question archive 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 = $6
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 = $6
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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 = $6.88 per 100 pounds of watermelon. Assume that ???? is known to be $1.90 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 (in dollars). What is the margin of error (in dollars)? (For each answer, enter a number. 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 = 0.37 for the mean price per 100 pounds of watermelon. (Enter a number. 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 (in dollars). What is the margin of error (in dollars)? Hint: 1 ton is 2000 pounds. (For each answer, enter a number. Round your answers to two decimal places.)
lower limit $
upper limit $
margin of error $
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a) margin of error, E = 0.49
lower limit = 6.39
upper limit = 7.37
b) n = 71
c) margin of error, E = 146.44
Lower limit = 1917.56
Upper limit= 2210.44
Step-by-step explanation
a) x? = 6.88
σ = 1.9
n= 41
df = n-1= 40
90% CI==> α= 0.1
critical value, Zc = 1.645 (from z-table or excel command: =-NORM.S.INV(α/2))
standard error, S.E=s/√n= 1.9 / √ 41 = 0.2967
margin of error, E = Zc × S.E= 1.645 × 0.2967 = 0.49
CI: x?±E
lower limit: x?-E= 6.88 - 0.49 = 6.39
upper limit: x?+E= 6.88 + 0.49 = 7.37
b) σ = 1.9
M.E= 0.37
90% CI ==> α= 0.1
critical value, Zc= 1.645 (from z-table or excel command: =-NORM.S.INV(α/2))
sample size, n = ( Zc × σ/E)2
= ( 1.645 × 1.9 / 0.37 )2
= 71.35683163
= 71
c) given 1 ton = 2000 pounds
so, 15 tons = 30000 pounds i.e., 300 hundred pounds
then, mean price, x? = 300 ( 6.88 ) = $ 2064
standard deviation, σ = 300 ( 1.9 ) = $ 570
n = 41
90% CI==> α= 0.1
critical value, Zc = 1.645
standard error, S.E=s/√n= 570 / √ 41 = 89.0190
margin of error, E = Zc × S.E= 1.645 × 89.0190 = 146.44
CI : x? ± E = 2064 ± 146.4400
Lower limit : 2064 - 146.44 = 1917.56
Upper limit : 2064 + 146.44 = 2210.44
