Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.98 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 $1
upper limit $2
margin of error $3
(b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.35 for the mean price per 100 pounds of watermelon. (Round up to the nearest whole number.)
4 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 $5
upper limit $6
margin of error $7

Answer 1

A) Upper limit = 7.39

Lower limit = 6.37

Margin of error = 0.51

B)Sample size = 87

C) Upper limit = 2218.50

Lower limit = 1909.50

Margin of error = 154.50

Explanation:

We are given;

x¯ = $6.88 per 100 pounds

σ = $1.98 per 100 pounds

n = 45

A) We want to find a 90% Confidence interval. It means the significance level is; 1 - 0.9 = 0.1

Critical z-value for significance level of 0.1 is z = 1.645

Now, margin of error is;

MOE = zσ/√n

MOE = (1.645 × 1.98)/√40

MOE ≈ 0.51

Confidence interval is;

CI = x¯ ± MOE

CI = 6.88 ± 0.51

CI = ((6.88 + 0.51), (6.88 - 0.51))

CI = (7.39, 6.37)

Upper limit = 7.39

Lower limit = 6.37

B) We are given E = 0.35

Thus;

E = zσ/√n

Rearranging, we have;

n = (zσ/E)²

n = (1.645 × 1.98/0.35)²

n ≈ 87

Sample size = 87

C)We are told that the farm brings 15 tons of watermelon to market.

Also, 1 ton = 2000 pounds

Thus;

15 tons = 15 × 2000 pounds = 30000 pounds

Now this is per 100 pounds.

Thus, we can write as: 30000/100 = 300 pounds

So;

x¯ = 300 × 6.88

x¯ = 2064

σ = 1.98 × 300

σ = 594

Critical z-value for significance level of 0.1 is z = 1.645

Now, margin of error is;

MOE = zσ/√n

MOE = (1.645 × 594)/√40

MOE ≈ 154.50

Confidence interval is;

CI = x¯ ± MOE

CI = 2064 ± 154.50

CI = (2064 + 154.50, 2064 - 154.50)

CI = (2218.5, 1909.5)

Upper limit = 2218.50

Lower limit = 1909.50