### Watermelon Crop Pricing AnalysisUnderstanding the market price for watermelon crops is essential for farmers to make informed decisions. This exercise details how to find confidence intervals and required sample sizes for price estimation.**Scenario**: In the third week of July, a random sample of 43 farming regions gave a sample mean price $ \bar{x} = \$6.88 $ per 100 pounds of watermelon. The standard deviation $ \sigma $ is known to be \$1.90 per 100 pounds.#### (a) 90% Confidence Interval for Population Mean PriceCalculate a 90% confidence interval for the population mean price (per 100 pounds) that farmers receive for their watermelon crop. Determine the margin of error. Record your answers to two decimal places.- **Lower Limit**: $____- **Upper Limit**: $____- **Margin of Error**: $____#### (b) Sample Size for Maximal ErrorDetermine the sample size necessary for a 90% confidence level with a maximal error of estimate $ E = 0.39 $ for the mean price per 100 pounds of watermelon. The sample size should be rounded up to the nearest whole number.- **Sample Size**: ____ farming regions#### (c) Confidence Interval for Market Cash ValueA farm brings 15 tons of watermelon to market. Calculate a 90% confidence interval for the population mean cash value of this crop. Note: 1 ton equals 2000 pounds. Round your answers to two decimal places.- **Lower Limit**: $____- **Upper Limit**: $____- **Margin of Error**: $____These calculations help in understanding the variability and expected range of market prices for the watermelon crops, thereby aiding in agricultural economics and decision-making processes.

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Answered: What price do farmers get for their…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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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 = Assume that o is known to be $1.92 per 100 pounds. $6.88 per 100 pounds of watermelon. (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 2$ 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. (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…
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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 o 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 = 0.37 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 %24
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