In: Statistics and Probability
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 = 1.79 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.)
σ is known n is large σ is unknown normal distribution of uric acid uniform distribution of uric acid
(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 0.95. There is not enough information to make an interpretation. There is a 5% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient. The probability that this interval contains the true average uric acid level for this patient is 0.05. 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) 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
a) At 95% confidence interval the critical value is z* = 1.96
The 95% confidence interval is
+/- z* *
= 5.35 +/- 1.96* 1.79/
= 5.35 +/- 1.06
= 4.29, 6.41
Lower limit = 4.29
Upper limit = 6.41
Margin of error = z* *
= 1.96* 1.79/
= 1.06
b) is known
normal distribution of uric acid
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) Margin of error = 1.16
or, z* * = 1.16
or, 1.96* 1.79/ = 1.16
or, n = (1.96 * 1.79/1.16)^2
or, n = 10