**Question:**A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test.What is the 99% confidence interval for the mean score, µ, of all students taking the test? *Round to one decimal place.***Answer:**[ ] < µ < [ ]**Explanation:**To solve for the 99% confidence interval, use the following steps:1. Determine the sample mean (x̄), which is 29.5.2. Find the standard deviation (σ), which is 5.2.3. Determine the sample size (n), which is 59.4. Find the critical value (z) for a 99% confidence interval. For the 99% confidence level, the z-score is approximately 2.576.5. Calculate the standard error (SE) of the mean using the formula: SE = σ / √n SE = 5.2 / √59 ≈ 0.67626. Calculate the margin of error (ME) using the formula: ME = z * SE ME ≈ 2.576 * 0.6762 ≈ 1.74217. Construct the confidence interval using the formula: CI = x̄ ± ME CI = 29.5 ± 1.7421Therefore, the 99% confidence interval for the mean score is approximately:27.8 < µ < 31.3

Given information- Sample size (n) = 59 Mean x̅ = 29.5 Standard deviation (s) = 5.2 Significance…

Answered: A group of 59 randomly selected…

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 group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. What is the 99% confidence interval for the mean score,, of all students taking the test? Round to one decimal place. <μ<