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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

User Stmax
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Answer: 8.323 (approximate)

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Step-by-step 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*(s)/(√(n))\\\\E \approx 1.645*(32)/(√(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.

User Equality
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