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

Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 38 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.86 per 100 pounds.

 

 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

First we need to name some of the variables we are given:

n = 38 the sample size

x = 6.88 the sample mean

𝝈 = 1.86 the population standard deviation

We need to find a 90% confidence interval for the mean price per 100 lbs of watermelon:

𝞪 = 1-.90 = .10

𝞪/2 = .05

z(𝞪/2) = -1.64

-z(𝞪/2) = 1.64

This can give us a probability expression:

P(-1.645 < z < 1.645) = .90

The margin of error is calculated with the formula: E = z(𝞪/2)(𝝈/√n)

E = 1.645(1.86/√38) = $.4963(300) = $148.89   

Then to calculate the upper and lower limit we add and subtract E from x:

lower limit = 6.88 - .4963 = $7.38(300) = $2214

upper limit = 6.88 + .4963 = $6.38(300) = $1914

Note: We multiply the E, upper and lower limits by 300 because 15 tons is 30000lbs and we need the price per 100 lbs, so we divide 30000/100 and get 300.