We appreciate your visit to Jack is a college athlete who has been weighing himself weekly on the same scale in the athletic center for the past few years Jack. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Final answer:
The standard error of the mean (SE) for Jack's weight is 1.333 lbs. The margin of error (m) for the 95% confidence interval is 2.61 lbs.
Explanation:
To calculate the standard error of the mean (SE), we divide the standard deviation (σ) by the square root of the sample size (n). In this case, the standard deviation is 4 lbs and the sample size is 9, so the SE = 4 / √9 = 4 / 3 = 1.333 lbs.
The margin of error (m) for a 95% confidence interval is calculated by multiplying the critical value (z*) by the standard error (SE). The critical value for a 95% confidence interval is approximately 1.96. Therefore, the margin of error (m) = 1.96 * 1.333 = 2.61 lbs.
Learn more about Standard Error of the Mean here:
https://brainly.com/question/14524236
#SPJ11
Thanks for taking the time to read Jack is a college athlete who has been weighing himself weekly on the same scale in the athletic center for the past few years Jack. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
