Here is the formula for split-half reliability:

\[
r_{xx} = \frac{N \sum X_1 X_2 - \sum X_1 \sum X_2}{\sqrt{\left(N \sum X_1^2 - (\sum X_1)^2\right) \left(N \sum X_2^2 - (\sum X_2)^2\right)}}
\]

### Substituting the values into the formula:

Given:
- \( N = 10 \) (number of participants)
- \( \sum X_1 = 68 \)
- \( \sum X_2 = 64 \)
- \( \sum X_1 X_2 = 460 \)
- \( \sum X_1^2 = 474 \)
- \( \sum X_2^2 = 424 \)

We substitute these values into the formula step by step:

1. **Numerator**:
   \[
   \text{Numerator} = N \sum X_1 X_2 - \sum X_1 \sum X_2 = (10)(460) - (68)(64) = 4600 - 4352 = 248
   \]

2. **Denominator**:
   The denominator is calculated using two parts:
   \[
   \text{Denominator part 1} = N \sum X_1^2 - (\sum X_1)^2 = (10)(474) - (68)^2 = 4740 - 4624 = 116
   \]
   \[
   \text{Denominator part 2} = N \sum X_2^2 - (\sum X_2)^2 = (10)(424) - (64)^2 = 4240 - 4096 = 144
   \]
   Now, calculate the full denominator:
   \[
   \text{Denominator} = \sqrt{(116)(144)} = \sqrt{16704} \approx 129.26
   \]

3. **Final Calculation**:
   \[
   r_{xx} = \frac{248}{129.26} \approx 1.92
   \]

The split-half reliability is \( r_{xx} \approx 1.92 \), which seems unusually high. This may suggest there's an issue with the scoring or test consistency across the two halves. Typically, reliability values range between 0 and 1. Would you like to explore why this might have occurred?

Learn how to calculate split-half reliability using the formula with step-by-step instructions. This problem involves statistical analysis and reliability assessment, using specific data points: N = 10, ΣX₁ = 68, ΣX₂ = 64, ΣX₁X₂ = 460, ΣX₁² = 474, ΣX₂² = 424. Suitable for advanced undergraduate students and researchers.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation