Mathematics > EXAM > MATH 225N Week 6 Assignment Confidence Interval for Mean – Population Standard Deviation Known (Su (All)

MATH 225N Week 6 Assignment Confidence Interval for Mean – Population Standard Deviation Known (Summer 2020) Chamberlain.

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The population standard deviation for the number of corn kernels on an ear of corn is 94 kernels. If we want to be 90%confident that the sample mean is within 17 kernels of the true population mean,... what is the minimum sample size that should be taken? z0.101.282 z0.051.645 z0.0251.960. z0.012.326. z0.0052.576 Use the table above for the z-score. Correct! You nailed it. 21 83 329 699 Answer Explanation Correct answer: 83 The formula for sample size is n=z2σ2EBM2 In this formula, z=zα2=z0.05=1.645 because the confidence level is 90%. From the problem, we know that σ=94 and EBM=17. Therefore, n=z2σ2EBM2=(1.645)2(94)2172≈82.74 Use n=83 to ensure that the sample size is large enough. Find the sample size required to estimate a population mean with a given confidence level Question Suppose the scores of a standardized test are normally distributed. If the population standard deviation is 2 points, what minimum sample size is needed to be 90% confident that the sample mean is within 1 point of the true population mean? z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576 Use the table above for the z-score, and be sure to round up to the nearest integer. Yes that's right. Keep it up! 11 test scores Answer Explanation Correct answers:  11 test scores The formula for sample size is n=z2σ2EBM2 In this formula, z=zα2=z0.05=1.645 because the confidence level is 90%. From the problem, we know that σ=2 and EBM=1. Therefore, n=z2σ2EBM2=(1.645)2(2)212≈10.82 Use n=11 to ensure that the sample size is large enough. Also, the sample size formula shown above is sometimes written using an alternate format of n=(zσE)2. In this formula, E is used to denote margin of error and the entire parentheses is raised to the exponent 2. Therefore, the margin of error for the mean can be denoted by "EBM" or by "E". Either formula for the sample size can be used and these formulas are considered as equivalent. Calculate and interpret the confidence interval for a population mean with a known standard deviation Question The length, in words, of the essays written for a contest are normally distributed with a population standard deviation of 442 words and an unknown population mean. If a random sample of 24 essays is taken and results in a sample mean of 1330 words, find a 99% confidence interval for the population mean. z0.10 z0.05 z0.025 z0.01 z0.005 1.28 2 1.64 5 1.96 0 2.32 6 2.57 6 You may use a calculator or the common z values above. Round the final answer to two decimal places. Perfect. Your hard work is paying off � (1097.59, 1562.41) Answer Explanation Correct answers:  (1097.59, 1562.41) Confidence intervals are written as (pointestimate−marginof error,pointestimate+marginof error) The point estimate is the sample mean, x¯, and the margin of error is marginof error=(zα2)(σn‾√) Substituting the given values σ=442, n=24, and zα2=2.576 for a confidence level of 99%, we have marginof error=(2.576)(44224‾‾‾√)≈(2.576)(90.223)≈232.41 With x¯=1330 and a margin of error of 232.41, the confidence interval is (1330−232.41,1330+232.41)(1097.59,1562.41). So we estimate with 99% confidence that the true population mean is between 1097.59 and 1562.41 words. Calculate and interpret the confidence interval for a population mean with a known standard deviation Question The lengths, in inches, of adult corn snakes are normally distributed with a population standard deviation of 8 inches and an unknown population mean. A random sample of 25 snakes is taken and results in a sample mean of 58 inches. Identify the parameters needed to calculate a confidence interval at the 99% confidence level. Then find the confidence interval. z0.10z0. 10 z0.05z0. 05 z0.025z0. 025 z0.01z0. 01 z0.005z0. 005 1.282 1.645 1.960 2.326 2.576 You may use a calculator or the common z values above. Round the final answer to two decimal places. Correct answers:  158  28  325  42.576  553.88  662.12 Confidence intervals are written as (point estimate−margin of error, point estimate+margin of error) The point estimate is the sample mean, x¯, and the margin of error is margino f error=(zα2)(σn‾√) Substituting the given values σ=8, n=25, and zα2=2.576 for a confidence level of 99%, we have margin of error=(2.576)(825‾‾‾√)≈(2.576)(1.6)≈4.12 With x¯=58 and a margin of error of 4.12, the confidence interval is (58−4.12,58+ [Show More]

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