The data shows the weights of samples of two different sodas. Test the claim that the weights of the two sodas have the same standard deviation using a "count five" test. x 0.7837 0.8111 0.7858 0.8123 0.8157 0.8149 0.7914 0.7893 0.8056 0.7805 y 0.8494 0.8029 0.7868 0.8453 0.8141 0.8156 0.8244 0.7854 0.7821 0.8493 a. For the first sample, find the mean absolute deviation (MAD) of each value. Do the same for the second sample. Find the MAD of each x value. Recall that the MAD of the sample value is x−x. x 0.7837 0.8111 0.7858 0.8123 0.8157 0.8149 0.7914 0.7893 0.8056 0.7805 MAD nothing nothing nothing nothing nothing nothing nothing nothing nothing nothing (Round to four decimal places as needed.) Find the MAD of each y value. y 0.8494 0.8029 0.7868 0.8453 0.8141 0.8156 0.8244 0.7854 0.7821 0.8493 MAD nothing nothing nothing nothing nothing nothing nothing nothing nothing nothing (Round to four decimal places as needed.) b. Let c1 be the number of MAD values in the first sample that are greater than the largest MAD value in the second sample. Also, let c2 be the number of MAD values in the second sample that are greater than the largest MAD value in the first sample. c1=nothing c2=nothing c. If the sample sizes are equal, use a critical value of 5. If not, use the formula below to find the critical value. log(α/2)logn1n1+n2 The critical value is nothing. (Type an integer or decimal rounded to one decimal place as needed.) d. If c1 is greater than or equal to the critical value, then conclude that σ21>σ22. If c2 is greater than or equal to the critical value, then conclude that σ22>σ21. Otherwise, fail to reject the null hypothesis that σ21=σ22. Which conclusion can be drawn? Choose the correct answer below. A. reject σ21=σ22 in favor of σ21>σ22 B. fail to reject σ21=σ22 C. reject σ21=σ22 in favor of σ22>σ21

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 data shows the weights of samples of two different sodas. Test the claim that the weights of the two sodas have the same standard deviation using a "count five" test.

x	0.7837	0.8111	0.7858	0.8123	0.8157	0.8149	0.7914	0.7893	0.8056	0.7805
y	0.8494	0.8029	0.7868	0.8453	0.8141	0.8156	0.8244	0.7854	0.7821	0.8493

a. For the first sample, find the mean absolute deviation (MAD) of each value. Do the same for the second sample.

Find the MAD of each x value. Recall that the MAD of the sample value is

x−x.

x	0.7837	0.8111	0.7858	0.8123	0.8157	0.8149	0.7914	0.7893	0.8056	0.7805
MAD	nothing	nothing	nothing	nothing	nothing	nothing	nothing	nothing	nothing	nothing

(Round to four decimal places as needed.)

Find the MAD of each y value.

y	0.8494	0.8029	0.7868	0.8453	0.8141	0.8156	0.8244	0.7854	0.7821	0.8493
MAD	nothing	nothing	nothing	nothing	nothing	nothing	nothing	nothing	nothing	nothing

(Round to four decimal places as needed.)

b. Let

be the number of MAD values in the first sample that are greater than the largest MAD value in the second sample. Also, let

be the number of MAD values in the second sample that are greater than the largest MAD value in the first sample.

c1=nothing

c2=nothing

c. If the sample sizes are equal, use a critical value of 5. If not, use the formula below to find the critical value.

log(α/2)logn1n1+n2

The critical value is

nothing.

(Type an integer or decimal rounded to one decimal place as needed.)

d. If

is greater than or equal to the critical value, then conclude that

σ21>σ22.

is greater than or equal to the critical value, then conclude that

σ22>σ21.

Otherwise, fail to reject the null hypothesis that

σ21=σ22.

Which conclusion can be drawn? Choose the correct answer below.

reject σ21=σ22 in favor of σ21>σ22

fail to reject σ21=σ22

reject σ21=σ22 in favor of σ22>σ21

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