Question

A simple random sample from a population with a normal distribution of 102 body temperatures has x = 98.70°F and s= = 0.62°F. Construct a 98% 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. 1°F<o</°F (Round to two decimal places as needed.)

Answer 1

We have given here,                  
                  
          
Sample standard deviation=0.62      
Sample size =102          
                  
Degree of freedom =n-1=101      
Level of significance given =1-0.98=0.02      
Chi square critical value for lower tail=70.901  
                  
Chi square critical value for upper tail=136.971  
$\sqrt{\frac{(n-1)S^{2}}{\chi _{1-(\alpha /2)}^{2}}}<\sigma <\sqrt{\frac{(n-1)S^{2}}{\chi _{(\alpha /2)}^{2}}}$

$= \sqrt{\frac{( 101 )* 0.3844 }{ 136.971 }}<\sigma<\sqrt{\frac{( 101 )* 0.3844 }{ 70.901 }}$

$0.53<\sigma<0.74$