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 eight blood tests for uric acid.  The mean concentration was 5.35 mg/dL. The distribution of uric acid in healthy adult males can be assumed to be normal, with a population standard deviation of 1.85 mg/dL. Construct a 98% confidence interval estimation of the population mean concentration of uric acid in this patient's blood. Write a sentence summarizing your results.

Answer 1

Solution :

Given that,

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

Margin of error = E = Z/2    * ( /n)

= 2.326 * ( 1.85/   8)

= 1.5214
At 98% confidence interval estimate of the population mean
is,

- E < < + E

5.35- 1.5214 <   < 5.35 +1.5214

3.8286 <   < 6.8714

( 3.8286 , 6.8714 )