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In Exercise 8.2-10, growth data are given for plants in

Probability and Statistical Inference | 9th Edition | ISBN: 9780321923271 | Authors: Robert V. Hogg, Elliot Tanis, Dale Zimmerman ISBN: 9780321923271 41

Solution for problem 18E Chapter 8.4

Probability and Statistical Inference | 9th Edition

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Probability and Statistical Inference | 9th Edition | ISBN: 9780321923271 | Authors: Robert V. Hogg, Elliot Tanis, Dale Zimmerman

Probability and Statistical Inference | 9th Edition

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Problem 18E

In Exercise 8.2-10, growth data are given for plants in normal air and for plants in CO2-enriched air. Those data are repeated here:

In this exercise, we shall test the null hypothesis that the medians are equal, namely, H0: mX = mY, against the alternative hypothesis H1: mX < mY. You may select the significance level. However, give the approximate p-value or state clearly why you arrived at a particular conclusion, for each of the tests. Show your work.

(a) What is your conclusion from the Wilcoxon test?

(b) What was your conclusion from the t test in Exercise 8.2-10?

(c) Write a comparison of these two tests.

Step-by-Step Solution:
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Koshar Amy Brogan February 22, 24, & 26, 2016 Week 7 Fair Division Dividing Up “Stuff” Equally  Equitable: both players believe that s/he received the same factorial part of the total value  Envy-free: neither player would be happier with what the other received  Pareto-optimal: no other allocation, arrived at by any means, can make one party better off without making the other person worse off o Important to property economists Adjusted Winner Procedure Points system where both parties go into separate rooms and give points to items they want more than others. Then the third-person party uses this system to decide who gets what. Example 1: Heather vs. Sandy: Business Split Heather Sandy Property 40 25 Equipment 10 5 Artwork 25 50 Website 25 20 1. Look at each item and see who has the most points for it. Heather Sandy Property 40 * 25 Equipment 10 * 5 Artwork 25 50 * Website 25 * 20 2. Add up the total points for “awarded” items Heather Sandy 40 10 50 25 75 50 3. If the points are not even, look to see what item can switch sides based on point value and equal want Property: 40/25 > 1.25 Equipment: 10/5 = 2 Website: 25/20 = 1.25 * closest to 1 In this case, just moving the item from side to the other will make them uneven again, and no progress can be made. 4. Figuring out the portion of divide. P1 – v1x = P2 + v2x (Starting points) – (value 1 of switching item) (x = some portion) = (Starting points) + (value 2 of item)x 75-25x = 50+20x the values of the switching item is based on the individual value of the object 25 = 45x x = 5/9 The portion of the item (the website) moving towards the second party 5. Who gets what: Heather: property, equipment, 4/9 of website Sandy: artwork, 5/9 of website Disadvantages of this system:  Possible an item can’t be split  Difficulty gauging how much points to give items for the parties Example 2: Equal Valued Items What do we do when the parties have given the same amount of points to the same item The same process as before actually. 1. Look at each item and see who has the most points for it. 2. Add up the total points for “awarded” items 3. If the points are not even, look to see what item can switch sides based on point value and equal want 4. Figuring out the portion of divide. 5. Who gets what: Harry Lloyd Dog Van 10 10 Parrot 10 13 Scooter 10 12 Samsonite Case 20 20 Lamborghini 50 45 1. Harry Lloyd Dog Van 10 10 Parrot 10 13 * Scooter 10 12 * Samsonite Case 20 20 Lamborghini 50 * 45 2. Harry Lloyd 10 13 12 20 50 50 25 In this case, we have an extra half-step. For the items that were awarded the same points we are going to give them to one of the two parties. Let’s give them to Lloyd to make it closer to even. Harry Lloyd 10 13 12 20 50 50 55 3. We can’t really make ratios for this problem since the items are weighted the same, so we can start moving over items that have the least amount of points. Here, the Dog Van. 4. Our equation will read: 50 + 10x = 55 – 10x 20x = 5 x = 5/20 = ¼ Harry will be gaining ¼ of the Dog Van, and Lloyd will be keeping ¾ of the Dog Van. 5. Harry: Lamborghini, ¼ of the Dog Van Lloyd: Parrot, Scooter, Samsonite case, and ¾ of the Dog Van Knaster Inheritance Procedure What if there are more than two parties And what if there is only one item that can’t be split and nobody wants to sell it and split the money There are similar steps as before, but without ratios and an equation. Again, away from each other, the parties value the item as to how much they would pay for it. Then the person who bids the most begins the act of buying the others’ shares. Example 1: Dan, Ed, and Bill inheriting their parents’ house. All interior contents are already taken care of. 1. How much is each party willing to pay for the item (the house) Dan Ed Bill $120,000 $105,000 $81,000 2. Who bid the highest Dan did, so at this step he has the house, but what about the other two 3. Buying the other shares: 3.A: split each of the bids by the number of parties. Dan Ed Bill $120,000 $105,000 $81,000 Divide by 3 Divide by 3 Divide by 3 40,000 35,000 27,000 3.B: put (n-1/n) shares of the highest bidder into the “Kitty” 2/3 in this case, two shares out of three people 40,000 + 40,000 = 80,000  use this to pay for the other shares 3.C 80,000 - 35,000 - 27,000 18,000 3.D: Split the remaining amount among the parties 18,000 / 3 = 6,000  each party, including Dan, gets 6,000 more. 4. Who gets what: Dan: house, pays $74,000  (80,000 – 6,000) Ed: 35,000 + 6,000 = $41,000 Bill: 27,000 + 6,000 = $33,000 Check: 41,000 + 33,000 = $74,000 = Dan’s pay-out Perks  If everyone gives an honest bid, everyone will end up better off. Drawbacks  Income discrepancies: one party may make more income than the others and thus be able to bid more, cutting out any chance the others have at the item Example 2: Thomas, Nikola, Benjamin, and Alexander look to inherit land in southern Ohio 1. Bids Thomas Nikola Benjamin Alexander $44,000 $50,000 $22,000 $31,000 2. Nikola bid the highest 3. Shares of each (dividing by 4 in this example) Thomas Nikola Benjamin Alexander $44,000 $50,000 $22,000 $31,000 Divide by 4 11,000 12,500 5,500 7,750 Kitty = ¾ = 12,500 + 12,500 + 12,500 = 37,500 37,500 - 11,000 - 5,000 - 7,750 13,250  divide by 4  each gets 3,312.50 4. Who gets what Thomas Nikola Benjamin Alexander 11,000 + 3,312.50 50,000 – 3,312.50 5,500 + 3,312.50 7,750 + 3,312.50 $14,312.50 Land, pays $34,187.50 $8,812.50 $11,062.50 Check: 14,312.50 + 8,812.50 + 11,062.50 = $34,187.50 Practice: Adjusted Winner Procedure Joey and Chandler: Roommate Split Joey Chandler Foosball Table 20 32 White dog statue 20 10 Duck and chick 25 10 Working recliner 20 30 Entertainment center 15 18 Knaster Inheritance Procedure Mary, Sara, Marcia, Nessa, and Diane all look to inherit their grandmother’s wedding ring. Their bids are as follows: Mary Sara Marcia Nessa Diane Ring $1,000 $500 $900 $100 $1,200 AWP 1. Who has the most points for an item Joey Chandler Foosball Table 20 32 * White dog statue 20 * 10 Duck and chick 25 * 10 Working recliner 20 30 * Entertainment center 15 18 * 2. Add up the points Joey Chandler 32 20 25 30 18 45 80 3. Ratios rd Foosball: 32/20 = 1.6 *** 3 Recliner: 30/20 = 1.5 ** 2nd Center: 18/15 = 6/5 = 1.2 * closest to 1 Just moving the entertainment center will not even out the points (60 to 62), though it is close 4. Portion of Divide We’ll let Joey keep the entertainment center and split the second closest item, the working recliner 60 + 20x = 62 – 30x 50x = 2 x = 1/25 = .04 We knew this would be a small number since they were so close after step 3 5. Who gets what Joey: White dog statue, duck and chick, the entertainment center, 4% of the working recliner 20+25+20(.04)+15 = 60.8 Chandler: Foosball table, 96% of the working recliner 32+30(.96) = 60.8 KIP Mary Sara Marcia Nessa Diane Ring $1,000 $500 $900 $100 $1,200 Divide by 5 Mary Sara Marcia Nessa Diane 200 100 180 20 240 Kitty: 4/5 = 240x4 = 960 or 1,200 – 240 = 960 960 - 200 - 100 - 180 - 20 460  divide by 5 = 92 Mary Sara Marcia Nessa Diane Ring Bids $1,000 $500 $900 $100 $1,200 200 + 92 100 + 92 180 + 92 20 + 92 960 – 92 End Totals $292 $192 $272 $112 Ring, Pay $868 Check: 292+192+272+112 = $868

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Chapter 8.4, Problem 18E is Solved
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Textbook: Probability and Statistical Inference
Edition: 9
Author: Robert V. Hogg, Elliot Tanis, Dale Zimmerman
ISBN: 9780321923271

This full solution covers the following key subjects: exercise, conclusion, test, hypothesis, plants. This expansive textbook survival guide covers 59 chapters, and 1476 solutions. The answer to “?In Exercise 8.2-10, growth data are given for plants in normal air and for plants in CO2-enriched air. Those data are repeated here: In this exercise, we shall test the null hypothesis that the medians are equal, namely, H0: mX = mY, against the alternative hypothesis H1: mX < mY. You may select the significance level. However, give the approximate p-value or state clearly why you arrived at a particular conclusion, for each of the tests. Show your work.(a) What is your conclusion from the Wilcoxon test?(b) What was your conclusion from the t test in Exercise 8.2-10?(c) Write a comparison of these two tests.” is broken down into a number of easy to follow steps, and 105 words. This textbook survival guide was created for the textbook: Probability and Statistical Inference , edition: 9. Probability and Statistical Inference was written by and is associated to the ISBN: 9780321923271. The full step-by-step solution to problem: 18E from chapter: 8.4 was answered by , our top Statistics solution expert on 07/05/17, 04:50AM. Since the solution to 18E from 8.4 chapter was answered, more than 359 students have viewed the full step-by-step answer.

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In Exercise 8.2-10, growth data are given for plants in