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  = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.92 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 (in dollars). What is the margin of error (in dollars)? (For each answer, enter a number. 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 = 0.29 for the mean price per 100 pounds of watermelon. (Enter a number. 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 (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

Answer: 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 = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.92 per 100 pounds.

Solution:

x̄ = 6.88

σ = 1.92

n = 41

(a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop (in dollars). What is the margin of error (in dollars)?

At 90% confidence interval, α = 0.10

Z critical =Zα/2 = 1.645

Margin of error, E = Zα/2 * σ/√n

E = 1.645 * 1.92/√41

Margin of error E = 0.4933

the 90% confidence interval for mean price:

CI = x̄ ± E

CI = 6.88 ± 0.4933

CI = (6.3867, 7.3733)

Therefore, the 90% confidence interval for mean price is

lower limit = $ 6.39

upper limit = $ 7.37

margin of error, E = $ 0.49

(b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.29 for the mean price per 100 pounds of watermelon.

At 90% confidence interval,

Zα/2 = 1.645

Margin of error E = 0.29

Sample size n:

n = (Zα/2 * σ / E)^2

n = (1.645 * 1.92 / 0.29)^2

n = 118=6146

n = 119

Therefore, farming regions = 119.

(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.

15 tons of watermelon to market,

15 * 2000/100 = 300 (per 100 pounds)

x̄ = 6.88 * 300 = 2064

σ = 1.92 * 300 = 576

At 90% confidence interval, α = 0.10

Z critical =Zα/2 = 1.645

Margin of error, E = Zα/2 * σ/√n

E = 1.645 * 576/√41

Margin of error E = 147.9777

the 90% confidence interval for mean price:

CI = x̄ ± E

CI = 2064 ± 147.9777

CI = (1916.0223, 2211.9777)

Therefore the 90% confidence interval for mean price is

lower limit = $ 1916.02

upper limit = $ 2211.98

margin of error = $ 147.98