• Question 21>The table below gives results from random samples of weights (in pounds) of men and women.PopGender1Men379.5Women3910.9At a = 0.10, test the claim that weights of men are more consistent than weights of women.1. The alternate hypothesis is Ha:O s1 > 82Ο σι > σ2O s1 < 82Ο σι < σ2. This is a left tailed test with:cordfn =and dfa =3a. The STS (to 2 decimals) is F =3b. The P-value (to 3 decimals) is:4a. At a = 0.10, the decision is:O Accept HoO Do not reject HoO Accept HaO Reject HoMacBook AirF3F44>F5F6F7F8F9%2$4.%23*RY965 4b. The conclusion is:O There is sufficient evidence to support the claim that weights of men are more consistent thanweights of womenO There is insufficient evidence to support the claim that weights of men are more consistent thanweights of womenO There is sufficient evidence to reject the claim that weights of men are more consistent thanweights of womenO There is insufficient evidence to reject the claim that weights of men are more consistent thanweights of womenSubmit QuestionMacBook Air吕0F2F311F4F5F6F7F8F9

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Description: • Question 21>The table below gives results from random samples of weights (in pounds) of men and women.PopGender1Men379.5Women3910.9At a = 0.10, test the claim that weights of men are more consistent than weights of women.1. The alternate hypothesi

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The table below gives results from random samples of weights (in pounds) of men and women. Pop Gender Men 37 9.5 Women 39 10.9 At a = 0.10, test the claim that weights of men are more consistent than weights of women. 1. The alternate hypothesis is Ha: O s1 > 82 Ο σι > σ2 O s1 < 82 Ο σι σ2 2. This is a left tailed test with: dfn= and dfa = 3a. The STS (to 2 decimals) is F %3D 3b. The P-value (to 3 decimals) is: 4a. At a = 0.10, the decision is: O Accept Ho Do not reject Ho Accept Ha Reject Ho

The table below gives results from random samples of weights (in pounds) of men and women. Pop Gender Men 37 9.5 Women 39 10.9 At a = 0.10, test the claim that weights of men are more consistent than weights of women. 1. The alternate hypothesis is Ha: O s1 > 82 Ο σι > σ2 O s1 < 82 Ο σι σ2 2. This is a left tailed test with: dfn= and dfa = 3a. The STS (to 2 decimals) is F %3D 3b. The P-value (to 3 decimals) is: 4a. At a = 0.10, the decision is: O Accept Ho Do not reject Ho Accept Ha Reject Ho

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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