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In Exercise use analysis of variance for | Ch 12.2 - 14BSC

Elementary Statistics | 12th Edition | ISBN: 9780321836960 | Authors: Mario F. Triola ISBN: 9780321836960 18

Solution for problem 14BSC Chapter 12.2

Elementary Statistics | 12th Edition

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Elementary Statistics | 12th Edition | ISBN: 9780321836960 | Authors: Mario F. Triola

Elementary Statistics | 12th Edition

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Problem 14BSC

In Exercise? ?use analysis of variance for the indicated test. Poplar Tree Weights? Weights (kg) of poplar trees were obtained from trees planted in a sandy and dry region. The trees were given different treatments identified in the table below. The data are from a study conducted by researchers at Pennsylvania State University and were provided by Minitab, Inc. Use a 0.05 significance level to test the claim that the four treatment categories yield poplar trees with the same mean weight. Is there a treatment that appears to be most effective in the sandy and dry region? N F I o e r T r r r t i Fertilize e i g r and a l a Irrigatio t i t n m z i e e o n r n t 0 0 0 1.07 . . . 2 9 9 4 2 6 1 0 1 . . . 1.63 6 0 4 9 7 3 1 0 1 . . . 1.39 2 5 2 3 6 6 0 1 1 . . . 0.49 9 7 5 9 4 7 1 1 0 . . . 0.95 8 1 7 0 3 2

Step-by-Step Solution:

Solution 14BSC Step 1 The given problem explain about Poplar Tree Weights. We have to check assumptions: 1. The Shapiro-wilk statistic values are First one is .585, second one is .996 , third one is .790 and fourth one is .943. Hence, .585, .996, .790 and .943 for no treatment, fertilizer, irrigation and fertilizer plus irrigation respectively. Then these values are greater than the level of significance 0.05 so the data is found to be normal. Tests of Normality S h Kol a mog p oro i v-S r mir o nov - W i l k Stat d Sig. Statistic df Sig. istic f No .200 .188 5 .928 5 .585 treatment * .200 Fertilizer .147 5 .996 5 .996 * .200 Irrigation .182 5 .957 5 .790 * Fertilizer .200 and .160 5 .982 5 .943 * Irrigation 2. Then we know that mean and standard deviations from below table : Descriptive Statistics Std. N Mean Deviatio n No treatment 5 1.1900 .62574 Fertilizer 5 .8840 .62492 Irrigation 5 1.1880 .34666 Fertilizer and 5 1.1060 .43598 Irrigation From the above table the standard deviations values are different. Here the the standard deviations 1st,2nd,3rd and 4th is 0.6257 , 0.6249 , 0.3466 , 0.4359. 3. The data comes from a well planned study. Hence, the samples are randomly selected 4. These are independent samples of each other. 5. The samples are categorized according to one factor, tree weight.

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Chapter 12.2, Problem 14BSC is Solved
Step 3 of 3

Textbook: Elementary Statistics
Edition: 12
Author: Mario F. Triola
ISBN: 9780321836960

Elementary Statistics was written by and is associated to the ISBN: 9780321836960. The full step-by-step solution to problem: 14BSC from chapter: 12.2 was answered by , our top Statistics solution expert on 03/15/17, 10:30PM. Since the solution to 14BSC from 12.2 chapter was answered, more than 546 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Elementary Statistics, edition: 12. This full solution covers the following key subjects: Trees, were, poplar, treatment, weights. This expansive textbook survival guide covers 121 chapters, and 3629 solutions. The answer to “In Exercise? ?use analysis of variance for the indicated test. Poplar Tree Weights? Weights (kg) of poplar trees were obtained from trees planted in a sandy and dry region. The trees were given different treatments identified in the table below. The data are from a study conducted by researchers at Pennsylvania State University and were provided by Minitab, Inc. Use a 0.05 significance level to test the claim that the four treatment categories yield poplar trees with the same mean weight. Is there a treatment that appears to be most effective in the sandy and dry region? N F I o e r T r r r t i Fertilize e i g r and a l a Irrigatio t i t n m z i e e o n r n t 0 0 0 1.07 . . . 2 9 9 4 2 6 1 0 1 . . . 1.63 6 0 4 9 7 3 1 0 1 . . . 1.39 2 5 2 3 6 6 0 1 1 . . . 0.49 9 7 5 9 4 7 1 1 0 . . . 0.95 8 1 7 0 3 2” is broken down into a number of easy to follow steps, and 198 words.

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In Exercise use analysis of variance for | Ch 12.2 - 14BSC