What happens to the standard error of the mean as you increase your sample size from 10 to 40 measurements?

Question

BlackTom AI · Accepted Answer

When considering how the standard error of the mean (SEM) behaves as sample size increases, we need to recall that SEM = s / sqrt(n), where s is the sample standard deviation and n is the sample size. 
Option 1: 'The precision of your mean estimate improves substantially' — As n grows from 10 to 40, the denominator sqrt(n) increases from sqrt(10) to sqrt(40), reducing SEM by a factor of sqrt(40/10) = sqrt(4) = 2. This means the mean estimate becomes noticeably more precise, so this statement aligns with the mathematical relationship.
Option 2: 'The precision of your mean estimate decreases' — This would imply SEM increases with n, which contradicts the formula SEM = s / sqrt(n). Unless s were to increase dramatically with n (which is not typically assumed in standard SEM discussion for a fixed population variance), precision does not decrease simply due to larger n.
Option 3: 'Individual measurements become more accurate' — Increasing sample size does not change the accuracy of each individual measurement; it changes the precision of the estimated mean. Measurement accuracy is a property of the measurement instrument and protocol, not a direct consequence of sampling more observations.
Option 4: 'The precision of your mean estimate remains unchanged' — This would hold only if n did not affect SEM, which conflicts with the inverse square root relationship. Increasing n reduces SEM, so precision does not stay the same.
In typical sampling theory, expanding the sample from 10 to 40 reduces the standard error by a factor of 2, reflecting greater precision in the mean estimate. The other statements either misinterpret the effect of sample size on SEM, confuse measurement accuracy with sampling precision, or ignore the known relationship between n and SEM.

What happens to the standard error of the mean as you increase your sample size from 10 to 40 measurements?单项选择题

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