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Bot GPT-4o To test the hypothesis of independence between the shift worked and the reported symptoms, we need to state the null and alternative hypotheses.
**Null Hypothesis (H₀):** The shift (Morning, Evening, Night) is independent of the type of symptom reported (Influenza, Headache, Weakness, Shortness of Breath).
**Alternative Hypothesis (H₁):** The shift (Morning, Evening, Night) is not independent of the type of symptom reported (Influenza, Headache, Weakness, Shortness of Breath).
Next, we need to organize our data into a contingency table:
| Symptom | Morning Shift | Evening Shift | Night Shift | Row Total |
|------------------------|---------------|---------------|-------------|-----------|
| Influenza | 20 | 18 | 12 | 50 |
| Headache | 15 | 38 | 7 | 60 |
| Weakness | 17 | 20 | 9 | 46 |
| Shortness of Breath | 4 | 7 | 14 | 25 |
| Column Total | 56 | 83 | 42 | 181 |
In order to test the hypothesis of independence, we will perform a Chi-Square test for independence. This test compares the observed frequencies with the expected frequencies, which are calculated under the assumption that the null hypothesis is true.
The expected frequency for each cell in the table can be calculated using the formula:
\[ E_{ij} = \frac{(Row \, Total \, for \, Row \, i) \cdot (Column \, Total \, for \, Column \, j)}{Grand \, Total} \]
After calculating the expected frequencies, we use the following formula to calculate the Chi-Square test statistic:
\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \]
Where \(O_{ij}\) represents the observed frequency and \(E_{ij}\) represents the expected frequency in the \(i\)-th row and \(j\)-th column.
The degrees of freedom (df) for this test is given by:
\[ \text{df} = (number \, of \, rows - 1) \times (number \, of \, columns - 1) = (4 - 1) \times (3 - 1) = 3 \times 2 = 6 \]
Once we calculate the Chi-Square test statistic, we can compare it with the critical value from the Chi-Square distribution table at the \( \alpha = 0.01 \) level of significance with 6 degrees of freedom, or we can use the P-value method to make our decision.
To simplify calculations and verify the results, it is often useful to employ statistical software or tools designed for these types of calculations. If the computed \(\chi^2\) value obtained from the sample is greater than the critical value from the Chi-Square distribution table, or if the P-value is less than 0.01, we reject the null hypothesis.
Let's proceed with these calculations in detail. Calculate the expected frequencies first:
\[ E_{11} (\text{Influenza, Morning Shift}) = \frac{50 \cdot 56}{181} \approx 15.47 \]
\[ E_{12} (\text{Influenza, Evening Shift}) = \frac{50 \cdot 83}{181} \approx 22.93 \]
\[ E_{13} (\text{Influenza, Night Shift}) = \frac{50 \cdot 42}{181} \approx 11.60 \]
\[ E_{21} (\text{Headache, Morning Shift}) = \frac{60 \cdot 56}{181} \approx 18.57 \]
\[ E_{22} (\text{Headache, Evening Shift}) = \frac{60 \cdot 83}{181} \approx 27.52 \]
\[ E_{23} (\text{Headache, Night Shift}) = \frac{60 \cdot 42}{181} \approx 13.91 \]
\[ E_{31} (\text{Weakness, Morning Shift}) = \frac{46 \cdot 56}{181} \approx 14.24 \]
\[ E_{32} (\text{Weakness, Evening Shift}) = \frac{46 \cdot 83}{181} \approx 21.09 \]
\[ E_{33} (\text{Weakness, Night Shift}) = \frac{46 \cdot 42}{181} \approx 10.68 \]
\[ E_{41} (\text{Shortness of Breath, Morning Shift}) = \frac{25 \cdot 56}{181} \approx 7.74 \]
\[ E_{42} (\text{Shortness of Breath, Evening Shift}) = \frac{25 \cdot 83}{181} \approx 11.46 \]
\[ E_{43} (\text{Shortness of Breath, Night Shift}) = \frac{25 \cdot 42}{181} \approx 5.8 \]
Next, calculate the \(\chi^2\) statistic using:
\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \]
For example, the contribution to the Chi-Square statistic from the first cell (Influenza, Morning Shift):
\[ \frac{(20 - 15.47)^2}{15.47} \approx 1.30 \]
Apply this formula to each cell and sum these values to get the total Chi-Square statistic. Determine the p-value using the Chi-Square distribution with 6 degrees of freedom to make a conclusion about your hypothesis.
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