For a sample mean, which combination of sample standard deviation $s$ and sample size $n$ will produce the largest value for the standard error $\frac{s}{\sqrt{n}}$?

• A. A small sample standard deviation $s$ and a large sample size $n$
• B. A large sample standard deviation $s$ and a large sample size $n$
• C. A small sample standard deviation $s$ and a small sample size $n$
• D. A large sample standard deviation $s$ and a small sample size $n$

Answer: (D)

Verified step by step guidance

1. Recall the formula for the standard error of the sample mean, which is given by: \[ SE = \frac{S}{\sqrt{n}} \] where \(S\) is the sample standard deviation and \(n\) is the sample size.
2. Understand that the standard error increases as the sample standard deviation \(S\) increases because it is in the numerator of the formula.
3. Recognize that the standard error decreases as the sample size \(n\) increases because it is in the denominator under the square root.
4. To produce the largest standard error, you want the numerator \(S\) to be as large as possible and the denominator \(\sqrt{n}\) to be as small as possible, meaning a large sample standard deviation and a small sample size.
5. Compare the given options by substituting the ideas about \(S\) and \(n\) into the formula to see which combination yields the largest value for \(SE\).