A machine is set to cut steel rods of mean length 120 cm. A random sample of 49 rods has a mean length of 118.6 cm and a sample standard deviation of 4.2 cm (calculated with divisor n − 1). To test \(H_0: \mu = 120\) against \(H_1: \mu \ne 120\), it is given that the table value of t at 5% level of significance (two-tailed) for 48 degrees of freedom is 2.011. Which of the following is correct?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → t = −2.33 (approx.), and \(H_0\) is rejected
The standard error is
\( \displaystyle\frac{s}{\sqrt{n}} = \frac{4.2}{\sqrt{49}} = \frac{4.2}{7} = 0.6 \)
So
\( t = \displaystyle\frac{118.6 - 120}{0.6} = \frac{-1.4}{0.6} \approx -2.33 \)
The test is two-tailed, so \(H_0\) is rejected when |t| is more than 2.011:
\(|t| = 2.33 \gt 2.011\)
So \(H_0\) is rejected: the mean length of the rods differs significantly from 120 cm.
Why the other options are incorrect
Option (2) t = −2.33 (approx.), and \(H_0\) is not rejected — Incorrect. |t| = 2.33 is more than 2.011, so \(H_0\) is rejected.
Option (3) t = −0.33 (approx.), and \(H_0\) is not rejected — Incorrect. −0.33 is −1.4 ÷ 4.2, dividing by s instead of by the standard error 0.6.
Option (4) t = −16.33 (approx.), and \(H_0\) is rejected — Incorrect. −16.33 comes from dividing s by n = 49 instead of by \(\sqrt{49} = 7\).