Table of t-Distribution Areasdf1 1.0002345670.2540.20Area inright tail1.3760.816 1.0610.765 0.9780.7410.9410.727 0.9200.718 0.906071108060.151.9631.3861.2501.1901.1561.13411100.100.053.0781.8861.638 2.3531.5332.1321.476 2.0151.4401415Table VI1.9431.805t-DistributionArea in Right Tail0.0250.026.314 12.7062.920 4.303 4.8493.1822.7762.5712.4472.3650.010.00515.894 31.821 63.6576.965 9.9253.482 4.541 5.8412.9993.747 4.6042.7573.3654.0322.612251743.1432.0083.7072.4000.00250.001127.321 318.30914.08922.3277.45310.2155.5987.1734.7735.8934.31740205.20847850.0005636.61931.59912.9248.6106.8695.9595408 A simple random sample of size n is drawn. The sample mean, x, is found to be 17.9, and the sample standard deviation, s, is found to be 4.4.Click the icon to view the table of areas under the t-distribution.(a) Construct a 95% confidence interval about u if the sample size, n, is 35.Lower bound:; Upper bound(Use ascending order. Round to two decimal places as needed.)

Given, Sample mean, = 17.9Sample standard deviation, s = 4.4(a) Confidence level, C = 95% =…

Answered: A simple random sample of size n is…

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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Cell Phone Lifetimes A recent study of the lifetimes of cell phones found the average is 24 months. The standard deviation is 4.2 months. If a company provides its 33 employees with a cell phone, find the probability that the mean lifetime of these phones will be less than 23.8 months. Assume cell phone life is a normally distributed variable, the sample is taken from a large population, and the correction factor can be ignored. Use a TI-83 Plus/TI-84 Plus calculator. Round your answer to at least four decimal places. P(X< 23.8)=
Tina catches a 14-pound bass. She does not know the population mean or standard deviation. So she takes a sample of five friends and they say the last bass they caught was 9, 12, 13, 10, and 10 pounds. Find the t and calculate a 95% (α = .05) confidence interval.
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The distribution of study times for first year students follows a Normal distribution with standard deviation σσ = 54 minutes. Suppose a sample of 487 students found that the mean study time for these students was x¯x¯= 173 minutes per week. Use this to find a 88.28% confidence interval to estimate the mean study time for all students. We are 88.28% confident that the mean study time for first year students is between minutes_____ and ____ minutes. What is the margin of error for your confidence interval?
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