Question:
A simple random sample of size 'n' will be drawn from a class of 125 students, and mean mathematics score of the sample will be compute
D. If the standard error of the sample mean for "with replacement sampling" is twice as much as the standard error of the sample mean for "without replacement sampling", then the value of n is
A simple random sample of size 'n' will be drawn from a class of 125 students, and mean mathematics score of the sample will be compute
D. If the standard error of the sample mean for "with replacement sampling" is twice as much as the standard error of the sample mean for "without replacement sampling", then the value of n is
D. If the standard error of the sample mean for "with replacement sampling" is twice as much as the standard error of the sample mean for "without replacement sampling", then the value of n is
Show Hint
The Standard Error for sampling without replacement is always smaller than with replacement because of the Finite Population Correction (FPC). In this problem, the FPC term must be $1/4$ for the ratio to be 2.
Updated On: Jun 6, 2026
- 79
- 94
- 63
- 33
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The Correct Option is B
Solution and Explanation
We compare the Standard Error (SE) formulas for Simple Random Sampling With Replacement (SRSWR) and Without Replacement (SRSWOR).
Step 1: \color{redState the SE Formulas
Let $N = 125$ (population size) and $n$ be the sample size.
$SE_{WR} = \sqrt{\frac{\sigma^{2}}{n}}$.
$SE_{WOR} = \sqrt{\frac{\sigma^{2}}{n} \cdot \left(\frac{N-n}{N-1}\right)}$.
Step 2: \color{redSet up the Ratio
Given: $SE_{WR} = 2 \times SE_{WOR}$.
$\sqrt{\frac{\sigma^{2}}{n}} = 2 \times \sqrt{\frac{\sigma^{2}}{n} \cdot \left(\frac{N-n}{N-1}\right)}$.
Step 3: \color{redAlgebraic Simplification
Square both sides:
$\frac{\sigma^{2}}{n} = 4 \times \frac{\sigma^{2}}{n} \cdot \left(\frac{N-n}{N-1}\right)$.
$1 = 4 \times \frac{N-n}{N-1}$.
$N - 1 = 4N - 4n$.
Step 4: \color{redSolve for n
$4n = 3N + 1$.
Substitute $N = 125$:
$4n = 3(125) + 1 = 375 + 1 = 376$.
$n = \frac{376}{4} = 94$.
The value of $n$ is 94.
Step 1: \color{redState the SE Formulas
Let $N = 125$ (population size) and $n$ be the sample size.
$SE_{WR} = \sqrt{\frac{\sigma^{2}}{n}}$.
$SE_{WOR} = \sqrt{\frac{\sigma^{2}}{n} \cdot \left(\frac{N-n}{N-1}\right)}$.
Step 2: \color{redSet up the Ratio
Given: $SE_{WR} = 2 \times SE_{WOR}$.
$\sqrt{\frac{\sigma^{2}}{n}} = 2 \times \sqrt{\frac{\sigma^{2}}{n} \cdot \left(\frac{N-n}{N-1}\right)}$.
Step 3: \color{redAlgebraic Simplification
Square both sides:
$\frac{\sigma^{2}}{n} = 4 \times \frac{\sigma^{2}}{n} \cdot \left(\frac{N-n}{N-1}\right)$.
$1 = 4 \times \frac{N-n}{N-1}$.
$N - 1 = 4N - 4n$.
Step 4: \color{redSolve for n
$4n = 3N + 1$.
Substitute $N = 125$:
$4n = 3(125) + 1 = 375 + 1 = 376$.
$n = \frac{376}{4} = 94$.
The value of $n$ is 94.
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