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 x = NO₃⁻ wet deposition (g N/m²) and y = lichen (% dry weight):

x	0.05	0.10	0.11	0.12	0.31	0.37	0.42	0.58	0.68	0.68	0.73	0.85	0.92
y	0.44	0.53	0.53	0.50	0.56	0.55	1.03	0.82	0.86	0.98	0.89	1.01	1.68

The regression equation is lichen N = 0.367 + 0.947 no3 depo
Predictor	Coef	Stdev	t-ratio	p
Constant	0.36709	0.09781	3.75	0.003
no3 depo	0.9473	0.1806	5.24	0.000
s = 0.1908   R-sq = 71.4%   R-sq(adj) = 68.8%

Analysis of Variance
SOURCE	DF	SS	MS	F	P
Regression	1	1.0010	1.0010	27.50	0.000
Error	11	0.4004	0.0364
Total	12	1.4014

(a) What are the least square estimates of β₀ and β₁? (Round your answers to three decimal places.)

0	=
1	=

(b) Predict lichen N for an NO₃⁻ deposition value of 0.8. (Round your answer to three decimal places.)
% dry weight

(c) What is the estimate of σ? (Round your answer to four decimal places.)
=  % dry weight

(d) What is the total sum of squares? (Round your answer to four decimal places.)


How much of it can be explained by the model relationship? (Round your answer to one decimal place.)
%

Answer 1

Answer:

0 =given in table as constant coeff = 0.367

1 = no3 dep coeff = 0.947

(b)

at x = 0.8, y(hat) = 0.367 + 0.947* 0.8 = 1.1246 y(hat) = 1.1246

Given in table, SSError = 0.4004 Thus var(e)= SSE/(n-2) = 0.40041/11 = 0.0364 var(e) = 0.0364 Standard Error se(e) = sqrt(var(e)) = 0.1907 sigma cap = se(e) = 0.1907

(d) Total variation given in table = SS total = 1.4014

(e) how much can be explained by the model relationship as R-sq = 71.4%

it means 71.4 of the variation can be explained by the estimated linear model

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