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motikmotik
3 years ago
11

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 o

f 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
Mathematics
1 answer:
Dafna11 [192]3 years ago
6 0

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 ]

   

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