To investigate whether employment status is distributed according to the provided percentages, we can perform a chi-square goodness-of-fit test. The observed values are from the contingency table in the image, and we will calculate the expected values using the given percentages. Here is the procedure and solution:

Step 1: Observed Values (from the contingency table)

The total sum of individuals is 104, and the observed values for employment status are:

• Casual: 23
• Full-time: 20
• Part-time: 31
• Unemployed: 30

Step 2: Expected Values (based on provided percentages)

We use the percentages given for each employment status to calculate the expected counts:

• Casual (30% of 104) = $0.3 \times 104 = 31.2$
• Full-time (50% of 104) = $0.5 \times 104 = 52.0$
• Part-time (10% of 104) = $0.1 \times 104 = 10.4$
• Unemployed (10% of 104) = $0.1 \times 104 = 10.4$

Step 3: Chi-square Test Statistic Calculation

The chi-square test statistic is calculated using the formula: $\chi^2 = \sum \frac{(O - E)^2}{E}$ where $O$ is the observed value, and $E$ is the expected value.

Let's calculate each component:

1. Casual: $\frac{(23 - 31.2)^2}{31.2} = \frac{(-8.2)^2}{31.2} = \frac{67.24}{31.2} = 2.1551$
2. Full-time: $\frac{(20 - 52.0)^2}{52.0} = \frac{(-32)^2}{52.0} = \frac{1024}{52.0} = 19.6923$
3. Part-time: $\frac{(31 - 10.4)^2}{10.4} = \frac{20.6^2}{10.4} = \frac{424.36}{10.4} = 40.8058$
4. Unemployed: $\frac{(30 - 10.4)^2}{10.4} = \frac{19.6^2}{10.4} = \frac{384.16}{10.4} = 36.9423$

Adding these values: $\chi^2 = 2.1551 + 19.6923 + 40.8058 + 36.9423 = 99.5955$

Step 4: Degrees of Freedom

The degrees of freedom for a chi-square goodness-of-fit test is $\text{df} = k - 1$, where $k$ is the number of categories. Here, $k = 4$, so: $\text{df} = 4 - 1 = 3$

Step 5: $p$-value Calculation

Using the chi-square distribution table or a calculator with $\chi^2 = 99.5955$ and 3 degrees of freedom, we find that the $p$-value is extremely small, essentially $p < 0.001$, indicating strong evidence against the null hypothesis.

Final Answers

1. Degrees of Freedom: 3
2. Test Statistic: 99.60
3. $p$-value: <0.001