Overproduction of uric acid in the body can be an indication

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 ? = 1.77 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? (Use 2 decimal places.)

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

the distribution of uric acid is uniform ? is unknown ? is known n is large the distribution of uric acid is normal

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

95% of the intervals created using this method will contain the true average uric acid level for this patient. The probability that this interval contains the true average uric acid level for this patient is 0.95.     The probability that this interval contains the true average uric acid level for this patient is 0.05. 5% of the intervals created using this method will contain the true average uric acid level for this patient.

(d) Find the sample size necessary for a 95% confidence level with maximal error of estimate E = 1.18 for the mean concentration of uric acid in this patient\'s blood.

Solution

(a) The degree of freedom =n-1=8-1=7

Given a=1-0.95=0.05, t(0.025, df=7) =2.36 (from student t table)

So the margin of error = t*s/vn

=2.36*1.77/sqrt(8)

=1.48

So the lower limit is

xbar - t*s/vn = 5.35- 2.36*1.77/sqrt(8) =3.873137

So the upper limit is

xbar +t*s/vn=5.35+ 2.36*1.77/sqrt(8) =6.826863

--------------------------------------------------------------------------------------------------

(b)the distribution of uric acid is normal

--------------------------------------------------------------------------------------------------

(c)95% of the intervals created using this method will contain the true average uric acid level for this patient.

--------------------------------------------------------------------------------------------------

(d) Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So n=(Z*s/E)^2

=(1.96*1.77/1.18)^2

=8.6436

Take n=9