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 fifteen 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 = 1.95 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.)

lower limit   ?
upper limit   ?
margin of error   ?

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

1. normal distribution of uric acid

2. σ is unknown

3. uniform distribution of uric acid

4. σ is known

5. n is large

(c) 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

Solution :

Given that,

Point estimate = sample mean = = 5.35

Population standard deviation =    = 1.95

Sample size = n = 15

a) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96

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

E = 1.96 * (1.95 /  15 )

E = 0.99

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

  ± E

5.35 ± 0.99

( 4.36, 6.34)  

lower limit = 4.36

upper limit = 6.34

margin of error = 0.99

b) σ is known

normal distribution of uric acid

c) margin of error = E = 1.16

sample size = n = [Z/2* / E] 2

n = [1.96 *1.95 / 1.16]2

n = 10.85

Sample size = n = 11 blood tests