Question

Q11. Use the given information to find the number of degrees of​ freedom, the critical values

chi Subscript Upper L Superscript 2χ2L

and

chi Subscript Upper R Superscript 2χ2R​,

and the confidence interval estimate of

sigmaσ.

It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution.

Platelet Counts of Women

90​%

​confidence;

n=40​,

s=65.5.

The confidence interval estimate of sigmaσ is:

_____less than<sigmaσless than<_____

​(Round to one decimal place as​ needed.)

Answer 1

Sample Size,   n=   40  
Sample Standard Deviation,   s=   65.5000  
Confidence Level,   CL=   0.90  
          
          
Degrees of Freedom,   DF=n-1 =    39  
alpha,   α=1-CL=   0.1  
alpha/2 ,   α/2=   0.05  
Lower Chi-Square Value=   χ²_1-α/2 =   25.6954  
Upper Chi-Square Value=   χ²_α/2 =   54.5722  
          
confidence interval for variance is          
lower bound=   (n-1)s²/χ²_α/2 =   3066.02  
          
          
upper bound=   (n-1)s²/χ²_1-α/2 =   6511.66  
          

      
confidence interval for std dev is       
lower bound=   √(lower bound variance)= √3066.02 = 55.4
      
      
upper bound=   √(upper bound of variance= √6511.66 = 80.7