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# In Exercise use analysis of variance for | Ch 12.2 - 13BSC ISBN: 9780321836960 18

## Solution for problem 13BSC Chapter 12.2

Elementary Statistics | 12th Edition

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Problem 13BSC

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 rich and moist 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? N F I o e r T r r r t i Fertilizer e i g and a l a Irrigatio t i t n m z i e e o n r n t 1 0 0 0.85 . . . 2 9 0 1 4 7 0 0 0 . . . 1.78 5 8 6 7 7 6 0 0 0 . . . 1.47 5 4 1 6 6 0 0 0 0 . . . 2.25 1 5 8 3 8 2 1 1 0 . . . 1.64 3 0 9 0 3 4

Step-by-Step Solution:

Solution 13BSC 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 .428, second one is .464 , third one is .196 and fourth one is .910. Hence, .428, .464, .196 and .910 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 a Kolm p ogor i ov-S r mirn o ov a - W i l k Stati d Stati D Sig. Sig. stic f stic f No treatment .246 5 .200 * .903 5 .428 Fertilizer .250 5 .200 * .909 5 .464 Irrigation .247 5 .200 * .851 5 .196 Fertilizer and .201 5 .200 * .976 5 .910 Irrigation 2. Then we know that mean and standard deviations from below table. Descriptive Statistics N Mean Std. Deviation No treatment 5 .7540 .49166 Fertilizer 5 .7760 .24419 Irrigation 5 .5180 .40770 Fertilizer and 1.598 5 .50889 Irrigation 0 From the above table the standard deviations values are different. Here the the standard deviations 1st,2nd,3rd and 4th is 0.49166 , 0.2441 , 0.4077 , 0.5088. 3. The samples are randomly selected it is mentioned in the question. 4. The samples are independent of each other. 5. The samples are categorized according to one factor weight. Step 2 From the given problem the claim that the four treatment categories yield poplar trees with the same mean weight. The null hypothesis is that the four samples are from populations with the same mean 1 2 =3 . 4 Then ANOVA for the data set table in below. ANOVA Weight Sum of df Mea F Sig. Squares n Squa re Between 3.346 3 1.115 6.141 .006 Groups Within 2.906 16 .182 Groups Total 6.253 19 The above table is SPSS output for one-way ANOVA table.

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##### ISBN: 9780321836960

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