Question

A hospital emergency room has collected a sample of 40 to estimate the mean number of visits per day. The sample standard deviation was found to be 32. Using a 90 percent confidence level, what is its margin of error

Answer 1

Answer:  8.323  (approximate)

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Explanation:

s = 32 = sample standard deviation

n = 40 = sample size

Despite not knowing the population standard deviation (sigma), we can still use the Z distribution because n > 30. When n is this large, the student T distribution is approximately the same (more or less) compared to the standard Z distribution. The Z distribution is nicer to work with.

At 90% confidence, the z critical value is roughly z = 1.645. This can be found using either a Z table or a calculator.

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We have these values:

• z = 1.645 (approximate)
• s = 32
• n = 40

Plug these values into the margin of error formula.

$E = z*\frac{s}{\sqrt{n}}\\\\E \approx 1.645*\frac{32}{\sqrt{40}}\\\\E \approx 8.32311480156318\\\\E \approx 8.323$

The margin of error is roughly 8.323

I rounded to 3 decimal places because z = 1.645 is rounded to 3 decimal places. If your teacher wants some other level of precision, then be sure to follow those instructions.