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 σ is known to be $2.00 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 = 0.43 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 | $ ____ |
a)
xbar= | 6.880 | |||||
σ = | 2.00 | |||||
n= | 41.000 | |||||
standard errror of mean = | σ_{x}=σ/√n= | 0.3123 | ||||
for 90 % CI value of z= | 1.645 | |||||
margin of error E=z*std error = | 0.51 | |||||
lower confidence bound=sample mean-margin of error= | 6.3662 | |||||
Upper confidence bound=sample mean +margin of error= | 7.3938 |
from above
lower limit= | 6.37 | |
upper limit= | 7.39 | |
margin of error= | 0.51 |
b)
required sample size n=(zσ/E)^{2 } = | 59 |
c)
multiplying part a) by 15*20 =300 |
lower limit= | 1909.87 | |||
upper limit= | 2218.13 | |||
margin of error= | 154.13 |
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