For one-sample t-tests of \(H_0: \mu = \mu_0\) at 5% level of significance, some table values of t are
| Degrees of freedom | 9 | 14 | 19 |
| One-tailed | 1.833 | 1.761 | 1.729 |
| Two-tailed | 2.262 | 2.145 | 2.093 |
Match List-I with List-II
| List-I (Alternative hypothesis, sample size) | List-II (Reject \(H_0\) if) | ||
| (A) | \(H_1: \mu \ne \mu_0\), n = 10 | (I) | t > 1.761 |
| (B) | \(H_1: \mu \gt \mu_0\), n = 15 | (II) | |t| > 2.093 |
| (C) | \(H_1: \mu \ne \mu_0\), n = 20 | (III) | t < −1.833 |
| (D) | \(H_1: \mu \lt \mu_0\), n = 10 | (IV) | |t| > 2.262 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
The degrees of freedom are n − 1. The sign in \(H_1\) decides the tail: ≠ is two-tailed (reject for large |t|), > is right-tailed (reject for large positive t), < is left-tailed (reject for large negative t).
(A) \(H_1: \mu \ne \mu_0\), n = 10
df = 9, two-tailed value 2.262, so this matches (IV) |t| > 2.262.
(B) \(H_1: \mu \gt \mu_0\), n = 15
df = 14, one-tailed value 1.761 in the right tail, so this matches (I) t > 1.761.
(C) \(H_1: \mu \ne \mu_0\), n = 20
df = 19, two-tailed value 2.093, so this matches (II) |t| > 2.093.
(D) \(H_1: \mu \lt \mu_0\), n = 10
df = 9, one-tailed value 1.833 in the left tail, so this matches (III) t < −1.833.