The marks of a random sample of 6 students in a quiz are 16, 18, 21, 21, 22, 22. The sample standard deviation is taken as \(s = \sqrt{\displaystyle\frac{\sum (x - \bar{x})^2}{n - 1}}\). For testing the claim that the mean mark of all students is 17.5, the value of the t-statistic is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2.5
The sample mean is
\( \displaystyle \bar{x} = \frac{16 + 18 + 21 + 21 + 22 + 22}{6} = \frac{120}{6} = 20 \)
The deviations from 20 are −4, −2, 1, 1, 2, 2, so
\(\sum (x - \bar{x})^2 = 16 + 4 + 1 + 1 + 4 + 4 = 30\)
\( \displaystyle s = \sqrt{\frac{30}{5}} = \sqrt{6} \)
The standard error is
\( \displaystyle \frac{s}{\sqrt{n}} = \frac{\sqrt{6}}{\sqrt{6}} = 1 \)
So the one-sample t-statistic is
\( \displaystyle t = \frac{20 - 17.5}{1} = 2.5 \)
Why the other options are incorrect
Option (2) 2.74 (approximately) — Incorrect. This uses divisor n = 6 (\(s^2 = 5\)) instead of n − 1 = 5.
Option (3) 1.02 (approximately) — Incorrect. This is 2.5 ÷ \(\sqrt{6}\), dividing by s instead of by the standard error 1.
Option (4) 0.42 (approximately) — Incorrect. This is 2.5 ÷ 6, dividing by the variance \(s^2\) instead of by the standard error.