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 eleven 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 = 1.89 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) Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.10 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 =    = 1.89

Sample size = n =11

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96* (1.89 /  11 )

= 1.12

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

- E < < + E

5.35 - 1.12 <   < 5.35+ 1.12

4.23 <   < 6.47

( 4.23 ,  6.47 )

lower limit =4.23

upper limit =6.47

margin of error = 1.28

Solution :

Given that,

standard deviation =   =1.89

Margin of error = E = 1.10

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size = n = [Z_/2* / E] ²

n = ( 1.96* 1.89/ 1.10 )²

n =11

Sample size = n =11