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
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
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.3: Measures Of Spread
Problem 26PFA
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Question
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.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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