Question

Final t nottesuo Chapter 8 Estimation Question 19dinuaib Iarmon A 97092 Wa bas 9703istuqmo cs Diagnostic Tests: Plasma Volume Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma (n) volume is influenced by the overall health and physical activity of the individual. Suppose that a random sample of 45 male firefighters are tested and that they had a plasma volume sample mean of -37.5 ml/kg (millimeters plasma per kilogram body ml/kg for the distribution of blood plasma DogeI0 - 7.50 (b) Assume that O (a) Find a 99% confidence interval for the population mean blood plasma volume in male firefighters. (b) What is the margin of error? em iT 9gulliv iqoH:oloo (c) Find the sample size necessary for a 99% confidence level with the maximum margin of berror E 2.50 for the mean plasma volume in male firefighters. udinizib vllmIon oze togolosodoA:1 lomoHoowo2) mm 0 3o YliaovinUyHA mbA olibo.s n Ouestion 20

Answer 1

n = sample size = 45

Sample mean = $\bar{x}=37.5$

Population standard deviation = $\sigma =7.50$

Here population standard deviation is known so we use here z interval

a) Confidence level = c = 0.99

99% confidence interval for population mean blood plasma volume in male fire fighters is

$\bar{x}-z_c*\frac{\sigma }{\sqrt{n}}<\mu<\bar{x}+z_c*\frac{\sigma }{\sqrt{n}}$

where zc is z critical value for (1+c)/2 = (1+0.99)/2 = 0.995

zc = 2.58                    (From statistical table of z values)

$37.5-2.58*\frac{7.50 }{\sqrt{45}}<\mu<37.5+2.58*\frac{7.50 }{\sqrt{45}}$

$37.5-2.8845<\mu<37.5+2.8845$

$34.6155<\mu<40.3845$                          (Round to 4 decimal)

99% confidence interval for population mean blood plasma volume in male fire fighters is (34.6155, 40.3845)

b) Margin of error (E)

$E=z_c*\frac{\sigma }{\sqrt{n}}$

$E=2.58*\frac{7.50}{\sqrt{45}}$

E = 2.88              (Round to 2 decimal)

Margin of error is 2.88

c) Margin of error = 2.50

$n=(\frac{z_c*\sigma }{E})^2$

$n=(\frac{2.58*7.50 }{2.50})^2$

n = 59.9076

n ~ 60              (Round to nearest integer)

Sample size = 60