$\color{red}\rule{10pt}{10pt} \color{pink}\rule{10pt}{10pt}\color{yellow}\rule{10pt}{10pt}\color{blue}\rule{10pt}{10pt}\color{green}\rule{10pt}{10pt}\color{royalblue}\rule{10pt}{10pt}\color{black}\rule{10pt}{10pt}\color{white}\rule{10pt}{10pt}\color{purple}\rule{10pt}{10pt}\color{magenta}\rule{10pt}{10pt}\color{orange}\rule{10pt}{10pt}\color{brown}\rule{10pt}{10pt}\color{lime}\rule{10pt}{10pt}$

5 0

3 years ago

Ten pairs of data yield r = 0.003 and the regression equation =2+3x Also, the mean = 5 What is the best predicted value of y for

Olenka [21]

Answer:

$\hat y(2)=2 +3(2) =8$

C 8.0

Step-by-step explanation:

Assuming the linear model y=mx+b where m is the slope and b the intercept.

For this case the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

$b=\bar y -m \bar x$

After the calculations we see that m=3 and b=2 from the info given by the linear model.

For this case we have the equation obtained by least squares given by:

$\hat y =2 +3x$

Where 2 represent the intercept and 3 the slope. We are interested on the best predicted value of y when x=2.

If we see our linear model we have the equation in terms of y and x. So we can replace directly the value of x=2 into the equation and see what we got:

$\hat y(2)=2 +3(2) =8$

5 0

3 years ago