Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 45 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.96 per 100 pounds.

(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

xbar=				6.880
σ =				1.96
n=				45.000
standard errror of mean =			σx=σ/√n=	0.2922
for 90 % CI value of z=				1.645
margin of error E=z*std error                            =						0.48
lower confidence bound=sample mean-margin of error=						6.3994
Upper confidence bound=sample mean +margin of error=						7.3606

c)

since there are 15*20 pouunds, multiplying above with 300 :

lower limit=				1919.82
upper limit=				2208.18
margin of error=				144.18