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A data set includes 110110 body temperatures of healthy adult humans having a mean of 98.098.0degrees°f and a standard deviation of 0.740.74degrees°f. construct a 9999​% confidence interval estimate of the mean body temperature of all healthy humans. what does the sample suggest about the use of 98.6degrees°f as the mean body​ temperature?

User YouEyeK
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I assume that says
n=110 and
\bar x = 98.0 and
\sigma = 0.74.

It doesn't really tell us if this is the standard deviation of the individual samples or of the sample average. Since they're related by a factor of
√(n), around ten, some common sense tells us this is the standard deviation of the individual samples.

So we can conclude our sample mean has a standard deviation


s = ( \sigma )/( \sqrt n ) = (.74)/( √(110)) = 0.0705562

and that
x=98.6 is


z = ( x- \bar x)/(s) = (98.6 - 98.0)/(0.0705562 ) = 8.5

That's a huge z score, 8.5 standard deviations away. 110 is big enough we can just use the normal distribution and not worry about t distributions.

We can conclude the 98.6 is very unlikely to be the true mean or close to the true mean of human temperatures.

User Theyuv
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