To test \(H_0: \mu = 25\) against \(H_1: \mu \ne 25\), a random sample of 12 items gives the value of the t-statistic as t = 2.6. The table value of t at 5% level of significance (two-tailed) for 11 degrees of freedom is 2.201. Then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(H_0\) is rejected, because the calculated |t| is greater than the table value
For a sample of 12, the degrees of freedom are 12 − 1 = 11, so the given table value 2.201 is the right one.
In a two-tailed test, \(H_0\) is rejected when |t| is more than the table value:
\(|t| = 2.6 \gt 2.201\)
So \(H_0\) is rejected at 5% level: the mean differs significantly from 25.
Why the other options are incorrect
Option (1) \(H_0\) is not rejected, because the calculated t is greater than the table value — Incorrect. A calculated |t| greater than the table value leads to rejection, not acceptance.
Option (2) \(H_0\) is rejected, because the calculated |t| is less than the table value — Incorrect. |t| = 2.6 is not less than 2.201.
Option (3) no decision can be made, because the table value for 12 degrees of freedom is needed — Incorrect. The degrees of freedom are n − 1 = 11, which is the value given.