Weights (kg) of poplar trees were obtained from trees planted in arch and moist

Question

Weights (kg) of poplar trees were obtained from trees planted in arch and moist region. The trees were given different treatments identified in the accompanying table. 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? Determine the null and alternative hypotheses. H_0: Find the F test statistics. (Round to four decimal places as) Find the P-value using the F test statistics. P - Value = (Round to four decimal places as needed)

Explanation / Answer

  

Null Hypothesis H0:mew1 = mew2=mew3=mew4 i.e there is no significance difference between the mean weight of the treatments

Alternative hypothesis H1: at least two of them not equal to zero(two-tailed test)

Descriptives

  
Treatment N   Mean   Std. Deviation   Std. Error
No Treatment 5 0.7280 0.54623 0.24428
Fertilizer 5 0.6700 0.25700 0.11493   
Irregation 5 0.5060 0.38142 0.17058   
Fertilizer and Irrergation 5   1.2700 0.27973 0.12510
Total 20 0.7935 0.45877 0.10258   

       ANOVA one -way classification table
Treatment
Sum of Squares   df   Mean Square   F Sig.
Between Groups   1.646 3 0.549 3.732   .033
Within Groups 2.353 16 0.147      
Total 3.999 19         

the critical value of F at 0.05 level with(3,16) degree of freedom is 3.238

therefore calculated value > table value (3.732>3.238)

therefore H0 is rejected i.e we may conclude that there is significance difference between the mean weight of the treatments

P-value: 0.033036

therefore The result is significant at p(0.033036 )< .05

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