Question 1206154
## Setting up the Chi-Squared Test
**Understanding the Problem:**
We want to test if the proportion of donations at Christmas differs from normal weekends. This is a classic chi-squared test for independence.

**Null Hypothesis (H₀):** The proportions of donations are the same at Christmas and normal weekends.
**Alternative Hypothesis (H₁):** The proportions of donations are different at Christmas and normal weekends.

**Setting up the Contingency Table:**

| Donation Amount | Normal Weekend | Christmas | Total |
|---|---|---|---|
| Over $200 | 0.14 | 0.28 | 0.42 |
| $100-$200 | 0.26 | 0.49 | 0.75 |
| Under $100 | 0.60 | 0.23 | 0.83 |
| Total | 1.00 | 1.00 | 1.80 |

**Calculating Expected Frequencies:**
To calculate the expected frequencies, we multiply the row total by the column total and divide by the grand total. For example, the expected frequency for the top-left cell is (0.42 * 1.00) / 1.80 = 0.2333.

**Calculating the Chi-Squared Test Statistic:**
The chi-squared test statistic is calculated as:
```
χ² = Σ [(Observed Frequency - Expected Frequency)² / Expected Frequency]
```

Calculating this for each cell and summing them up, we get:

```
χ² = [(0.14 - 0.2333)²/0.2333] + [(0.26 - 0.4167)²/0.4167] + ... + [(0.23 - 0.4611)²/0.4611]
```

**Calculating the Test Statistic:**
Using a calculator or statistical software, we find the chi-squared test statistic to be approximately **0.282**. 

**Note:** To complete the hypothesis test, we would also need to determine the degrees of freedom (df) and compare the calculated chi-squared statistic to the critical value or calculate the p-value. However, the question only asks for the test statistic, so we've provided that. 

The degrees of freedom for a chi-squared test of independence is calculated as (rows - 1) * (columns - 1). In this case, df = (3-1) * (2-1) = 2.