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 ? = 1.87 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.)

? is knownnormal distribution of uric acid? is unknownn is largeuniform distribution of uric acid

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

There is not enough information to make an interpretation.The probability that this interval contains the true average uric acid level for this patient is 0.95.     There is a 95% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient.The probability that this interval contains the true average uric acid level for this patient is 0.05.There is a 5% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient.

(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)lower limit =sample mean -z*std error =5.35-1.96*1.87/sqrt(8)=4.05

upper limit=sample mean +z*std error =5.35+1.96*1.87/sqrt(8)=6.65

margin of error =z*std error =1.96*1.87/sqrt(8)=1.30

b)

? is known normal distribution of uric acid

c)There is a 95% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient.

d)

for 95 % CI value of z=	1.960
standard deviation =	1.87
margin of error E =	1.16
required sample size n=(z/E)2    =	10.0