Solution Found!
In Exercise use analysis of variance for
Chapter 12, Problem 14BSC(choose chapter or problem)
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
Questions & Answers
QUESTION:
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
ANSWER: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.