Problem

The displayed bootstrap mean is centered at \(15\) with standard error \(2\); a benchmark has standard error \(1\). Compute the \(\text{one}\text{-}\text{standard}\text{-}\text{error}~\text{interval}\) and compare precision and sampling variability.

Neutral comparison scaffold for a bootstrap mean centered at 15 with standard error 2 and a benchmark standard error of 1; it withholds computed endpoints and any precision conclusion.

Big Picture

What this problem is really about

Separate location from spread: the bootstrap center estimates the parameter, while the standard error describes sample-to-sample variability. A one-standard-error interval moves one standard error in each direction from the center; its full width is twice that spread. When comparable estimates are judged on the same scale, the smaller standard error indicates greater precision.

Answer choices

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The one-SE interval is [13, 17]. Its larger standard error means lower precision and greater sampling variability than the benchmark.

Use the full width 17 - 13 = 4 as the standard error, giving [11, 19]. This makes the bootstrap estimate less precise than the benchmark.

Treat 2 as the center and 15 as the standard error, giving a one-SE interval of [-13, 17]. Its larger standard error means lower precision than the benchmark.

The one-SE interval is [13, 17]. Its larger standard error means higher precision and less sampling variability than the benchmark.

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The displayed bootstrap mean is centered at \(15\) with standard error \(2\); a benchmark has standard error \(1\). Compute the \(\text{one}\text{-}\text{standard}\text{-}\text{error}~\text{interval}\) and compare precision and sampling variability.

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