Question
In order to determine whether the proportion of defectives produced by workers was the same for the day, evening and night shift, the following data were recorded.
Skof / Shift
Dag / Day
Aand / Evening
Nag / Night
Defektiewe / Defectives
24
17
15
Nie - defektiewes / Nondefectives
405
280
150
Use a 0.025 level of significance to determine if the proportion of defectives is the same for all three shifts. The table value is:
0.051
7.38
14.44
1.24
Skof / Shift
Dag / Day
Aand / Evening
Nag / Night
Defektiewe / Defectives
24
17
15
Nie - defektiewes / Nondefectives
405
280
150
Use a 0.025 level of significance to determine if the proportion of defectives is the same for all three shifts. The table value is:
0.051
7.38
14.44
1.24
Answers
GPT-4o mini
To determine if the proportion of defectives produced by the workers is the same for the day, evening, and night shifts, we can use the Chi-Square test for homogeneity. Here's how we will do this step-by-step:
1. **Set Up the Hypotheses**:
- Null Hypothesis (H0): The proportion of defectives is the same for all three shifts.
- Alternative Hypothesis (H1): The proportion of defectives is not the same for all three shifts.
2. **Create a Contingency Table**:
\[
\begin{array}{|c|c|c|c|c|}
\hline
\text{Shift} & \text{Defectives} & \text{Non-defectives} & \text{Total} \\
\hline
\text{Day} & 24 & 405 & 429 \\
\text{Evening} & 17 & 280 & 297 \\
\text{Night} & 15 & 150 & 165 \\
\hline
\text{Total} & 56 & 835 & 891 \\
\hline
\end{array}
\]
3. **Calculate Expected Values**:
The expected count for each cell in a contingency table can be calculated using the formula:
\[
E_{ij} = \frac{(row \ total) \times (column \ total)}{grand \ total}
\]
Calculating for each cell:
- For Day Defectives:
\[
E_{Day Def} = \frac{429 \times 56}{891} \approx 32.17
\]
- For Day Non-defectives:
\[
E_{Day Non-def} = \frac{429 \times 835}{891} \approx 396.83
\]
- For Evening Defectives:
\[
E_{Even Def} = \frac{297 \times 56}{891} \approx 22.46
\]
- For Evening Non-defectives:
\[
E_{Even Non-def} = \frac{297 \times 835}{891} \approx 274.54
\]
- For Night Defectives:
\[
E_{Night Def} = \frac{165 \times 56}{891} \approx 10.18
\]
- For Night Non-defectives:
\[
E_{Night Non-def} = \frac{165 \times 835}{891} \approx 154.82
\]
4. **Calculate the Chi-Square Statistic**:
\[
\chi^2 = \sum \frac{(O - E)^2}{E}
\]
Where O is the observed frequency and E is the expected frequency.
Calculating for each cell:
- For Day Defectives:
\[
\frac{(24 - 32.17)^2}{32.17} \approx 1.947
\]
- For Day Non-defectives:
\[
\frac{(405 - 396.83)^2}{396.83} \approx 0.030
\]
- For Evening Defectives:
\[
\frac{(17 - 22.46)^2}{22.46} \approx 1.279
\]
- For Evening Non-defectives:
\[
\frac{(280 - 274.54)^2}{274.54} \approx 0.093
\]
- For Night Defectives:
\[
\frac{(15 - 10.18)^2}{10.18} \approx 2.296
\]
- For Night Non-defectives:
\[
\frac{(150 - 154.82)^2}{154.82} \approx 0.089
\]
Calculating the total:
\[
\chi^2 \approx 1.947 + 0.030 + 1.279 + 0.093 + 2.296 + 0.089 \approx 5.734
\]
5. **Degrees of Freedom**:
The degrees of freedom for the Chi-Square test for homogeneity is calculated as:
\[
df = (rows - 1)(columns - 1) = (3 - 1)(2 - 1) = 2
\]
6. **Determine the Critical Value**:
At a significance level of \( \alpha = 0.025 \) and \( df = 2 \), we refer to the Chi-Square distribution table. The critical value for \( df = 2 \) at \( \alpha = 0.025 \) is approximately 7.378 (close to 7.38 based on the options provided).
7. **Decision Rule**:
If the calculated \( \chi^2 \) statistic > critical value, we reject the null hypothesis.
Since \( 5.734 < 7.38 \), we do not reject the null hypothesis.
**Conclusion**:
There is not enough evidence to conclude that the proportion of defectives is different between the day, evening, and night shifts at the 0.025 significance level.
Thus, the answer you are looking for is:
**7.38** (the critical value).
1. **Set Up the Hypotheses**:
- Null Hypothesis (H0): The proportion of defectives is the same for all three shifts.
- Alternative Hypothesis (H1): The proportion of defectives is not the same for all three shifts.
2. **Create a Contingency Table**:
\[
\begin{array}{|c|c|c|c|c|}
\hline
\text{Shift} & \text{Defectives} & \text{Non-defectives} & \text{Total} \\
\hline
\text{Day} & 24 & 405 & 429 \\
\text{Evening} & 17 & 280 & 297 \\
\text{Night} & 15 & 150 & 165 \\
\hline
\text{Total} & 56 & 835 & 891 \\
\hline
\end{array}
\]
3. **Calculate Expected Values**:
The expected count for each cell in a contingency table can be calculated using the formula:
\[
E_{ij} = \frac{(row \ total) \times (column \ total)}{grand \ total}
\]
Calculating for each cell:
- For Day Defectives:
\[
E_{Day Def} = \frac{429 \times 56}{891} \approx 32.17
\]
- For Day Non-defectives:
\[
E_{Day Non-def} = \frac{429 \times 835}{891} \approx 396.83
\]
- For Evening Defectives:
\[
E_{Even Def} = \frac{297 \times 56}{891} \approx 22.46
\]
- For Evening Non-defectives:
\[
E_{Even Non-def} = \frac{297 \times 835}{891} \approx 274.54
\]
- For Night Defectives:
\[
E_{Night Def} = \frac{165 \times 56}{891} \approx 10.18
\]
- For Night Non-defectives:
\[
E_{Night Non-def} = \frac{165 \times 835}{891} \approx 154.82
\]
4. **Calculate the Chi-Square Statistic**:
\[
\chi^2 = \sum \frac{(O - E)^2}{E}
\]
Where O is the observed frequency and E is the expected frequency.
Calculating for each cell:
- For Day Defectives:
\[
\frac{(24 - 32.17)^2}{32.17} \approx 1.947
\]
- For Day Non-defectives:
\[
\frac{(405 - 396.83)^2}{396.83} \approx 0.030
\]
- For Evening Defectives:
\[
\frac{(17 - 22.46)^2}{22.46} \approx 1.279
\]
- For Evening Non-defectives:
\[
\frac{(280 - 274.54)^2}{274.54} \approx 0.093
\]
- For Night Defectives:
\[
\frac{(15 - 10.18)^2}{10.18} \approx 2.296
\]
- For Night Non-defectives:
\[
\frac{(150 - 154.82)^2}{154.82} \approx 0.089
\]
Calculating the total:
\[
\chi^2 \approx 1.947 + 0.030 + 1.279 + 0.093 + 2.296 + 0.089 \approx 5.734
\]
5. **Degrees of Freedom**:
The degrees of freedom for the Chi-Square test for homogeneity is calculated as:
\[
df = (rows - 1)(columns - 1) = (3 - 1)(2 - 1) = 2
\]
6. **Determine the Critical Value**:
At a significance level of \( \alpha = 0.025 \) and \( df = 2 \), we refer to the Chi-Square distribution table. The critical value for \( df = 2 \) at \( \alpha = 0.025 \) is approximately 7.378 (close to 7.38 based on the options provided).
7. **Decision Rule**:
If the calculated \( \chi^2 \) statistic > critical value, we reject the null hypothesis.
Since \( 5.734 < 7.38 \), we do not reject the null hypothesis.
**Conclusion**:
There is not enough evidence to conclude that the proportion of defectives is different between the day, evening, and night shifts at the 0.025 significance level.
Thus, the answer you are looking for is:
**7.38** (the critical value).