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**Research question 1:** We wish to investigate whether employment status in the Sydney region is distributed according to the table below.

employment status | Casual | Full-time | Part-time | Unemployed

Percentage        |     30 |        50 |        10 |         10

QUESTIONS:

Report the number of the degrees of freedom [integer]:

Answer 1 Question 1

Report the value of the test statistic [use at least 2dp]:

Answer 2 Question 1

Report the 

*𝑝*

p

-value [use at least 3dp]:

Answer 3 Question 1





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

We wish to investigate whether employment status in the Sydney region is distributed according to the table below. 

employment status | Casual | Full-time | Part-time | Unemployed

Percentage | 30 | 50 | 10 | 10

Analyze whether employment status in Sydney follows a specified distribution using a chi-square goodness-of-fit test. This example includes observed and expected values for casual, full-time, part-time, and unemployed statuses. Learn to calculate the chi-square test statistic and p-value to determine statistical significance.

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