A population has standard deviation 20. Random samples of size 64 are drawn from it. Which of the following statements are correct?
(A) The standard error of the sample mean is 2.5.
(B) If the sample size is doubled to 128, the standard error becomes 1.25.
(C) For random samples of size 16, the standard error of the sample mean is 5.
(D) The mean of the sampling distribution of the sample mean is equal to the population mean.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
The standard error of the mean is \( SE = \displaystyle\frac{\sigma}{\sqrt{n}} \) with \(\sigma = 20\).
n = 64: \( SE = 20/8 = 2.5 \)
n = 128: \( SE = 20/\sqrt{128} \approx 1.77 \)
n = 16: \( SE = 20/4 = 5 \)
The sample means of all possible samples are centred at the population mean \(\mu\).
So SE = 2.5 for n = 64, about 1.77 for n = 128, 5 for n = 16, and the sampling distribution of \(\bar{x}\) has mean \(\mu\).
Checking the statements
(A) The standard error of the sample mean is 2.5. — Correct. \(20/\sqrt{64} = 20/8 = 2.5\).
(B) If the sample size is doubled to 128, the standard error becomes 1.25. — Incorrect. \(20/\sqrt{128} \approx 1.77\); halving the standard error needs four times the sample size (n = 256).
(C) For random samples of size 16, the standard error of the sample mean is 5. — Correct. \(20/\sqrt{16} = 20/4 = 5\).
(D) The mean of the sampling distribution of the sample mean is equal to the population mean. — Correct. The sample means are centred at \(\mu\).