Overproduction of uric acid in the body can be an indication of cell breakdown.
ID: 3064806 • Letter: O
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.t Over a period of months, an adult male patient has taken nine 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 -1.95 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 is unknown O normal distribution of uric acid is known O n is large (c) Interpret your results in the context of this problem O The probability that this interval contains the true average uric acid level for this patient is 0.05 O There is a 95% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient. O 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. O The probability that this interval contains the true average uric acid level for this patient is 0.95 (d) Find the sample size necessary for a 95% confidence level with maximal margin of error E-1.14 for the mean concentration of uric acid in this patient's blood. (Round your answer up to the nearest whole number.) blood testsExplanation / Answer
PART A.
TRADITIONAL METHOD
given that,
standard deviation, =1.95
sample mean, x =5.35
population size (n)=9
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 1.95/ sqrt ( 9) )
= 0.65
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.65
= 1.27
III.
CI = x ± margin of error
confidence interval = [ 5.35 ± 1.27 ]
= [ 4.08,6.62 ]
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DIRECT METHOD
given that,
standard deviation, =1.95
sample mean, x =5.35
population size (n)=9
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 5.35 ± Z a/2 ( 1.95/ Sqrt ( 9) ) ]
= [ 5.35 - 1.96 * (0.65) , 5.35 + 1.96 * (0.65) ]
= [ 4.08,6.62 ]
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interpretations:
1. we are 95% sure that the interval [4.08 , 6.62 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
PART B.
sigma is known,
normal distribution of uric acid
PART C.
there is 95% chance that the confidence interval is one of the interval containing the population average
uric and level for this patient
PART D.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 1.95
ME =1.14
n = ( 1.96*1.95/1.14) ^2
= (3.82/1.14 ) ^2
= 11.24 ~ 12
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