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 seven 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 = 1.85 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?

lower limit

upper limit

margin of error

(b) What conditions are necessary for your calculations?

σ is known

uniform distribution of uric acid

σ is unknown

normal distribution of uric acid

n is large

(c) Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.04 for the mean concentration of uric acid in this patient's blood.

Answer 1

Solution

Given that,

= 5.35

= 1.85

n = 7

a ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

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

= 1.960 * (1.85 / 7 )

= 1.37

Margin of error =1.37

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

- E < < + E

5.35 - 1.37 < < 5.35 + 1.37

3.98 < < 6.72

lower limit = 3.98

upper limit = 6.72

b ) σ is unknown

normal distribution of uric acid

c ) margin of error = E = 1.04

At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025 1.960

Z/2 = Z0.025 =

Sample size = n = ((Z/2 * ) / E)2

= (( 1.960 *1.85) / 1.04)2

= 12

Sample size = 12