Overproduction of uric acid in the body can be an indication of cell breakdown.

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 nine 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 = 1.95 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.)

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

n is large

is known

uniform distribution of uric acid

normal distribution of uric acid

is unknown

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

There is not enough information to make an interpretation.

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.

The probability that this interval contains the true average uric acid level for this patient is 0.95.

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 = 1.02 for the mean concentration of uric acid in this patient's blood. (Round your answer up to the nearest whole number.)
blood tests

Explanation / Answer

here std error of mean =std deviaiton/(n)1/2 =1.95/(9)1/2 =0.65

for 95% CI ; z =1.96

hence  lower limit =mean -z*std error =4.08

upper limit =mean +z*std error =6.62

b)

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)

sample szie n=(z*std deviaiton/E)2 ~ 15

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.