For an independent sale-price sample, the sample standard deviation is $20,000 and n=16. Estimate the standard error of the mean as s/√n. Report to the nearest dollar.
Correct Answer
D) $5,000
20000/√16=5000. The standard error describes uncertainty in the sample mean under the stated independent-sample model, not the spread of individual prices.
Why This Is the Correct Answer
Option D is correct because the standard error of the mean is s / √n = $20,000 / √16 = $20,000 / 4 = $5,000. This figure describes uncertainty in the sample mean under the stated independent-sample model, not the spread of individual prices.
Why the Other Options Are Wrong
Option A: $1,250
Option A, $1,250, comes from dividing by n instead of by the square root of n: $20,000 / 16 = $1,250. The correct calculation is $20,000 / √16 = $20,000 / 4 = $5,000.
Option B: $20,000
Option B, $20,000, is just the sample standard deviation, which measures the spread of individual prices. The standard error divides it by √16 = 4, giving $5,000.
Option C: $80,000
Option C, $80,000, multiplies the standard deviation by the square root of n ($20,000 × 4) instead of dividing by it. The standard error is $20,000 / √16 = $20,000 / 4 = $5,000, smaller than the standard deviation, not larger.
Root n tames error
Standard error equals s over root n, so quadrupling the sample only halves the error. Divide by the square root, not by n itself.
How to use: Use this to avoid the common slip of dividing the standard deviation by n.
Exam Tip
Take the square root of n and divide by it. The standard error of the mean is smaller than the standard deviation whenever n is greater than 1, and s divided by n is a classic distractor.
Common Mistakes to Avoid
- -Dividing the standard deviation by n instead of by the square root of n
- -Multiplying by the square root of n instead of dividing
- -Reporting the standard deviation as the standard error
Concept Deep Dive
Analysis
This question tests the standard error of the mean, computed as s divided by the square root of n. With s = $20,000 and n = 16, the square root of 16 is 4, so the standard error is $5,000. The key trap is dividing by n instead of by its square root.
Background Knowledge
The standard error of the sample mean equals the standard deviation divided by the square root of the sample size. It shrinks as the sample grows, reflecting greater precision in estimating the mean. Standard deviation describes individual observations, while standard error describes the estimated mean.
Real-World Application
An appraiser summarizing 16 verified sales finds a standard deviation of $20,000. Reporting a standard error of $5,000 tells the reader how precisely the sample mean is estimated under the stated model. The appraiser does not confuse that figure with how much individual sale prices vary.
More Statistics Questions
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Assessment-to-sale ratios are 0.94, 0.96, 0.98, 1, 1.01. Compute the arithmetic mean. Report as a decimal ratio to three places.
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