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 make patient has taken nine blood tests for uric acid. The mean concentration was x= 5.35 mg/dl. This distribution of uric acid in healthy adult males can be assumed to be normal with SD= 1.93 mg/dl.

(a) Find a 95% confidence interval for the population mean concentration of uric acid in this patients blood. What is the margin of error?

lower limit:

upper limit:

margin of error:

(b) What conditions are nessady for your calculations? (select all that apply)

- SD is known

- SD is unknown

- uniform distribution or uric acid

- n is large

- normal distribution of uric acid

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

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 5.35

Population standard deviation =    = 1.93

Sample size = n = 9

a) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_0.025 = 1.96

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

E = 1.96 * (1.93 /  9)

E = 1.26

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

  ± E

5.35 ± 1.26

( 4.09, 6.61)  

lower limit = 4.09

upper limit = 6.61

margin of error = 1.26

b) SD 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