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	$
upper limit	$
margin of error	$

(b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.27 for the mean price per 100 pounds of watermelon. (Round up to the nearest whole number.)

(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

Part a

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

We are given

Xbar = 6.88

σ = 1.98

n = 40

Confidence level = 90%

Critical Z value = 1.6449

(by using z-table)

Confidence interval = Xbar ± Z*σ/sqrt(n)

Confidence interval = 6.88 ± 1.6449*1.98/sqrt(40)

Confidence interval = 6.88 ± 1.6449*0.3131

Confidence interval = 6.88 ± 0.5149

Lower limit = 6.88 - 0.5149 = 6.37

Lower limit = 6.37

Upper limit = 6.88 + 0.5149 = 7.39

Upper limit = 7.39

Margin of error = Z*σ/sqrt(n)

Margin of error = 1.6449*1.98/sqrt(40)

Margin of error = 0.5149

Margin of error = 0.51

(b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.27 for the mean price per 100 pounds of watermelon.

The sample size formula is given as below:

n = (Z*σ/E)^2

We are given

Confidence level = 90%

Z = 1.6449

(by using z-table)

σ = 1.98

E = 0.27

n = (1.6449*1.98/0.27)^2

n = 145.5063

n = 146

Required ample size = 146

(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?

1 ton = 2000 pounds

15 ton = 15*2000 = 30000 pounds

Xbar for 100 pounds = 6.88

Xbar for 30000 pounds = (30000/100)*6.88 = 300*6.88 = 2064

σ for 30000 pounds = (30000/100)*1.98 = 300*1.98 = 594

n = 40

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

We are given

Xbar = 2064

σ = 594

n = 40

Confidence level = 90%

Critical Z value = 1.6449

(by using z-table)

Confidence interval = Xbar ± Z*σ/sqrt(n)

Confidence interval = 2064± 1.6449*594/sqrt(40)

Confidence interval = 2064± 1.6449*93.9196

Confidence interval = 2064± 154.4841

Lower limit = 2064 - 154.4841= 1909.52

Lower limit = 1909.52

Upper limit = 2064 + 154.4841 = 2218.48

Upper limit = 2218.48

Margin of error = Z*σ/sqrt(n)

Margin of error = 1.6449*594/sqrt(40)

Margin of error = 154.4841

Margin of error = 154.4841