Question

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1degreesf and a standard deviation of 0.56degreesf. construct a 99​% confidence interval estimate of the mean body temperature of all healthy humans. what does the sample suggest about the use of 98.6degreesf as the mean body​ temperature?

Answer 1

Answer:

The 99% confidence interval estimate for the mean is 97.9576 ≤ μ ≤ 98.246

A) This suggests that the mean body temperature could very possibly be 98.6 °F

Step-by-step explanation:

 The number of body temperatures, n = 103

The mean body temperature, $\bar x$ = 98.1

The standard deviation, s = 0.56

Confidence interval required = 99%

Confidence interval, CI is given by

$CI=\bar{x}\pm z\frac{s}{\sqrt{n}}$

Plugging in the values we get;

z = 2.56 at 99%

$CI=98.1}\pm 2.56\times \frac{0.56}{\sqrt{103}}$

Therefore, we have

$\mu_{min} = 97.9579 , \ \mu_{max} = 98.242$,

The statistical result suggest that at 99% confidence level, the sample mean temperature is likely to be 98.1 °F