Total plasma volume is important in determining the required plasma component in

Question

Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. Suppose that a random sample of 50 male firefighters are tested and that they have a plasma volume sample mean of x = 37.5 ml/kg (milliliters plasma per kilogram body weight). Assume that = 7.20 ml/kg for the distribution of blood plasma.

(a) Find a 99% confidence interval for the population mean blood plasma volume in male firefighters. What is the margin of error? (Round your answers to two decimal places.)

lower limit = _____

upper limit = _____

margin of error = _____

(b) What conditions are necessary for your calculations? (Select all that apply.)

- n is large

- the distribution of weights is uniform

- the distribution of weights is normal

- is known

- is unknown

(c) Find the sample size necessary for a 99% confidence level with maximal margin of error E = 2.50 for the mean plasma volume in male firefighters. (Round up to the nearest whole number.)

_____= male firefighters

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 =    0.005          
X = sample mean =    37.5          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    7.2          
n = sample size =    50          
              
Thus,              

Lower bound =    34.87720363       [ANSWER]      
Upper bound =    40.12279637       [ANSWER]      
Margin of Error E =    2.62279637   [ANSWER]      

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

B)

We used z because n is large and sigma is known, so

- n is large
- is known [ANSWER]

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

c)

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 =    7.2  
E = margin of error =    2.5  
      
Thus,      
      
n =    55.03248637  
      
Rounding up,      
      
n =    56   [ANSWER]

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