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 = $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 = 0.41 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	$

Answer 1

Solution :

Given that mean x-bar = 6.88 , standard deviation σ = 1.96 , n = 41

(a)
=> for 90% confidence interval, Z = 1.645

=> Margin of error E = Z*σ/sqrt(n)

= 1.645*1.96/sqrt(41)

= 0.5035

= 0.50 (rounded)

=> The 90% confidence interval for the population mean price is

=> x-bar +/- E

=> 6.88 +/- 0.50

=> (6.38 , 7.38)

=> Lower limit = 6.38

=> Upper limit = 7.38

=> Margin of error E = 0.50

(b)
=> Margin of error E = 0.41

=> for 90% confidence interval, Z = 1.645

=> Sample size n = (Z*σ/E)^2

= (1.645*1.96/0.41)^2

= 61.8410

= 62 (nearest whole number)

(c) n = 15 tons = 15*2000 = 30000

=> for 90% confidence interval, Z = 1.645

=> Margin of error E = Z*σ/sqrt(n)

= 1.645*1.96/sqrt(30000)

= 0.0186

= 0.02 (rounded)

=> The 90% confidence interval for the population mean price is

=> x-bar +/- E

=> 6.88 +/- 0.02

=> (6.86 , 6.9)

=> Lower limit = 6.86

=> Upper limit = 6.9

=> Margin of error E = 0.02