**Chi-Square Distribution in Estimating Confidence Intervals****Objective:** Use the given information to find the number of degrees of freedom, the critical values $ \chi^2_L $ and $ \chi^2_R $, and the confidence interval estimate of $ \sigma $. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution.**Given:** - Sample context: Nicotine in menthol cigarettes- Confidence level: 98%- Sample size ($ n $): 21- Sample standard deviation ($ s $): 0.26 mg**Instructions:** 1. Click the icon to view the table of Chi-Square critical values.2. Calculate the degrees of freedom: $ df = n - 1 $.---**Input Field:** - $ df = $ [Type a whole number.]**Explanation:** To estimate the population variance and standard deviation, find the degrees of freedom and the corresponding critical values from the Chi-Square distribution table for the given confidence level. This will allow us to calculate the confidence interval for the population standard deviation $ \sigma $. **Note:** The critical values $ \chi^2_L $ and $ \chi^2_R $ are essential in determining the bounds of the confidence interval.

We have given that Sample size n=21 , s= 0.26 , We find 98% confidence interval

Answered: Jse the given information to find the…

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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The “Descriptive Statistics” function in Excel … calculates the confidence level using both the t and the z distributions calculates the confidence level using the z distributions calculates the confidence level using the t distributions calculates the confidence level using the chi-square distributions
Use the sample information x=36, o=7, n=17 calculate the following confidence intervals for u assuming the sample is from a normal population 90 percent confidence b.95 percent confidence
Construct a 90% confidence interval to estimate the population mean using the data below. What assumptions need to be made about this population? x=38 s=9.9 n=24 Click here to view page 1 of the critical t-score table. Click here to view page 2 of the critical t-score table. Question content area bottom Part 1 The 90%confidence interval for the population mean is from a lower limit of here to an upper limit of (Round to two decimal places as needed.) Part 2 What assumptions need to be made about this population? A. The population distribution is skewed to one side. B. The population size is larger than 30. C. The population follows the Student's t-distribution. D. The population follows the normal probability distribution.
Use the given information to find the number of degrees of freedom, the critical values x and x, and the confidence interval estimate of a. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Nicotine in menthol cigarettes 80% confidence; n= 26, s=0.23 mg. E Click the icon to view the table of Chi-Square critical values. df =(Type a whole number.) Enter your answer in the answer box and then click Check Answer. 3 parts remaining Clear All Check Anwer MacBook Air 44 4) esc F5 F7 F6 19 10 F2 F3 %23 2$ % & * delete 2 4 5 6. 7 8 Q T Y tab A S F G K caps lock
Planning engineers want to construct a confidence interval for the standard deviation of the runtime of a certain manufacturing process (0), assuming that the runtimes will be approximately Normal. They make 25 random observations and find that, on average, the process takes 290 seconds, with a sample standard deviation of 35 seconds. Click here to view page 1 of the table of critical values of the chi-squared distribution. Click here to view page 2 of the table of critical values of the chi-squared distribution. With 90%, the engineers can conclude that o will be between and| (Round to two decimal places including any zeros.)
Traumatic Brain Injuries have been shown to affect brain volume. A psychologist measures the brain volume of N=8 TBI patients at 2 months and 12 months after their injuries. Evaluate the results of her data using α = .052tail. **Attached Image** a. Show how the 95% confidence interval was calculated. b. State the conclusion in APA format.
Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value t/2, (b) find the critical value z,/2, or (c) state that neither the normal distribution nor the t distribution applies. 230, x 32.6 hg, s = 7.9 hg. The confidence level is 90% Here are summary statistics for randomly selected weights of newborn girls: n Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. (Round to two decimal places as needed.) В. Za/2 (Round to two decimal places as needed.) C. Neither the normal distribution nor the t distribution applies.
n1=16 x‾1=124 s1=35 n2=12 x‾2=86 s2=31 Use this data to find the 98% confidence interval for the true difference between the population means. Assume that the population variances are not equal and that the two populations are normally distributed. Copy Data Step 2 of 3 : Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places. Step 3 of 3 : Please also find the high and low endpoints and round to four decimal points.
INFERENCE FOR SINGLE POPULATION PROPORTION (CONFIDENCE INTERVAL) 11. A random sample of 400 electronic components manufactured by a certain process is tested, and 30 are found to be defective. i. Let p be the proportion of defective components. Find a 95% confidence interval for p. ii. How many components must be sampled so that the 95% confidence interval will specify the proportion of defective components to within ±0.02? Source: Statistics for Engineers and Scientists, 3rd Edition, William, C. Navidi.
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The data below are the ages of a random sample of 8 men in a bar. It is known that the population of ages are normally distributed with a standard deviation of 7. Determine the 90% confidence interval estimate of the population mean. 49, 68, 25, 36, 56, 38, 27, 47 Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. Confidence Interval =
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