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 σ = 8.00 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.)

σ is unknown

the distribution of weights is uniform

σ is known

n is large

the distribution of weights is normal

(c) Interpret your results in the context of this problem.

The probability that this interval contains the true average blood plasma volume in male firefighters is 0.01.

99% of the intervals created using this method will contain the true average blood plasma volume in male firefighters.     

The probability that this interval contains the true average blood plasma volume in male firefighters is 0.99.

1% of the intervals created using this method will contain the true average blood plasma volume in male firefighters.

(d) Find the sample size necessary for a 99% confidence level with maximal margin of error E = 2.70 for the mean plasma volume in male firefighters. (Round up to the nearest whole number.)
male firefighters

Answer 1

from above

lower limit =34.58 (try 34.59 if this comes wrong due to rounding of z)

upper limit =40.42 (try 40.41 if this comes wrong due to rounding of z)

margin of error =2.92( try 2.91 if this comes wrong due to rounding of z)

b)

σ is known

n is large

v)

99% of the intervals created using this method will contain the true average blood plasma volume in male firefighters.   

d)

for 99 % CI value of z=	2.580
standard deviation σ=	8.00
margin of error E =	2.7
required sample size n=(zσ/E)2                                         =	59.0