Question

A number of studies have shown lichens (certain plants composed of an alga and a fungus) to be excellent bioindicators of air pollution. An article gives the following data (read from a graph) on xNO3 wet deposition (g N/m2) and y lichen (% dry welght): x 0.05 0.10 0.11 0.12 0.31 0.37 0.42 0.58 .68 0.68 0.73 0.85 0.92 ri 0.48 0.53 0.4g 0.48 0.63 0.47 1.04 0.88 0.86 1.01 0.87 1.02 1.70 The regression equation is lichen N0.364 + 0.968 no3 depo Predictor Constant no3 depo Coef 0.9675 R-sq t-ratio Stdev 0.36403 0.10278 0.1898 3.540.005 5.10 0.000 s-0.2005 70.3% - Analysis of Variance SOURCE Regression Error Total MS 1 1.04421.0442 25.9 0.000 11 .4421 0.0402 12 1.4863 (a) What are the least square estimates of Po and B? (Round your answers to three decimal places.) (b) Predict lichen N for an NO3 deposition value of 0.6. (Round your answer to three decimel places.) % dry weight (c) what is the estimate of σ? (Round your answer to four decimal places.) % dry weight (d) What is the value of total variation?(Round your ansiver to four decimal places.) How much of it can be explained by the model relationship? (Round your answer to one decimal place.)

Answer 1

ALl answers are given in the table. a) As = 0.364 0.968 b) Predicted value for 0.6 is given by regression equation: y=b0 + bl@ y = (0.364) + (0.968) * (0.6) = 0.9448 Predicted value: y= 0.9448 c) σ= s-0.2005 d) SST 1.4863 e) % explaind-R-squared = 70.3% oThank youes