Determine the margin of error for a confidence interval to estimate the population proportion for the following confidence levels witha sample proportion equal to 0.38 and n = 100.a. 90%b. 95%c. 98%Click the icon to view a portion of the Cumulative Probabilities for the Standard Normal Distribution table....a. The margin of error for a confidence interval to estimate the population proportion for the 90% confidence level is .080(Round to three decimal places as needed.)b. The margin of error for a confidence interval to estimate the population proportion for the 95% confidence level is(Round to three decimal places as needed.)

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Answered: Determine the margin of error for a…

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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Determine whether the normal distribution can be used to compare the following population proportions. n₁ = 33, n₂ = 46, 1 = 0.818,2 0.913 Step 1 of 2: Calculate the four values n₁₁, n₁ (1-1), n22, and n2 (1-2). Round your answers to three decimal places, if necessary.
You work for a soft-drink company in the quality control division. You are interested in the standard deviation of one of your production lines as a measure of consistency. The product is intended to have a mean of 12 ounces, and your team would like the standard deviation to be as low as possible. You gather a random sample of 17 containers. Estimate the population standard deviation at a 98% level of confidence. 11.94 11.93 12.04 11.99 12.06 11.92 11.98 12.11 11.83| 11.88 12.05 11.93 11.96 12.2 11.92 11.82 11.97 (Data checksum: 203.53) Note: Keep as many decimals as possible while making these calculations. If possible, keep all answers exact by storing answers as variables on your calculator or computer. a) Find the sample standard deviation: b) Find the lower and upper x critical values at 98% confidence: Lower: Upper: c) Report your confidence interval for o: (
The high price of medicines is a source of major expense for those seniors in the United States who have to pay for these medicines themselves. A random sample of 1600 seniors who pay for their medicines showed that they spent an average of $4300 last year on medicines with a standard deviation of $500. Make a 90% confidence interval for the corresponding population mean. Round your answers to cents. i to $ i %24
The grade point averages (GPA) for 12 randomly selected college students are shown on the right. Complete parts (a) through (c) below. Assume the population is normally distributed. 2.4 3.4 2.7 1.8 0.9 4.0 2.2 1.1 3.6 0.4 2.1 3.1 (a) Find the sample mean. x=_____ (Round to two decimal places as needed.) (b) Find the sample standard deviation. s=_____ (Round to two decimal places as needed.) (c) Construct a 90% confidence interval for the population mean μ. A 90% confidence interval for the population mean is (__, __) (Round to two decimal places as needed.)
The grade point averages (GPA) for 12 randomly selected college students are shown on the right. Complete parts (a) through (c) 25 3.2 2.6 O below. 1.7 0.7 4.0 Assume the population is normally distributed. 2.3 1.3 3.8 0.4 2.3 3.3 (a) Find the sample mean. x- (Round to two decimal places as needed.)
You work for a soft-drink company in the quality control division. You are interested in the standard deviation of one of your production lines as a measure of consistency. The product is intended to have a mean of 12 ounces, and your team would like the standard deviation to be as low as possible. You gather a random sample of 16 containers. Estimate the population standard deviation at a 90% level of confidence. 11.98 12.11 11.88 11.99 11.89 12.02 11.84 12.05 12.13 11.94 12.31 12 12.07 12.02 12.04 12.04 (Data checksum: 192.31) Keep as many decimals as possible while making these calculations. a) Find the sample standard deviation: b) Find the lower and upper χ2χ2 critical values at 90% confidence:Lower: Upper: c) Report your confidence interval for σσ: (? , ? )
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You work for a soft-drink company in the quality control division. You are interested in the standard deviation of one of your production lines as a measure of consistency. The product is intended to have a mean of 12 ounces, and your team would like the standard deviation to be as low as possible. You gather a random sample of 17 containers. Estimate the population standard deviation at a 95% level of confidence. 11.94 11.99 11.97 11.95 11.85 12.1 12.07 12.16 12.13 12.1 12.01 12.1 11.95 12 11.93 11.96 12.1 (Data checksum: 204.31) Note: Keep as many decimals as possible while making these calculations. If possible, keep all answers exact by storing answers as variables on your calculator or computer. Round final answer to 4 decimal places. a) Find the sample standard deviation: b) Find the lower and upper x² critical values at 95% confidence: Lower: Upper: c) Report your confidence interval for o: (
The grade point averages (GPA) for 12 randomly selected college students are shown on the right. Complete parts (a) through (c) below. Assume the population is normally distributed. (a) Find the sample mean. x=(Round to two decimal places as needed.) Clear all 2.3 3.1 2.9 1.6 0.9 4.0 2.4 1.3 3.6 0.1 2.4 3.3 E € 413 PM
For the specified margin of error and confidence level, obtain a sample size that will ensure a margin of error of at most the one specified. margin of error=0.05;confidence level=95%
The grade point averages (GPA) for 12 randomly selected college students are shown on the right. Complete parts (a) through (c) below. 2.4 3.2 2.6 D 1.9 0.8 4.0 Assume the population is normally distributed. 2.2 1.5 3.6 0.5 2.3 3.4 (a) Find the sample mean. X = x= (Round to two decimal places as needed.)

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