Question

The data show systolic and diastolic blood pressure of certain people. Find the regression equation, letting the first variable be the independent (x) variable. Find the best predicted diastolic pressure for a person with a systolic reading of 113. use a significance level of 0.05.
Systolic| 150 129 142 112 134 122 126 120

Diastolic| 88 96 106 80 98 63 95 64

a. What is the regression equation?

^y = __ + __x (Round to two decimal places as needed.)

b. What is the best predicted value?

^y is about __ (Round to one decimal place as needed.

Answer 1

Answer:

a) $y=0.75 x -10.65$

b) $y= 0.749*113 - 10.652=73.985\ approx 74.0$

Step-by-step explanation:

We assume that the data is this one:

x: 150 129 142 112 134 122 126 120

y: 88 96 106 80 98 63 95 64

Part a

We want to find a linear model $y = mx+b$. Where:

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

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:

$\sum_{i=1}^n x_i =1035$

$\sum_{i=1}^n y_i =690$

$\sum_{i=1}^n x^2_i =134965$

$\sum_{i=1}^n y^2_i =61290$

$\sum_{i=1}^n x_i y_i =90064$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=134965-\frac{1035^2}{8}=1061.875$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=90064-\frac{1035*690}{8}=795.25$

And the slope would be:

$m=\frac{795.25}{1061.875}=0.749$

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

$\bar x= \frac{\sum x_i}{n}=\frac{1035}{8}=129.375$

$\bar y= \frac{\sum y_i}{n}=\frac{690}{8}=86.25$

And we can find the intercept using this:

$b=\bar y -m \bar x=86.25-(0.749*129.375)=-10.652$

So the line would be given by:

$y=0.75 x -10.65$

Part b

For this case w ejust need to replace x= 113 in the regression model and we got this:

$y= 0.749*113 - 10.652=73.985\ approx 74.0$