Chapter 9: Problem 50
in Fish In Exercise 9.21 , we see that the conditions are met for using the \(\mathrm{pH}\) of a lake in Florida to predict the mercury level of fish in the lake. The data are given in FloridaLakes. Computer output is shown for the linear model with several values missing: The regression equation is AvgMercury \(=1.53-0.152 \mathrm{pH}\) Predictor Constant \(\mathrm{pH}\) \(\begin{array}{rrrr}\text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ 1.5309 & 0.2035 & 7.52 & 0.000 \\ -0.15230 & \text { * (c) }^{* *} & -5.02 & 0.000\end{array}\) stant \(S=* *(b)^{* *} \quad R-S q=* *(a)^{* x}\) Analysis of Variance Source Regression Residual Erri Total \(\begin{array}{rrrrrr} & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { ion } & 1 & 2.0024 & 2.0024 & 25.24 & 0.000 \\ \text { Error } & 51 & 4.0455 & 0.0793 & & \\ & 52 & 6.0479 & & & \end{array}\) (a) Use the information in the ANOVA table to compute and interpret the value of \(R^{2}\). (b) Show how to estimate the standard deviation of the error term, \(s_{\epsilon}\). (c) Use the result from part (b) and the summary statistics below to compute the standard error of the slope, \(S E,\) for this model: $$ \begin{array}{lrrrrr} \text { Variable } & \text { N } & \text { Mean } & \text { StDev } & \text { Minimum } & \text { Maximum } \\ \text { pH } & 53 & 6.591 & 1.288 & 3.600 & 9.100 \\ \text { AvgMercury } & 53 & 0.5272 & 0.3410 & 0.0400 & 1.3300 \end{array} $$
Short Answer
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