STATISTICS.docx - Google Docs+A docs.google.com/document/d/1Xgqq34HC4kub_OY7HprnxliVKF2wTIZI/edit3. A study of 25 residents showed that they had lived at their homes for 9.3 years. The standarddeviation was 2 years. Find the 90% confidence interval (hint: the Tdistribution applies)3:24 AMP Type here to searchW36°F Mostly clear12/15/2021

We want to find 90% confidence interval

Answered: 3. A study of 25 residents showed that…

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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A medical doctor who wanted to study the heights of adult men and women took a random sample of 150 adult men and 120 adult women and found the following summary results. Mean Height Standard Deviation Adult Men 170cm 8cm Adult Women 164cm 6cm Find a 90% confidence interval for the difference of mean height of adult men and women.
In monitoring lead in the air after the explosion at the battery factory, it is found that the amounts of lead over a 8 day period had a standard error of 1.88. Find the margin of error that corresponds to a 95% confidence interval. (Round to 2 decimal places)
A 90 percent confidence interval is to be created to estimate the proportion of television viewers in a certain areawho favor moving the broadcast of the late weeknight news to an hour earlier than it is currently. Initially, theconfidence interval will be created using a simple random sample of 9,000 viewers in the area. Assuming that thesample proportion does not change, what would be the relationship between the width of the original confidenceinterval and the width of a second 90 percent confidence interval that is created based on a sample of only 1,000viewers in the area?(A) The second confidence interval would be 9 times as wide as the original confidence interval.(B) The second confidence interval would be 3 times as wide as the original confidence interval.(C) The width of the second confidence interval would be equal to the width of the original confidence interval.(D) The second confidence interval would be 1/3 as wide as the original confidence interval.(E) The second…
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A 30 month study is conducted to determine the difference in rates of accidents per month between two departments in an assembly plant. If a sample of 12 from the first department averaged 12.3 accidents per month with a standard deviation of 3.5, and a sample of 9 from the second department averaged 7.6 accidents per month with standard deviation of 3.4, find: The t-value for a 95% confidence interval is: A. 2.09302 B. 1.724718 C. 2.08596 D. 1.729133 E. none of the preceding
Use the data file A2Q2.xlsx to answer this question. A sample containing years tomaturity and yield for 40 corporate bonds is contained in the file A2Q2.xlsx.a. What is the sample mean years to maturity for corporate bonds and what is thesample standard deviation?b. Develop a 95% confidence interval for the population mean years to maturity.c. What is the sample mean yield on corporate bonds and what is the sample standard deviation?d. Develop a 95% confidence interval for the population mean yield on corporatebonds. Company Ticker Years to Maturity Yield HSBC 12.00 4.079 GS 9.75 5.367 C 4.75 3.332 MS 9.25 5.798 C 9.75 4.814 TOTAL 5.00 2.069 MS 5.00 4.739 WFC 10.00 3.682 TOTAL 10.00 3.270 TOTAL 3.25 1.748 BAC 9.75 4.009 RABOBK 9.75 3.203 GS 9.25 5.755 AXP 5.00 2.181 MTNA 5.00 4.366 MTNA 10.00 6.046 JPM 8.25 2.310 GE 28.00 5.030 LNC 10.00 4.163 BAC 8.00 3.699 FCX 15.00 4.030 GS 25.50 6.013 RABOBK 4.75 2.755 GE 24.75 5.008 HCN 8.00…

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3. A study of 25 residents showed that they had lived at their homes for 9.3 years. The standard deviation was 2 years. Find the 90% confidence interval (hint: the Tdistribution applies)