Question

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	$ ____

Answer 1

a)

xbar=				6.880
σ =				2.00
n=				41.000
standard errror of mean =			σx=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