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 fourteen 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. (Round your answers to two decimal places.)

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

is unknownn is largenormal distribution of uric aciduniform distribution of uric acid is known

(c) Give a brief interpretation of your results in the context of this problem.

There is a 5% 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 95% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient.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.

(d) Find the sample size necessary for a 95% confidence level with maximal error of estimate E = 1.04 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

a)for 95% CI, z=1.96

hence lower limit=sample mean -z*std deviation =1.6849

upper limit=sample mean +z*std deviation =9.0151

margin of error =z*std deviation =3.6651

b) is known

normal distribution of uric acid

c)

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

d)here as we know that sample size n=(z*std deviation/E)2 =~13

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