To test \(H_0: \mu = 120\) against \(H_1: \mu \ne 120\), a random sample of 25 observations gives a mean of 123 and a sample standard deviation of 6 (calculated with divisor n − 1). The table values of t (two-tailed) for 24 degrees of freedom are 2.064 at 5% level of significance and 2.797 at 1% level of significance. Which of the following statements are correct?
(A) The standard error of the mean is 0.24.
(B) The value of the t-statistic is 2.5.
(C) The number of degrees of freedom is 24.
(D) \(H_0\) is rejected at 5% level of significance but not at 1% level of significance.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (B), (C) and (D) only
The standard error is
\( \displaystyle \frac{s}{\sqrt{n}} = \frac{6}{5} = 1.2 \)
\( \displaystyle t = \frac{123 - 120}{1.2} = 2.5 \)
The degrees of freedom are 25 − 1 = 24. Since 2.064 < 2.5 < 2.797, \(H_0\) is rejected at 5% level but not at 1% level.
So SE = 1.2, t = 2.5, df = 24, and \(H_0\) is rejected at 5% but not at 1%.
Checking the statements
(A) The standard error of the mean is 0.24. — Incorrect. \(6/\sqrt{25} = 1.2\); 0.24 is 6 ÷ 25.
(B) The value of the t-statistic is 2.5. — Correct. t = 3 ÷ 1.2 = 2.5.
(C) The number of degrees of freedom is 24. — Correct. df = 25 − 1 = 24.
(D) \(H_0\) is rejected at 5% level of significance but not at 1% level of significance. — Correct. |t| = 2.5 is more than 2.064 but less than 2.797.