**Chi-Square Distribution and Confidence Interval Estimation in Statistics**In this example, we are tasked with finding the number of degrees of freedom, the critical chi-square values \( \chi^2_L \) and \( \chi^2_R \), and constructing the confidence interval estimate of \( \sigma \). We start by assuming that a simple random sample has been drawn from a normally distributed population.**Parameters:**- Confidence Level: 98%- Sample size (\( n \)): 25- Sample standard deviation (\( s \)): 0.28 mg**Procedure:**1. **Degrees of Freedom (\( df \)):** The degrees of freedom for a chi-square distribution in this context is \( n - 1 \). Therefore, with a sample size of 25, the degrees of freedom is \( 25 - 1 = 24 \).2. **Critical Values:** To determine the critical values \( \chi^2_L \) and \( \chi^2_R \), you would typically refer to a chi-square distribution table. These values are needed to construct the confidence interval for the population standard deviation.3. **Confidence Interval Estimation:** - The estimation process involves using the chi-square critical values and the sample standard deviation to calculate the range in which the true population standard deviation is expected to lie, with a 98% confidence level.**Interactive Element:**- **Chi-Square Critical Values Table:** An interactive element (icon) is included to view or calculate the critical values from a chi-square distribution table, which are needed to accurately determine the confidence interval for the population standard deviation.This practical example helps apply statistical concepts to real-world data, enhancing understanding of interval estimates and the use of chi-square distributions in hypothesis testing.

Answered: Image /qna-images/answer/3045ca31-161d-4839-8f7e-9f406b627e60.jpg

Use the given information to find the number of degrees of freedom, the critical values x? and x, and the confidence interval estimate of o. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Nicotine in menthol cigarettes 98% confidence; n= 25, s = 0.28 mg.

Use the given information to find the number of degrees of freedom, the critical values x? and x, and the confidence interval estimate of o. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Nicotine in menthol cigarettes 98% confidence; n= 25, s = 0.28 mg.

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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