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 six 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.)

uniform distribution of uric acid

σ is unknown

n is large normal distribution of uric acid

σ is known

(c) Interpret your results in the context of this problem.

The probability that this interval contains the true average uric acid level for this patient is 0.05.

The probability that this interval contains the true average uric acid level for this patient is 0.95.    

There is not enough information to make an interpretation.

There is a 95% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient.

There is a 5% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient.

(d) Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.00 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

a)

z value at 95% =1.96

Margin of error = z *(s/sqrt(n))
= 1.96 *(1.95/sqrt(6))
= 1.56

CI = mean +/- ME
= 5.35 +/- 1.56

= (3.79 , 6.91)

Lower limit = 3.79
Upper limit = 6.91

b)

n is large normal distribution of uric acid

σ is known

c)

There is a 95% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient.

d)

ME =1 , s = 1.95

z value at 95% = 1.96

ME = z *(s/sqrt(n))
1 = 1.96 *(1.95/sqrt(n))
n = (1.96 *1.95/1)^2
n = 15