Question

In an experiment with 15 observations on \( x \), the results \( \sum x^2 = 2830 \) and \( \sum x = 170 \) were available. One observation (20) was wrong and was replaced by 30. The corrected variance is:

• A. \( 9.3 \)
• B. \( 8.3 \)
• C. \( 188.6 \)
• D. \( 177.3 \)
• E. \( 78 \)

Hint:

When correcting data, remember that errors in \( \sum x^2 \) are corrected by squaring the individual values, not by squaring the total sum error.

Answer 1

Answer: (E)

Concept: Variance \( (\sigma^2) \) is calculated as \( \frac{\sum x^2}{n} - \left( \frac{\sum x}{n} \right)^2 \). When an observation is replaced, we must calculate the new "Corrected \( \sum x \)" and "Corrected \( \sum x^2 \)" by subtracting the wrong value and adding the correct one.

Step 1: Calculate corrected sums.
Corrected \( \sum x \): \[ \sum x_{\text{new}} = 170 - 20 + 30 = 180 \] Corrected \( \sum x^2 \): \[ \sum x^2_{\text{new}} = 2830 - 20^2 + 30^2 = 2830 - 400 + 900 = 3330 \]

Step 2: Calculate corrected mean and corrected variance.
Corrected Mean \( (\bar{x}_{\text{new}}) \): \[ \bar{x}_{\text{new}} = \frac{180}{15} = 12 \] Corrected Variance \( (\sigma^2_{\text{new}}) \): \[ \sigma^2_{\text{new}} = \frac{\sum x^2_{\text{new}}}{n} - (\bar{x}_{\text{new}})^2 \] \[ \sigma^2_{\text{new}} = \frac{3330}{15} - (12)^2 \] \[ \sigma^2_{\text{new}} = 222 - 144 = 78 \]