Question

A data set includes 40 pulse rates of men, and those pulse rates have a mean of 67.3 beats per minute and a standard deviation of 10.3 beats per minute. Construct a 99% confidence interval estimate of the standard deviation of the pulse rates of men.a) 8.4 beats per minute

Answer 1

Answer:

The 99% confidence interval of the  standard deviation of the pulse rates of men is  

       $[ 7.949 \  \ , \  14.387 ]$

Step-by-step explanation:

From the question we are told that

  The  sample size is  n  =  40

  The mean is  $\= x = 67.3 \ beats \ per \ minute$

  The standard deviation is  $s = 10.3 \ beats \ per \ minute$

Generally the degree of freedom is mathematically represented as

     $df = n- 1$

=>  $df = 40- 1$

=>  $df = 39$

From the question we are told the confidence level is  99% , hence the level of significance is    

      $\alpha = (100 - 99 ) \%$

=>   $\alpha = 0.01$

Generally from the chi distribution table the critical value  of   at a degree of freedom of is  

   $X^2 _{\frac{\alpha }{2} , 39} =  65.476$

Generally from the chi distribution table the critical value  of   at a degree of freedom of is  

   $X^2 _{1- \frac{\alpha }{2} , 39} =  19.996$

Generally the 99% confidence interval of the standard deviation of the pulse rates of men is mathematically represented as

  $s \sqrt{\frac{n- 1}{X^2_{1- \frac{\alpha }{2} , n-1} } }\  \ , \ s \sqrt{\frac{n- 1}{X^2_{\frac{\alpha }{2} , n-1} } }$

=>  $10.3 \sqrt{\frac{40- 1}{65.476} }\  \ , \ 10.3 \sqrt{\frac{40- 1}{19.996} }$

=>  $[ 7.949 \  \ , \  14.387 ]$