### Part A: Chi-Square Test for Independence
We need to test the hypothesis that family size and the level of education attained by the father are independent using the Chi-Square test of independence.
#### Step 1: Formulate Hypotheses
- Null hypothesis (\(H_0\)): Family size and the level of education are independent.
- Alternative hypothesis (\(H_1\)): Family size and the level of education are not independent.
#### Step 2: Construct the Observed Frequency Table
The observed frequency table is given as follows:
```
Number of Children
Education 0 – 1 2 – 3 Over 3 Row Totals
Primary 14 37 32 83
Secondary 19 42 17 78
College 12 17 10 39
Column Totals 45 96 59 200
```
#### Step 3: Calculate the Expected Frequency
The formula for expected frequency \( E_{ij} \) in the cell corresponding to the \( i \)-th row and the \( j \)-th column is:
\[ E_{ij} = \frac{(Row \, Total \, of \, i)(Column \, Total \, of \, j)}{Grand \, Total} \]
Calculate the expected frequencies:
- For Primary, 0-1:
\[ E_{11} = \frac{83 \times 45}{200} = 18.675 \]
- For Primary, 2-3:
\[ E_{12} = \frac{83 \times 96}{200} = 39.84 \]
- For Primary, Over 3:
\[ E_{13} = \frac{83 \times 59}{200} = 24.515 \]
Following the same method for each cell:
```
Expected frequency table:
Number of Children
Education 0 – 1 2 – 3 Over 3
Primary 18.675 39.84 24.515
Secondary 17.55 37.44 23.01
College 8.775 18.72 11.475
```
#### Step 4: Calculate the Chi-Square Statistic
\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \]
Where \( O_{ij} \) and \( E_{ij} \) are the observed and expected frequencies, respectively.
Calculate for each cell and sum up:
\[ \chi^2 = \frac{(14 - 18.675)^2}{18.675} + \frac{(37 - 39.84)^2}{39.84} + \cdots \]
\[ \chi^2 \approx 0.059 + 0.2 + 2.28 + 0.122 + 0.56 + 1.57 + 1.07 + 1.58 + 0.19 \approx 7.63 \]
#### Step 5: Determine the Degrees of Freedom
Degrees of freedom (df) = (number of rows - 1) \* (number of columns - 1)
\[ df = (3 - 1) \times (3 - 1) = 4 \]
#### Step 6: Determine the Critical Value and Compare
Using a Chi-Square table at \(\alpha = 0.05\) and \(df = 4\), the critical value is approximately 9.488.
#### Step 7: Conclusion
Since 7.63 < 9.488, you fail to reject the null hypothesis. There is not enough evidence to conclude that family size and the level of education attained by the father are dependent at the 0.05 significance level.
### Part B: Price Indices
#### Part B (i): Compute Price Relatives for Each Item in 2001 Using 1989 as the Base Period
The price relative is calculated as:
\[ PR = \left(\frac{Price \, in \, 2001}{Price \, in \, 1989}\right) \times 100 \]
For Item A:
\[ PR_A = \left(\frac{7.75}{7.50}\right) \times 100 = 103.33 \]
For Item B:
\[ PR_B = \left(\frac{1500}{630}\right) \times 100 = 238.10 \]
#### Part B (ii): Compute an Unweighted Aggregate Price Index for the Two Items in 2001 Using 1989 as the Base Period
The unweighted aggregate price index is:
\[ \frac{\sum PR}{Number \, of \, Items} = \frac{103.33 + 238.10}{2} \approx 170.72 \]
#### Part B (iii): Compute a Weighted Aggregate Price Index for the Two Items Using Laspeyres’ Method
Laspeyres' index uses base year quantities:
\[ \text{Laspeyres Index} = \frac{\sum (P_1 \cdot Q_0)}{\sum (P_0 \cdot Q_0)} \times 100 \]
For Item A:
\[ P_1 \cdot Q_0 = 7.75 \cdot 1500 = 11625 \]
\[ P_0 \cdot Q_0 = 7.50 \cdot 1500 = 11250 \]
For Item B:
\[ P_1 \cdot Q_0 = 1500 \cdot 2 = 3000 \]
\[ P_0 \cdot Q_0 = 630 \cdot 2 = 1260 \]
\[ \text{Laspeyres Index} = \frac{11625 + 3000}{11250 + 1260} \times 100 \approx 170.79 \]
#### Part B (iv): Compute a Weighted Aggregate Price Index for the Two Items Using Paasche’s Method
Paasche's index uses current year quantities:
\[ \text{Paasche Index} = \frac{\sum (P_1 \cdot Q_1)}{\sum (P_0 \cdot Q_1)} \times 100 \]
For Item A:
\[ P_1 \cdot Q_1 = 7.75 \cdot 1800 = 13950 \]
\[ P_0 \cdot Q_1 = 7.50 \cdot 1800 = 13500 \]
For Item B:
\[ P_1 \cdot Q_1 = 1500 \cdot 1 = 1500 \]
\[ P_0 \cdot Q_1 = 630 \cdot 1 = 630 \]
\[ \text{Paasche Index} = \frac{13950 + 1500}{13500 + 630} \times 100 \approx 172.02 \]
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
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.
(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
ii. Compute an unweighted aggregate price index for the two items in 2001 using 1989
as the base period
iii. Compute a weighted aggregate price index for the two items using the Laspeyre’s
method.
iv. Compute a weighted aggregate price index for the two items using the Paasche
method.
1 answer