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 ten 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 = 1.75 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.)
(c) Give a brief interpretation of your results in the context of this problem.
(d) Find the sample size necessary for a 95% confidence level with maximal error of estimate 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 tests
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2.What price do farmers get for their watermelon crops? In the third week of July, a random sample of 41 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that σis known to be $1.94 per 100 pounds.
(a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. What is the margin of error? (Round your answers to two decimal places.)
lower limit $ _________
upper limit $ _________
margin of error $_________
(b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.29 for the mean price per 100 pounds of watermelon. (Round up to the nearest whole number.)
_________ farming regions
(c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop. What is the margin of error? Hint: 1 ton is 2000 pounds. (Round your answers to two decimal places.)
lower limit $___________
upper limit $ _________
margin of error $ ________
Solution:- given that n = 10, x = 5.35, σ = 1.75, 95% Confidence interval for Z = 1.96
(a) 95% confidence interval for the population mean concentration of uric acid in this patient's blood : X +/- Z*(σ/sqrt(n))
lower limit : 5.35 - (1.96*1.75/sqrt(10)) = 4.27
upper limit : 5.35 + (1.96*1.75/sqrt(10)) = 6.43
Margin of error : Z*(σ/sqrt(n)) = (1.96*1.75/sqrt(10)) = 1.08
(b) option A.
(c) option B.
(d) n = (Z*σ/E)^2 = (1.96*1.75/1.14)^2 = 9.052 = 9
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2. Given that x = $6.88, σ = 1.94, n = 41 90% Confidence interval for Z = 1.645
(a) 90% Confidence interval for the population mean
lower limit : 6.88 - (1.645*1.94/sqrt(41)) = 6.38
upper limit : 6.88 + (1.645*1.94/sqrt(41)) = 7.38
Margin of error : (1.645*1.94/sqrt(41)) = 0.50
(b) n = (1.645*1.94/0.29)^2 = 121.09 = 121 (rounded)
(c) for 15tons = 300
lower limit = 6.38*300 = 1914
upper limit = 7.38*300 = 2214
margin of error = 0.50*300 = 150