Question

Find the regression equation, letting the first variable be the
predictor (x) variable.
Using the listed duration and interval after times, find the best predicted "interval after" time for an eruption with a duration of 253 seconds. How does it compare to an actual eruption with a duration of 253 seconds and an interval after time of 83 minutes?
Duration - 242 - 255 - 227 - 251 - 262 - 207 - 140
Interval After - 81 - 81 - 92 - 102 - 94 - 91

Answer 1

Answer:

$y=0.00673(253) +90.190=91.894$

And the difference is given by:

$r_i =91.894-83=8.894$

Step-by-step explanation

We assume that th data is this one:

x: 242-255 -227-251-262-207-140

y: 91- 81 -91 - 92 - 102 - 94 - 91

Find the least-squares line appropriate for this data.  

For this case we need to calculate the slope with the following formula:

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

Where:

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

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

So we can find the sums like this:

$\sum_{i=1}^n x_i =242+255+227+251+262+207+140=1584$

$\sum_{i=1}^n y_i =91+ 81 +91 + 92 + 102 + 94 + 91=642$

$\sum_{i=1}^n x^2_i =242^2 +255 ^2 +227^2 +251^2 +262^2 +207^2 +140^2=369212$

$\sum_{i=1}^n y^2_i =91^2 + 81 ^2 +91 ^2 + 92 ^2 + 102 ^2 + 94 ^2 + 91^2=59108$

$\sum_{i=1}^n x_i y_i =242*91 +255*81 +227*91 +251*92 +262*102 +207*94 +140*91=145348$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=369212-\frac{1584^2}{7}=10775.429$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=145348-\frac{1584*642}{7}=72.571$

And the slope would be:

$m=\frac{72.571}{10775.429}=0.00673$

Now we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{1584}{7}=226.286$

$\bar y= \frac{\sum y_i}{n}=\frac{642}{7}=91.714$

And we can find the intercept using this:

$b=\bar y -m \bar x=91.714-(0.00673*226.286)=90.190$

So the line would be given by:

$y=0.00673 x +90.190$

The prediction for 253 seconds is:

$y=0.00673(253) +90.190=91.894$

And the difference is given by:

$r_i =91.894-83=8.894$