Determine the finite and infinite sample size for a 5000 population. use percentage population of 12%, confidence interval of 0.02 and confidence level 90*
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A: From the provided information Confidence interval is (0.34, 0.52)
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Q: nutritionist
A: Please refer to the image below
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A: Margin of error(E)=8standard deviation()=55confidence level=95%
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A: Given,Sample size, n = 35Confidence level, C = 99% = 0.99
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A: Given, Confidence Interval is (0.68, 0.82) Upper limit U = 0.82 Lower limit L = 0.68
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- For each combination of sample size and sample proportion, find the approximate margin of error for the 95% confidence level. (Round the answers to three decimal places.) (a) n = 100, p̂ = 0.57. (b) n = 200, p̂ = 0.57. (c) n = 200, p̂ = 0.40.You want to estimate the mean weight of quarters in circulation. A sample of 50 quarters has a mean weight of 5.632 grams and a standard deviation of 0.056 gram. Use a single value to estimate the mean weight of all quarters. Also, find the 95% confidence interval for the average weight of all quarters. The estimate for the mean weight of all quarters is (Round to three decimal places as needed.) Find the 95% confidence interval for the mean weight of all quarters. grams <μ grams (Round to three decimal places as needed.) grams. GA recording company wants to estimate the mean length of musical cd’s being recorded. A random sample of 28 cd’s has a mean length of 48.47 minutes and standard deviation of 13.9 minutes. Use this sample data to construct the 95% confidence interval for the population mean length of cd’s. Show lower and upper values for interval using two-decimal accuracy.
- What is the point estimate of the population proportion, the margin of error for each confidence interval, and the number of individuals in the sample lower bound 0.201, upper bound 0.249, n = 1200What is the positive t-value for a confidence interval for the population mean with a sample size of 30 at a 90% confidence level? Express with 3 decimalsThe average weight of 31 randomly selected minivans was 4150 pounds. The sample standard deviation was 480 pounds. Find the 99 percent confidence interval of the true population mean weight of minivans.
- Find the critical value x²R corresponding to a sample size of 19 and a confidence level of 99 percent. X²R = (Round to 3 decimals)You want to estimate the mean weight of quarters in circulation. A sample of 60 quarters has a mean weight of 5.682 grams and a standard deviation of 0.061 gram. Use a single value to estimate the mean weight of all quarters. Also, find the 95% confidence interval for the average weight of all quarters. The estimate for the mean weight of all quarters is (Round to three decimal places as needed.) grams. Find the 95% confidence interval for the mean weight of all quarters. grams < p< grams (Round to three decimal places as needed.)Help please!
- y =99% , Confidence Level = 90% , n = 10 One-sided Tolerance IntervalThe heights of pine trees at a local tree farm are known to follow a normal distribution with a standard deviation o = 7 inches. A researcher takes a sample of 99 trees and finds they have a mean height of = 121 inches. Use this to find a 90% confidence interval for the height of all the pine trees at the tree farm. What is the margin of error? *Include 3 decimals places in your final answer* Confidence Interval: ( Margin of Error:Determine the sample size required to estimate a population mean to within 11 units given that the population standard deviation is 50. Use a confidence level of 95%. Sample Size =