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Example problem Degrees of freedom is n-1. Can you please explain to me why we use df=35 instead? I attached the question and my professor's solutions below. I need an explanation for that part please, thanks.

Example problem Degrees of freedom is n-1. Can you please explain to me why we use df=35 instead? I attached the question and my professor's solutions below. I need an explanation for that part please, thanks.

MATLAB: An Introduction with Applications
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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Degrees of freedom is n-1. Can you please explain to me why we use df=35 instead?

I attached the question and my professor's solutions below. I need an explanation for that part please, thanks.

Example 7.1.2 A speed reading course claims that at the conclusion of the course, participants will be able to read and comprehend a 1000 word passage in less than a minute. company randomly selects 40 of its employees and sends them to this course. At the conclusion, their average time to read the passage was 55 seconds with standard deviation 10 seconds. a) Does this indicate that the claim made by the course is true? Use a = .005. b) Find c for the corresponding CI. C c) Find the CI. d) Do the CI and hypothesis test give the same conclusion? Explain.
Transcribed Image Text:Example 7.1.2 A speed reading course claims that at the conclusion of the course, participants will be able to read and comprehend a 1000 word passage in less than a minute. company randomly selects 40 of its employees and sends them to this course. At the conclusion, their average time to read the passage was 55 seconds with standard deviation 10 seconds. a) Does this indicate that the claim made by the course is true? Use a = .005. b) Find c for the corresponding CI. C c) Find the CI. d) Do the CI and hypothesis test give the same conclusion? Explain.
•n=40 X = 55 a) Assumptions... SRSV normal Ho μ=60 H₁: μ260 5=10 fi CH pop - no, but since n = 40, it should be robust 1005 as df = 39 +2.725 Use df = 35 55-60 t = 10/140 = -3,162 p-value = P(t <-3.162): p-value <,005 -3.162 Since t= -3.162 is in the RR and the p-value <.005 = α, reject th. Conclude that the mean time to read the passage is 2 60 sec. =२ The claim appears to be true, 6) 1-sided test =) c=1-22=1-21.005) = +99 c) 99% CI: 55 ± 2.725(10/140) = 55 ±4.3) = (50.69, 59.31) We are 99% confident that the true mean time it takes participants of this course to read the passages between 50.69 sec and 59.31 se because 99% of possible sample's taken would produce an interval that contains d) Yes. Since M₂ = 60 is not in the CI, it rejects H₂: μ = 60. Since the Мо entire CI is 460, it indicates HSG00
Transcribed Image Text:•n=40 X = 55 a) Assumptions... SRSV normal Ho μ=60 H₁: μ260 5=10 fi CH pop - no, but since n = 40, it should be robust 1005 as df = 39 +2.725 Use df = 35 55-60 t = 10/140 = -3,162 p-value = P(t <-3.162): p-value <,005 -3.162 Since t= -3.162 is in the RR and the p-value <.005 = α, reject th. Conclude that the mean time to read the passage is 2 60 sec. =२ The claim appears to be true, 6) 1-sided test =) c=1-22=1-21.005) = +99 c) 99% CI: 55 ± 2.725(10/140) = 55 ±4.3) = (50.69, 59.31) We are 99% confident that the true mean time it takes participants of this course to read the passages between 50.69 sec and 59.31 se because 99% of possible sample's taken would produce an interval that contains d) Yes. Since M₂ = 60 is not in the CI, it rejects H₂: μ = 60. Since the Мо entire CI is 460, it indicates HSG00
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