We found x0.025 = 32.852 and x0.975 = 8.907. Recall the sample standard deviation s = 4, and sample size n = 20. Substitute these values to first find the lower bound for the confidence interval of the population variance, rounding the result to two decimal places. (n – 1)s2 2 X'al2 lower bound - (20 – 1)42 Now substitute these values to find the upper bound for the confidence interval of the population variance, rounding the result to two decimal places. (n – 1)s? upper bound - I X1- a/2 (20 – 1)42
We found x0.025 = 32.852 and x0.975 = 8.907. Recall the sample standard deviation s = 4, and sample size n = 20. Substitute these values to first find the lower bound for the confidence interval of the population variance, rounding the result to two decimal places. (n – 1)s2 2 X'al2 lower bound - (20 – 1)42 Now substitute these values to find the upper bound for the confidence interval of the population variance, rounding the result to two decimal places. (n – 1)s? upper bound - I X1- a/2 (20 – 1)42
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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Question
![Step 6
We found xo.025 = 32.852 and x6.975 = 8.907. Recall the sample standard deviation s = 4, and sample size
n = 20.
Substitute these values to first find the lower bound for the confidence interval of the population variance,
rounding the result to two decimal places.
(n – 1)s?
lower bound =
(20 – 1)42
Now substitute these values to find the upper bound for the confidence interval of the population variance,
rounding the result to two decimal places.
(n – 1)s?
upper bound =
X1- a/2
(20 - 1)42](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc754fcbc-47de-4905-bc20-92b693330107%2Ffa1cc51c-14b3-4287-a683-88b2241835f6%2F60ainqg_processed.png&w=3840&q=75)
Transcribed Image Text:Step 6
We found xo.025 = 32.852 and x6.975 = 8.907. Recall the sample standard deviation s = 4, and sample size
n = 20.
Substitute these values to first find the lower bound for the confidence interval of the population variance,
rounding the result to two decimal places.
(n – 1)s?
lower bound =
(20 – 1)42
Now substitute these values to find the upper bound for the confidence interval of the population variance,
rounding the result to two decimal places.
(n – 1)s?
upper bound =
X1- a/2
(20 - 1)42
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