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 thirteen 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 = 1.87 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 (d) Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.14 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

Solution :

Given that,

a) Point estimate = sample mean = = 5.35  

Population standard deviation = =1.87

Sample size = n = 13

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z_{0.025 = 1.960}

Z/2 = 1.960  

Margin of error = E = Z/2 * ( /n)

=1.960 * (1.87 /  13 )

=1.0165 = 1.02

At 95 % confidence interval estimate of the population mean is,

- E < < + E

5.35 - 1.02 <   < 5.35 + 1.02

4.33 <   < 6.37

( 4.33 , 6.37 )

Lower limit = 4.33

Upper limit = 6.37

Margin of error = 1.02

d)

Margin of error = E = 1.14

Z_/2 = 1.96

sample size = n = [Z_/2* / E] ²

n = [ 1.96* 1.87 / 1.14 ]²

n = 10

Sample size = n = 10