Assume that the helium porosity (in percentage) of coal samples taken from any p

Question

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.79. Compute a 95% Cl for the true average porosity of a certain seam if the average porosity for 22 specimens from the seam was 4.85. (Round your answers to two decimal places.) Compute a 98% Cl for true average porosity of another seam based on 18 specimens with a sample average porosity of 4.56. (Round your answers to two decimal places.) How large a sample size is necessary if the width of the 95% interval is to be 0.46? (Round your answer up to the nearest whole number.) What sample size is necessary to estimate true average porosity to within 0.22 with 99% confidence? (Round your answer up to the nearest whole number.)

Explanation / Answer

a)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    4.85          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    0.79          
n = sample size =    22          
              
Thus,              
Margin of Error E =    0.330113923          
Lower bound =    4.519886077          
Upper bound =    5.180113923          
              
Thus, the confidence interval is              
              
(   4.519886077   ,   5.180113923   ) [ANSWER]

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b)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.01          
X = sample mean =    4.56          
z(alpha/2) = critical z for the confidence interval =    2.326347874          
s = sample standard deviation =    0.79          
n = sample size =    18          
              
Thus,              
Margin of Error E =    0.433177107          
Lower bound =    4.126822893          
Upper bound =    4.993177107          
              
Thus, the confidence interval is              
              
(   4.126822893   ,   4.993177107   ) [ANSWER]

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c)

Hence, the margin of error is

E = width/2 = 0.46/2 = 0.23.

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.025  
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
s = sample standard deviation =    0.79  
E = margin of error =    0.23  
      
Thus,      
      
n =    45.3205  
      
Rounding up,      
      
n =    46   [ANSWER]

*******************

d)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.005  
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
s = sample standard deviation =    0.79  
E = margin of error =    0.22  
      
Thus,      
      
n =    85.55452415  
      
Rounding up,      
      
n =    86   [ANSWER]

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