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any value. 4. Find the 2 values for the following situations: (a) Confidence level 94% %3D (b) Confidence level = 79% (c) H= 100, o = 15, n = 64, confidence = 93.5% %3D %3D %3D (d) = 13, o = 2, n 1364, confidence = 86% %3D %3D

any value. 4. Find the 2 values for the following situations: (a) Confidence level 94% %3D (b) Confidence level = 79% (c) H= 100, o = 15, n = 64, confidence = 93.5% %3D %3D %3D (d) = 13, o = 2, n 1364, confidence = 86% %3D %3D

MATLAB: An Introduction with Applications
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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i, then explore how we might use this concept to determine if a data value falls into an ave: interval tells you about the valu range. 3. Give an explanation in your own words why we need to use technology like RStudio to find th. values for percentages like 88%, 79% or 92.57%. Contrast this with the values of which can found using our chart for nearly any value. 4. Find the values for the following situations: (a) Confidence level = 94% (b) Confidence level = 79% (c) u= 100, o = 15, n = 64, confidence = 93.5% (d) u = 13, o = 2, n = 1364, confidence = 86% 5. Find the values for the following situations: (a) n = 101 93% confidence (b) n = 21, confidence = 91% (c) n = 101, confidence = 82% (d) D.F. = 70 confidence = 98% (e) D.F. = 17 confidence = 77% 6. Find the margin of error for the following situations: Recall, any confidence interval (z, t, or p) can be written as ī - E<uSī+E Also giving E RightSide-LeftSide (a) za = 1.65 o = 1.38 n = 49 (b) = 2.014 s = 2.21 n = 41 (c) 11.7 < µ < 21.6 (d) 32 < µ < 38.1 (e) = 1.96 P = .84 n = 49 7. A random sample of 500 incoming students from University of Tennessee found that 35 freshmen. If their average GPA was 3.43 with a sample standard deviation of 1.81, find confidence interval for the percent of incoming students who are freshman at the Unive Tennessee. Determine if the information can be used to create a confidence interval. If r a detailed explanation as to why this information does not fit the definition.
Transcribed Image Text:i, then explore how we might use this concept to determine if a data value falls into an ave: interval tells you about the valu range. 3. Give an explanation in your own words why we need to use technology like RStudio to find th. values for percentages like 88%, 79% or 92.57%. Contrast this with the values of which can found using our chart for nearly any value. 4. Find the values for the following situations: (a) Confidence level = 94% (b) Confidence level = 79% (c) u= 100, o = 15, n = 64, confidence = 93.5% (d) u = 13, o = 2, n = 1364, confidence = 86% 5. Find the values for the following situations: (a) n = 101 93% confidence (b) n = 21, confidence = 91% (c) n = 101, confidence = 82% (d) D.F. = 70 confidence = 98% (e) D.F. = 17 confidence = 77% 6. Find the margin of error for the following situations: Recall, any confidence interval (z, t, or p) can be written as ī - E<uSī+E Also giving E RightSide-LeftSide (a) za = 1.65 o = 1.38 n = 49 (b) = 2.014 s = 2.21 n = 41 (c) 11.7 < µ < 21.6 (d) 32 < µ < 38.1 (e) = 1.96 P = .84 n = 49 7. A random sample of 500 incoming students from University of Tennessee found that 35 freshmen. If their average GPA was 3.43 with a sample standard deviation of 1.81, find confidence interval for the percent of incoming students who are freshman at the Unive Tennessee. Determine if the information can be used to create a confidence interval. If r a detailed explanation as to why this information does not fit the definition.
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