufacturing company is investigating whether there are differences in the quality of workmanship among three daily shifts. A total of 499 randomly selected tires were inspected, and each tire was classified as either perfect, satisfactory, or defective. The shift that produced each tire was also recorded. The categorical variables of interest are the shift and the condition of the tire. The summarized data is presented in the following two-way table: #### Two-Way Table | | Perfect | Satisfactory | Defective | Total | |-----------|---------|--------------|-----------|-------| | **Shift 1** | 106 | 124 | 2 | 232 | | **Shift 2** | 66 | 85 | 3 | 154 | | **Shift 3** | 36 | 74 | 3 | 113 | | **Total** | 208 | 283 | 8 | 499 | #### Analysis **A) Degree of Freedom Calculation** The degrees of freedom for this sample can be calculated using the formula for the chi-square test of independence: \[ DF = (r - 1) \times (c - 1) \] where \( r \) is the number of rows and \( c \) is the number of columns in the two-way table. \[ DF = \_\_\_\_ \] **B) Calculation of Expected Count** If shift and quality are independent, the expected count in row 1, column 1 (\( E_{1,1} \)) is calculated using the formula: \[ E_{1,1} = \frac{(\text{Row Total} \times \text{Column Total})}{\text{Grand Total}} \] With any expected count accurate to 2 decimal places, \[ E_{1,1} = \_\_\_\_ \] **C) Calculation of Component Value** The component value in row 1, column 1 of the table can be computed as follows: \[ \text{component}_{1,1} = \frac{(O_{1,1} - E_{1,1})^2}{E_{1,1}} \] where \( O_{1,1} \) is the observed count and \( E_{1,1
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