### Part (a): Chi-Square Test for Independence
Given data:
| Education | Number of Children |
|-----------|--------------------|
| | 0 - 1 | 2 - 3 | Over 3 |
| Primary | 14 | 37 | 32 |
| Secondary | 19 | 42 | 17 |
| College | 12 | 17 | 10 |
1. **Step 1: Formulate Hypotheses**
- Null Hypothesis (\( H_0 \)): Family size and level of education attained by the father are independent.
- Alternative Hypothesis (\( H_1 \)): Family size and level of education attained by the father are not independent.
2. **Step 2: Create the Contingency Table**
| | 0 - 1 | 2 - 3 | Over 3 | Row Sum |
|------------|-------|-------|--------|---------|
| Primary | 14 | 37 | 32 | 83 |
| Secondary | 19 | 42 | 17 | 78 |
| College | 12 | 17 | 10 | 39 |
| Column Sum | 45 | 96 | 59 | 200 |
3. **Step 3: Calculate Expected Frequencies**
\[ E_{ij} = \frac{(Row\ Tot_{i}) \times (Column\ Tot_{j})}{Grand\ Total} \]
For example, the expected frequency for Primary and 0-1 children (\( E_{11} \)) is:
\[ E_{11} = \frac{(83 \times 45)}{200} = \frac{3735}{200} = 18.675 \]
Calculating all expected frequencies:
| | 0 - 1 | 2 - 3 | Over 3 |
|------------|-------|-------|--------|
| Primary | 18.675| 39.84 | 24.545 |
| Secondary | 17.55 | 37.44 | 23.01 |
| College | 8.775 | 18.72 | 11.445 |
4. **Step 4: Calculate Chi-Square Statistic (\( \chi^2 \))**
\[ \chi^2 = \sum \frac{(Observed - Expected)^2}{Expected} \]
Perform the calculation for each cell:
For Primary and 0-1 children:
\[ \frac{(14 - 18.675)^2}{18.675} = 1.171 \]
Calculating for all cells and summing up:
\[ \chi^2 = (1.171 + 0.203 + 2.297 + 0.121 + 1.235 + 1.574 + 1.185 + 0.158 + 0.182) = 7.126 \]
5. **Step 5: Determine the Degrees of Freedom and Critical Value**
Degrees of freedom = (number of rows - 1) * (number of columns - 1)
\[ \text{df} = (3-1) * (3-1) = 2 * 2 = 4 \]
Using the Chi-Square distribution table, the critical value for 4 degrees of freedom at the 0.05 level of significance is approximately 9.488.
6. **Step 6: Compare and Draw Conclusions**
Since \( \chi^2_{calculated} = 7.126 \) is less than \( \chi^2_{critical} = 9.488 \), we fail to reject the null hypothesis.
**Conclusion:** There is not enough evidence to suggest that family size and level of education attained by the father are not independent.
### Part (b): Price Index Calculations
Given data:
| Item | Quantity 1989 | Quantity 2001 | Unit Price 1989 | Unit Price 2001 |
|------|---------------|---------------|-----------------|-----------------|
| A | 1500 | 1800 | 7.50 | 7.75 |
| B | 2 | 1 | 630 | 1500 |
1. **Step 1: Compute Price Relatives**
Price relative for Item A:
\[ \text{Price Relative for A} = \frac{7.75}{7.50} \times 100 = 103.33\% \]
Price relative for Item B:
\[ \text{Price Relative for B} = \frac{1500}{630} \times 100 = 238.10\% \]
2. **Step 2: Compute an Unweighted Aggregate Price Index**
\[ \text{Unweighted Aggregate Price Index} = \frac{Sum\ of\ Price\ Relatives}{Number\ of\ Items} \]
\[ \text{Unweighted Aggregate Price Index} = \frac{103.33 + 238.10}{2} = 170.72\% \]
3. **Step 3: Compute a Weighted Aggregate Price Index using Laspeyre’s Method**
\[ \text{Laspeyre's Index} = \frac{\sum (P_1 \times Q_0)}{\sum (P_0 \times Q_0)} \times 100 \]
Where \( P_1 \) and \( P_0 \) are prices in 2001 and 1989, and \( Q_0 \) is the quantity in 1989.
\[ \text{Laspeyre's Index} = \frac{(7.75 \times 1500) + (1500 \times 2)}{(7.50 \times 1500) + (630 \times 2)} \times 100 \]
\[ \text{Laspeyre's Index} = \frac{(11625) + (3000)}{(11250) + (1260)} \times 100 \]
\[ \text{Laspeyre's Index} = \frac{14625}{12510} \times 100 \approx 116.93\% \]
4. **Step 4: Compute a Weighted Aggregate Price Index using Paasche’s Method**
\[ \text{Paasche's Index} = \frac{\sum (P_1 \times Q_1)}{\sum (P_0 \times Q_1)} \times 100 \]
Where \( Q_1 \) is the quantity in 2001.
\[ \text{Paasche's Index} = \frac{(7.75 \times 1800) + (1500 \times 1)}{(7.50 \times 1800) + (630 \times 1)} \times 100 \]
\[ \text{Paasche's Index} = \frac{(13950) + (1500)}{(13500) + (630)} \times 100 \]
\[ \text{Paasche's Index} = \frac{15450}{14130} \times 100 \approx 109.34\% \]
**Conclusion:**
- Price relatives indicate changes in prices of individual items.
- Unweighted price index provides an average change in price.
- Weighted indices (Laspeyre’s and Paasche’s) provide more accurate measures by considering quantities, reflecting different perspectives on price changes.
(a) A random sample of 200 married men, all retired, were classified according to education
and number of children:
Education Number of Children
0 – 1 2 – 3 Over 3
Primary 14 37 32
Secondary 19 42 17
College 12 17 10
Step by step answer the all questions.
And Use the Chi –square to test the hypothesis that family size and level of education attained
by the father are independent. Use 0.05 level of significance. (10 Marks)
(b) The following table reports prices and usage quantities for two items in 1989 and 2001
Quantity Unit Price
Item 1989 2001 1989 2001
A 1500 1800 7.50 7.75
B 2 1 630 1500
i. Compute price relatives for each item in 2001 using 1989 as the base period
(2 Marks)
ii. Compute an unweighted aggregate price index for the two items in 2001 using 1989
as the base period (2 Marks)
iii. Compute a weighted aggregate price index for the two items using the Laspeyre’s
method. (3 Marks)
iv. Compute a weighted aggregate price index for the two items using the Paasche
method.
1 answer