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yarga [219]
3 years ago
13

Find the regression equation, letting the first variable be the

Mathematics
1 answer:
Sonbull [250]3 years ago
7 0

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

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