Porter Mccullough

2022-05-03

Table of values

$\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ y& 3& 6& 8& 9& 0\\ y& 4& 6& 1& 2& 4\end{array}$

and know that it is a simple linear regression model, what is the value of $n$? I think it is either $5$ or $10$ but am not sure which one.

$\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ y& 3& 6& 8& 9& 0\\ y& 4& 6& 1& 2& 4\end{array}$

and know that it is a simple linear regression model, what is the value of $n$? I think it is either $5$ or $10$ but am not sure which one.

Alexis Wolf

Beginner2022-05-04Added 13 answers

$\mathcal{F}(a,b)\equiv \frac{1}{2}\sum _{i=1}^{5}\sum _{\sigma =\pm}{(a{x}_{i}+b-{y}_{i\sigma})}^{2}$

$\begin{array}{rl}0& =\frac{{\mathrm{\partial}}^{}\mathcal{F}(a,b)}{\mathrm{\partial}{a}^{}}=\sum _{i=1}^{5}\sum _{\sigma =\pm}(a{x}_{i}+b-{y}_{i\sigma}){x}_{i}=\sum _{i=1}^{5}(2{x}_{i}^{2}a+2{x}_{i}b-{x}_{i}\sum _{\sigma =\pm}{y}_{i\sigma})\\ 0& =\frac{{\mathrm{\partial}}^{}\mathcal{F}(a,b)}{\mathrm{\partial}{b}^{}}=\sum _{i=1}^{5}\sum _{\sigma =\pm}(a{x}_{i}+b-{y}_{i\sigma})=\sum _{i=1}^{5}(2{x}_{i}a+2b-\sum _{\sigma =\pm}{y}_{i\sigma})\end{array}$

$\{\begin{array}{rcrcl}\stackrel{{\displaystyle \equiv \text{}{S}_{xx}}}{\stackrel{\u23de}{\left(2\sum _{i=1}^{5}x{i}^{2}\right)}}\text{}a& +& \stackrel{{\displaystyle \equiv \text{}{S}_{x}}}{\stackrel{\u23de}{\left(2\sum _{i=1}^{5}xi\right)}}\text{}b& =& \stackrel{{\displaystyle \equiv {S}_{xy}}}{\stackrel{\u23de}{\sum _{i=1}^{5}{x}_{i}\sum _{\sigma =\pm}{y}_{i\sigma}}}\\ \underset{{\displaystyle =\text{}{S}_{x}}}{\underset{\u23df}{\left(2\sum _{i=1}^{5}{x}_{i}\right)}}\text{}a& +& 10\phantom{\rule{thinmathspace}{0ex}}b& =& \underset{{S}_{y}}{\underset{\u23df}{\sum _{i=1}^{5}\sum _{\sigma =\pm}{y}_{i\sigma}}}\end{array}$

${S}_{xx}=110\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1em}{0ex}}{S}_{x}=30\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1em}{0ex}}{S}_{xy}=122\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1em}{0ex}}{S}_{y}=43$

$\begin{array}{rcrcl}55a& +& 15b& =& 61\\ 30a& +& 10& b=& 43\end{array}\}\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}\{\begin{array}{rclcr}a& =& \frac{61\times 10-43\times 15}{100}& =& -\phantom{\rule{thinmathspace}{0ex}}\frac{7}{20}\\ b& =& \frac{51\times 43-30\times 61}{100}& =& \frac{363}{100}\end{array}$

which yields

$y\left(x\right)=\frac{1}{100}(-35x+363)\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{2em}{0ex}}\u27f9\phantom{\rule{2em}{0ex}}\begin{array}{cccl}x& y& & \\ 1& \frac{82}{25}& =& 3.28\\ 2& \frac{293}{100}& =& 2.93\\ 3& \frac{129}{50}& =& 2.58\\ 4& \frac{223}{100}& =& 2.23\\ 5& \frac{47}{25}& =& 1.88\end{array}$

$\begin{array}{rl}0& =\frac{{\mathrm{\partial}}^{}\mathcal{F}(a,b)}{\mathrm{\partial}{a}^{}}=\sum _{i=1}^{5}\sum _{\sigma =\pm}(a{x}_{i}+b-{y}_{i\sigma}){x}_{i}=\sum _{i=1}^{5}(2{x}_{i}^{2}a+2{x}_{i}b-{x}_{i}\sum _{\sigma =\pm}{y}_{i\sigma})\\ 0& =\frac{{\mathrm{\partial}}^{}\mathcal{F}(a,b)}{\mathrm{\partial}{b}^{}}=\sum _{i=1}^{5}\sum _{\sigma =\pm}(a{x}_{i}+b-{y}_{i\sigma})=\sum _{i=1}^{5}(2{x}_{i}a+2b-\sum _{\sigma =\pm}{y}_{i\sigma})\end{array}$

$\{\begin{array}{rcrcl}\stackrel{{\displaystyle \equiv \text{}{S}_{xx}}}{\stackrel{\u23de}{\left(2\sum _{i=1}^{5}x{i}^{2}\right)}}\text{}a& +& \stackrel{{\displaystyle \equiv \text{}{S}_{x}}}{\stackrel{\u23de}{\left(2\sum _{i=1}^{5}xi\right)}}\text{}b& =& \stackrel{{\displaystyle \equiv {S}_{xy}}}{\stackrel{\u23de}{\sum _{i=1}^{5}{x}_{i}\sum _{\sigma =\pm}{y}_{i\sigma}}}\\ \underset{{\displaystyle =\text{}{S}_{x}}}{\underset{\u23df}{\left(2\sum _{i=1}^{5}{x}_{i}\right)}}\text{}a& +& 10\phantom{\rule{thinmathspace}{0ex}}b& =& \underset{{S}_{y}}{\underset{\u23df}{\sum _{i=1}^{5}\sum _{\sigma =\pm}{y}_{i\sigma}}}\end{array}$

${S}_{xx}=110\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1em}{0ex}}{S}_{x}=30\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1em}{0ex}}{S}_{xy}=122\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1em}{0ex}}{S}_{y}=43$

$\begin{array}{rcrcl}55a& +& 15b& =& 61\\ 30a& +& 10& b=& 43\end{array}\}\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}\{\begin{array}{rclcr}a& =& \frac{61\times 10-43\times 15}{100}& =& -\phantom{\rule{thinmathspace}{0ex}}\frac{7}{20}\\ b& =& \frac{51\times 43-30\times 61}{100}& =& \frac{363}{100}\end{array}$

which yields

$y\left(x\right)=\frac{1}{100}(-35x+363)\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{2em}{0ex}}\u27f9\phantom{\rule{2em}{0ex}}\begin{array}{cccl}x& y& & \\ 1& \frac{82}{25}& =& 3.28\\ 2& \frac{293}{100}& =& 2.93\\ 3& \frac{129}{50}& =& 2.58\\ 4& \frac{223}{100}& =& 2.23\\ 5& \frac{47}{25}& =& 1.88\end{array}$

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