Question

A simple random sample from a population with a normal distribution of 98 body temperatures has x=98.20°F and s=0.61°F. Construct a 99% confidence interval estimate of the standard deviation of body temperature of all healthy humans. Click the icon to view the table of Chi-Square critical values. °F

Answer 1

from the question;

we are to construct 99% confidence interval. this can be done using

`\bar{}x\text{ }\pm\text{ z}(\frac{s}{\sqrt[]{n}})`

where,

`\bar{x}\text{ = }98.20,\text{ s = 0.61, n = 98 z= 2.576}`

inserting values

`\begin{gathered} 98.20\text{ }\pm2.576\text{ }\frac{0.61}{\sqrt[]{98}} \\ 98.20\text{ }\pm\text{ 2.576}*0.0616 \\ =\text{ 98.20 }\pm\text{ }0.159 \\ =98.20\text{ + }0.159\text{ or 98.20 - 0.159} \\ =\text{ 98.359 0r 98.041} \end{gathered}`

therefore the 99% confident inter vale is between 98.041 to 98.359