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 $1.80 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?

(b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.33 for the mean price per 100 pounds of watermelon.

(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.

Answer 1

m = 6.88, s = 1.80 , n = 38

(a) Standard error, SE = s/√n = 1.80/√38= 0.2920

The 90% confidence interval for the mean is given by

= [m - 1.645 * SE, m + 1.645 * SE]

= [6.88 - 1.645 * 0.2920 , 6.88 + 1.645 * 0.2920]

= [$6.3997, $7.3603]

(b) n = z^2 * s^2 / e^2

= 1.645^2 * 1.80^2 / 0.33^2

= 80.51

= 81

(c) Margin of error = z * s / √n

= (1.645 * 1.8)/ √38

= 0.4803

90% confidence interval for the mean is given by [m - 1.645 * SE, m + 1.645 * SE]

= [6.88 * (15 * 2000/100) - 1.645 * 0.2920, 6.88 * (15 * 2000/100) + 1.645 * 0.2920]

= [ 2064 - 0.4803, 2064 + 0.4803 ]

= [$2063.52, $2064.48]