If in a test of 50 items with means of 30 and standard deviation of 6, which will be the reliability of the test?

Correct option is D

To calculate the reliability of a test, Kuder-Richardson Formula 21 (KR-21) is commonly used. The formula is as follows:
r=KK−1(1−M(K−M)K(SD2))r = \frac{K}{K - 1} \left(1 - \frac{M(K - M)}{K(SD^2)}\right)r=K−1K​(1−K(SD2)M(K−M)​)
Where:
· rrr: Reliability coefficient
· KKK: Number of items in the test
· MMM: Mean score
· SDSDSD: Standard deviation of the scores
Step-by-Step Calculation: 1. Given Values:
· K=50K = 50K=50 (Number of items)
· M=30M = 30M=30 (Mean score)
· SD=6SD = 6SD=6 (Standard deviation)
2. Compute SD2SD^2SD2 (Variance):
SD2=62=36SD^2 = 6^2 = 36SD2=62=36
3. Substitute into KR-21 Formula:
r=5050−1(1−30(50−30)50(36))r = \frac{50}{50 - 1} \left(1 - \frac{30(50 - 30)}{50(36)}\right)r=50−150​(1−50(36)30(50−30)​) r=5049(1−30(20)1800)r = \frac{50}{49} \left(1 - \frac{30(20)}{1800}\right)r=4950​(1−180030(20)​) r=5049(1−6001800)r = \frac{50}{49} \left(1 - \frac{600}{1800}\right)r=4950​(1−1800600​) r=5049(1−0.3333)r = \frac{50}{49} \left(1 - 0.3333\right)r=4950​(1−0.3333) r=5049×0.6667r = \frac{50}{49} \times 0.6667r=4950​×0.6667 r≈0.68r \approx 0.68r≈0.68
Correct Answer: (d) 0.68 Information Booster: 1. Reliability refers to the consistency of test scores. Higher reliability indicates that the test produces stable results across multiple administrations.
2. KR-21 is used for dichotomous items (e.g., correct/incorrect responses). It assumes all items measure the same construct.
3. A reliability coefficient (rrr) closer to 1 indicates higher reliability.