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 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.)

• σ is known
• σ is unknown
• n is large
• normal distribution of uric acid
• uniform distribution of uric acid

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

• The probability that this interval contains the true average uric acid level for this patient is 0.95.
• We are 95% confident that the true uric acid level for this patient falls within this interval.     
• The probability that this interval contains the true average uric acid level for this patient is 0.05.
• We are 5% confident that the true uric acid level for this patient falls within this interval.

(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

____________________________________________________________________________________________________________________________________

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    $ ________

Answer 1

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

======================

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