A hospital emergency room (ER) has collected a sample of n = 40 to estimate the mean wait time in the ER. It has found the population standard deviation is 32. Using a 90 percent confidence level, what is its margin of error?

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BlackTom AI · Accepted Answer

We start by identifying the formula for the margin of error when estimating a population mean with a known standard deviation: ME = z * (sigma / sqrt(n)).

First, determine the z-critical value for a 90% confidence level. The 90% two-tailed interval leaves 5% in each tail, so z ≈ 1.645.

Next, note the given values: sigma = 32 and n = 40. Compute the standard error: sigma / sqrt(n) = 32 / sqrt(40).

Calculate sqrt(40) ≈ 6.3249, so 32 / 6.3249 ≈ 5.059.

Now multiply by the z-critical value: 1.645 * 5.059 ≈ 8.32.

Thus the margin of error is about 8.32, which rounds to approximately ±8.3 visits.

Option by option analysis:
- Option 1: Approximately ±8.3 visits — This matches the computed ME (≈8.32). It is correct.
- Option 2: Approximately ±1.5 visits — This is far smaller than the calculated margin; it would imply either a much smaller standard error or a smaller z-value, which is inconsistent with the given data.
- Option 3: About ±1.3 visits — Similar issue as Option 2; the margin is not close to 1.3, indicating a miscalculation of standard error or z-score.
- Option 4: About ±9.9 visits — This is larger than the computed ME (≈8.32) and would overstate the uncertainty; it suggests overestimating either z or sigma relative to n.

In summary, the precise computation using the standard formula with the supplied sigma, n, and confidence level yields a margin of error around 8.32, aligning with Option 1.

OSC 2030-009 Fall 2025 SUBMIT: Graded Homework Module 7 (Chap 8)

A hospital emergency room (ER) has collected a sample of n = 40 to estimate the mean wait time in the ER. It has found the population standard deviation is 32. Using a 90 percent confidence level, what is its margin of error?

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