// What price do farmers get for their watermelon crops? In the third week of Ju
ID: 3295811 • Letter: #
Question
// What price do farmers get for their watermelon crops? In the third week of July, a random sample of 39 farming regions gave a sample mean of
x = $6.88 per 100 pounds of watermelon. Assume that is known to be $1.98 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 = 0.45 for the mean price per 100 pounds of watermelon. (Round up to the nearest whole number.)
(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: $
Thanks
Explanation / Answer
a)
margin of error = z-score * std/sqrt(n)
= 1.645 * 1.98/sqrt(39)
= 0.5216
lower bound = 6.88 -0.5216 = 6.3584
upper bound = 6.88 + 0.5216 = 7.4016
b)
margin of error = 0.45
=>
1.645 * 1.98/sqrt(n) = 0.45
=>
n = 52.39
c)
mean = 6.88 * 2000/100 * 15 = 2064
standard deviation = 1.98 * 2000/100 * 15 = 594
margin of error = 1.645 *594/sqrt(39) = 156.466
lower bound = 2064 - 156.466 = 1907.534
upper bound = 2064 + 156.466 = 2220.466
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