Question #279446

sample of soil were prepared with varying amount of inorganic phosphorus. corn plant grown ineach soil were harvested after 8 days and analysed fir phosphorous contents. the objective was to see to what extent additional phosphorous in the soil increased the concentration of phosphorous in the plants. the data for 8 samples are as follows



Inorganic soil phosphorous (PPM) 1,4, 5, 9, 13, 11, 23. 23



Organic soil phosphorous (PPM) 64, 71, 54, 81, 93, 76, 77, 95

1
Expert's answer
2021-12-15T17:33:02-0500

e2=800.43 Standard error, se =(E2/(n2))=(800.43/(92))=10.6933 Standard error for slope, se(b1) =se/SSxx=10.6933/734=0.3947 Null and alternative hypothesis:  Ho: β1=0; Ha: β10 Test statistic: t= b1/se(b1) =1.42/0.3947=3.59 df =n2=7p-value =T.DIST.2T(ABS(3.5898),7)=0.0089 Conclusion: p-value <α, Reject the null hypothesis.  There is enough evidence to conclude that soil P is a useful predictor of corn P.\begin{aligned} &\sum \mathrm{e}^{2}=800.43 \\ &\text { Standard error, se } \left.=\sqrt{( \mathrm{E}^{2} /(\mathrm{n}-2)}\right)=\sqrt{(800.43 /(9-2))}=10.6933 \\ &\text { Standard error for slope, se(b1) }=\operatorname{se} / \sqrt{S Sx x} =10.6933 / \sqrt{734}={0.3947} \\ &\text { Null and alternative hypothesis: } \\ &\text { Ho: } \beta_{1}=0 ; \text { Ha: } \beta_{1} \neq 0 \\ &\text { Test statistic: } \\ &t=\text { b1/se(b1) }=1.42 / 0.3947=3.59 \\ &\text { df }=n-2=7 \\ &p \text {-value }=T . D I S T .2 T(A B S(3.5898), 7)=0.0089 \\ &\text { Conclusion: } \\ &p \text {-value }<\alpha, \text { Reject the null hypothesis. } \\ &\text { There is enough evidence to conclude that soil } P \text { is a useful predictor of corn } P . \end{aligned}


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