Question

Overproduction of uric acid in the body can be an indication of cell breakdown. This may...

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 fourteen 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 ? = 1.83 mg/dl.

(a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. (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.)

? is known

uniform distribution of uric acid

n is large

? is unknown

normal distribution of uric acid


(c) Give a brief interpretation of your results in the context of this problem.

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 0.95.    

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.

There is not enough information to make an interpretation.


(d) Find the sample size necessary for a 95% confidence level with maximal error of estimate 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

Homework Answers

Answer #1

Data given is:

Sample mean, m = 5.35

Population standard deviation S = 1.83

Sample size n = 14

(a)

95% CI is:

m-(1.96*(S/(n^0.5))) < < m+(1.96*(S/(n^0.5)))

5.35-(1.96*(1.83/(14^0.5))) < < 5.35+(1.96*(1.83/(14^0.5)))

4.39 < < 6.31

(b)

The conditions are:

? is known

normal distribution of uric acid

(c)

The proper interpretation for the confidence interval is:

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, ME = 1

For a 95% CI, ME = (1.96*(S/(n^0.5)))

Put values:

1 = (1.96*(1.83/(n^0.5)))

Solve to get:

n = 12.86 = 13

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