7. A health care organization selected 25 healthy adults at random and measured

Question

7. A health care organization selected 25 healthy adults at random and measured their temperatures and found a mean of 96.72o F and a standard deviation of 1.39o F. The organization wishes to estimate the population mean at 90% confidence. You do not have any data available to you on the standard deviation of temperature for the population of all adults.

a. What is the formula for the margin of error for the mean in this sample?
b. At 90% confidence, what is the actual value of the margin of error?
c. What is the 90% confidence interval for the mean temperature of the population?

Explanation / Answer

Given n=25, xbar=mean=96.72, s=1.39
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(a) the formula for the margin of error for the mean in this sample is
s/n

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(b) Given =0.1, the critival value is |Z(0.05)|=1.645 (check normal table)

So the actual value of the margin of error is

Z*s/n = 1.645*1.39/sqrt(25) = 0.457

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(c) the 90% confidence interval for the mean temperature of the population is

xbar ± Z*s/n

--> 96.72 ± 0.457

--> (96.263, 97.177)

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