Question

(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.)

Answer 1

First we need to name some of the variables we are given:n = 38 the sample sizex = 6.88 the sample mean = 1.86 the population standard deviationWe need to find a 90% confidence interval for the mean price per 100 lbs of watermelon: = 1-.90 = .10/2 = .05z(/2) = -1.64-z(/2) = 1.64This can give us a probability expression:P(-1.645 < z < 1.645) = .90The margin of error is calculated with the formula: E = z(/2)(/√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) = $2214upper limit = 6.88 + .4963 = $6.38(300) = $1914Note: 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.

Explanation: