Question

# What price do farmers get for their watermelon crops? In the third week of July, a...

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 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.

(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 = 0.35 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)

Given,

Here at 90% CI, z value is 1.645

Interval = xbar +/- z*s/sqrt(n)

substitute values

= 6.88 +/- 1.645*1.96/sqrt(40)

= 6.88 +/- 0.51

= (6.37 , 7.39)

b)

Here at 90% CI, z value is 1.645

sample n = (z*s/E)^2

substitute values

= (1.645*1.96/0.35)^2

= 84.860944

= 85

c)

For 1 ton = 2000 pounds

15 tons = 30000 pounds

Now for 100 pounds , xbar = 6.88

For 30000 pounds , xbar = 6.88*30000/100 = 2064

standard deviation = 1.96*30000/100 = 588

Interval = xbar +/- z*s/sqrt(n)

substitute values

= 2064 +/- 1.645*588/sqrt(40)

= 2064 +/- 152.94

= (1911.06 , 2216.94)

lower limit = 1911.06

upper limit = 2216.94

margin of error = 152.94