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Question

Find an equation of the least squares linear regression line of y and x for the given data, and sketch both the line and the data. Predict the value of y corresponding to x=3.5.

x1234567y11.8244.579\begin{array}{ccccccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline y & 1 & 1.8 & 2 & 4 & 4.5 & 7 & 9 \end{array}

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Answered 2 years ago
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(a)\textbf{(a)} Recall that the linear regression line of yy on xx in the form y^=a+bx\hat y = a + bx with yy-intercept aa and slope bb has the following formula for the linear coefficients aa and bb:

a=(i=1nxi2)(i=1nyi)(i=1nxi)(i=1nxiyi)ni=1nxi2(i=1nxi)2b=ni=1nxiyi(i=1nxi)(i=1nyi)ni=1nxi2(i=1nxi)2\begin{align*} a &= \dfrac{\left(\sum_{i=1}^{n}x_i^2\right)\left(\sum_{i=1}^{n}y_i\right)-\left( \sum_{i=1}^{n}x_i \right) \left(\sum_{i=1}^{n}x_iy_i \right)}{n\sum_{i=1}^{n}x_i^2-\left( \sum_{i=1}^{n}x_i \right)^2} \\\\ b &= \dfrac{n\sum_{i=1}^{n}x_iy_i-\left( \sum_{i=1}^{n}x_i \right) \left(\sum_{i=1}^{n}y_i \right)}{n\sum_{i=1}^{n}x_i^2-\left( \sum_{i=1}^{n}x_i \right)^2} \end{align*}

We may use a spreadsheet in order to hasten the calculations. Based on this,

i=1nxi=28i=1nyi=29.3i=1nxiyi=154.1i=1nxi2=140\begin{align*} &\sum_{i=1}^{n}x_i = 28 &&\sum_{i=1}^{n}y_i = 29.3\\ &\sum_{i=1}^{n}x_iy_i = 154.1 &&\sum_{i=1}^{n}x_i^2 = 140 \end{align*}

And using the formulas, we can calculate that a=1.0857a=-1.0857 and b=1.3179b=1.3179. Hence, y^=1.0857+1.3179x\boxed{\hat y = -1.0857+1.3179x}.

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