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Using OLS, you could estimate both of these equations in one regression using a dummy variable. Let $\alpha_i=1$ if $i=2$ and $\alpha_i=0$ otherwise. Then the regression you estimate is (omitting error terms and indices for observation number) $$y_i=\beta x_i+\gamma\alpha_ix_i.$$ Just plug in the values to see that you can really recreate the two above ...


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In the first part, in order to fit the data to $y=b\, x^m$, you linearize the model to get $\log(y)=\beta=m\log(x)$ which corresponds to a linear regression. But, this gives you estimates of the parameters. From this points, you need to continue with nonlinear regression since what you want to minimize is the sum of the squared errors for the $y$'s and not ...


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For the $R^2$, 2 is plenty. If your data is simulated, then getting a value of $1$ isn't that weird. Leave it at $0.99$ if you don't like $1$ For the coefficients, it depends on the context. The key question to ask is: Does adding a digit provide any useful information to your audience? Personally, I don't like having more than 3. If more are needed, then ...


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I can present a deriviation with more steps: $Var (\hat \beta)= Var ((X'X)^{-1}X'y) \quad | $ replace $y = Xβ + u$ $= Var ((X'X)^{-1}(X'X)β + (X'X)^{-1}X'u )\quad | (X'X)^{-1}(X'X)=I$ $= Var (β + (X'X)^{-1}X'u )\quad | Var(b+Y)=Var(Y)$ $= Var ( (X'X)^{-1}X'u )\quad | \texttt{factoring out} \ (X'X)^{-1}X'$ $=(X'X)^{-1}X'Var(u)X(X'X)^{-1} \quad | ...


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The idea is: We have data (here it are the tuples: (monarch egg production,overwintering population)). We ask ourselves if there may exist a lineair relationship between the two. This question may look odd to you. Why would we care of a linear relationship? Well, it is important to get more insight in the thing you want to study. See this example: We ...


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For illustration purposes of Marty Cohen's answer, I generated $50$ data points ($i$ from $1$ to $50$), $x_i=i$, $y_i=12.34 \sin (x_i+2.345)+2(-1)^i$ (the error is quite large). The largest absolute value of the $y_i$'s is $14.3129$. From there, the sum of the $\phi_i$'s equals $-1275.08$ so $\phi =-25.5017$; adding $9\pi$, this gives $\phi=2.7726$. Using ...


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Here is a very simple-minded way. First, set $V = \max(|y_i|)$. Then, let $\phi_i =\arcsin(y_i/V)-x_i $. Finally, $\phi =\frac1{n}\sum \phi_i $. Note: If the data is over multiple sinusodial cycles, adjust $\phi_{i+1}$ so it differs from $\phi_i$ by less than $\pi$.


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It is certainly possible (in mostly silly cases). For example, suppose you sample the same $x$ value twice and get two experimental $y$ values that differ. One such case might be getting the points $(0, -1)$ and $(0, 1)$ if we sample $x=0$ twice. Then, any line passing through the origin has SSE=2 which is the minimum possible. But of course there are ...


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Your equation (9.2.3) isn't meant to be "solved". It is simply an example of crop yield $Y$ at time $X$. The regression method, which hasn't been discussed yet at that point in your text, attempts to find values for the parameters $\beta_i$ which minimize the error over the entire set of sample points $(x_k,y_k)$. It does not attempt to find parameters such ...


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For the first question, the model is $$y=\frac{1}{a+b~x^c}$$ which is nonlinear with respect to all parameters. I suppose that you have $n$ data points$(x_i,y_i)$ based on which you want to adjust parameters $a,b,c$ and, as usual, the problem with nonlinear regression is to start with reasonable values. For the time, generate values $z_i=\frac 1 {y_i}$ ...


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When you perform, say, a linear regression $y=a+bx$ based on the data points ($x_i,y_i$), if the $y_i$'s cover a large range, the largest values strongly contribute to define the parameters. So, instead of minimizing $$SSQ=\sum_{i=1}^n (a+bx_i-y_i)^2$$ weighted least-square method minimize $$SSQ=\sum_{i=1}^n w_i(a+bx_i-y_i)^2$$ where $w_i$ is a weight ...


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To see why first lets take an ordinary regression: $ y_i=\alpha + X_i'\beta +\epsilon_i \space \space \space \space $ (1) Therefore we have $\hat{y_i} = \hat{\alpha} + X_i'\hat{\beta}$ But by definition, $\hat{\alpha} = \bar{y}-\bar{X}'\hat{\beta}$ and thus $\hat{y_i} = \bar{y} + (X_i-\bar{X})'\hat{\beta} = \bar{y} + X_d'\hat{\beta}$ ...


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Hoping that I properly understood the first question : for each data point create a binary variable $z_i$ which is $1$ if the data point come from the first sample and $2$ if the data point come from the second sample. Since the model is multilinear, you now have to fit $$y=(a+bz)+(c+dz)x=\alpha+\beta z+\gamma x+\delta x z$$ so three regressors ($x,z,xz$).



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