In a harvest of 50 watermelons, farmers prefer at least 72% of watermelons to be classified as "large." With a confidence level of 99%, what is the margin of error? A. ±0.019% B. ±1% C. ±16.3% D. ±10.1%

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A: Solution: a. From the given information, confidence level is 0.99 and E=0.02.

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A: Given data isn=332x=36p^=xn=36332=0.108434Margin og error(E)=0.04Confidence level=95%

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A: Given information- Margin of error, E = 0.03 Confidence level, c = 90% Since nothing is known about…

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A: Given,population proportions (p*)=0.5margin of error (E)=0.04α=1-0.90=0.1α2=0.05Z0.05=1.645  (from…

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A: Find z-critical value:…

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A: The margin of error is 0.02, and confidence level is 99%.

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A: n = 150p^ = 0.2590% confidence level

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A: p = 0.19 n = 100 Margin of error = 0.02

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A: Given: e=0.01 a) Assume p=0.5 and q=0.5. From the Standard Normal Table, the value of z* for 90%…

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A: Given Information:Margin of error, E=0.02Confidence level, c=99%=0.99

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Q: 14. Find the sample size that will produce the maximum error of a 99% confidence interval for a…

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A: Assume sample proportion(p^)=0.5margin of error(E)=0.05α=1-0.99=0.01α2=0.005Z0.005=2.58    (from…

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A: The required sample size is calculated by using the formula:

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Q: If n=420 and ˆpp^ (p-hat) =0.19, find the margin of error at a 95% confidence level

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Q: With a given sample of n=15, x=8900, and s=500, construct 99 percent confidence intervals for…

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