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 bar = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $2.00 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 (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

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 6.88

Population standard deviation = = 2

Sample size = n = 30000

At 90% confidence level the z is ,

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645* ( 2/ 30000 )

= 0.02

At 90% confidence interval estimate of the population mean is,

- E < < + E

6.88 - 0.02 < < 6.88 + 0.02

6.86 < < 6.90

lower limit = 6.86

upper limit = 6.90

margin of error = 0.02