Question

Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma.† Over a period of months, an adult male patient has taken eight blood tests for uric acid. The mean concentration was x = 5.35 mg/dl. The distribution of uric acid in healthy adult males can be assumed to be normal, with σ = 1.91 mg/dl.

(a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. 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.)

normal distribution of uric acidn is largeσ is unknownσ is knownuniform distribution of uric acid

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

The probability that this interval contains the true average uric acid level for this patient is 0.05.The probability that this interval contains the true average uric acid level for this patient is 0.95.    We are 95% confident that the true uric acid level for this patient falls within this interval.We are 5% confident that the true uric acid level for this patient falls within this interval.

(d) Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.16 for the mean concentration of uric acid in this patient's blood. (Round your answer up to the nearest whole number.)
blood tests

Answer 1

a)

sample mean 'x̄=		5.350
sample size    n=		8
std deviation σ=		1.910
std error ='σx=σ/√n=1.91/√8=		0.6753

for 95 % CI value of z=		1.960
margin of error E=z*std error =		1.3235
lower bound=sample mean-E=		4.0265
Upper bound=sample mean+E=		6.6735

lower limit=		4.03
upper limit =		6.67
margin of error=		1.32

b)

normal distribution of uric acid

σ is known

c)

  We are 95% confident that the true uric acid level for this patient falls within this interval

d)

for95% CI crtiical Z          =	1.960	from excel:normsinv(0.975)
standard deviation σ=	1.910
margin of error E =	1.16
required n=(zσ/E)2 =	11	Rounding up