Question

algebra 2 unit 2 test part 2
name:
11
mohamed wants to know if the amount of candy he eats correlates to the amount of cavities he has when he goes to the dentist. he keeps track of the pieces of candy, x, he eats before each dentist appointment, and the amount of cavities he has, y.

x (pieces of candy)	y (cavities)
21	0
114	2
134	3

using the table, write the equation for the regression line of the data in slope - intercept form.

Explanation:

Step1: Calculate the means of x and y

Let $x_1 = 64$, $x_2=21$, $x_3 = 114$, $x_4=134$, $y_1 = 1$, $y_2 = 0$, $y_3=2$, $y_4 = 3$.
The mean of $x$, $\bar{x}=\frac{64 + 21+114+134}{4}=\frac{333}{4}=83.25$.
The mean of $y$, $\bar{y}=\frac{1 + 0+2+3}{4}=\frac{6}{4}=1.5$.

Step2: Calculate the slope (m)

The formula for the slope $m=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}$.
$(x_1-\bar{x})(y_1 - \bar{y})=(64 - 83.25)(1 - 1.5)=(-19.25)(-0.5)=9.625$.
$(x_2-\bar{x})(y_2 - \bar{y})=(21 - 83.25)(0 - 1.5)=(-62.25)(-1.5)=93.375$.
$(x_3-\bar{x})(y_3 - \bar{y})=(114 - 83.25)(2 - 1.5)=(30.75)(0.5)=15.375$.
$(x_4-\bar{x})(y_4 - \bar{y})=(134 - 83.25)(3 - 1.5)=(50.75)(1.5)=76.125$.
$\sum_{i = 1}^{4}(x_i-\bar{x})(y_i - \bar{y})=9.625+93.375+15.375+76.125 = 194.5$.
$(x_1-\bar{x})^2=(64 - 83.25)^2=(-19.25)^2 = 370.5625$.
$(x_2-\bar{x})^2=(21 - 83.25)^2=(-62.25)^2=3875.5625$.
$(x_3-\bar{x})^2=(114 - 83.25)^2=(30.75)^2 = 945.5625$.
$(x_4-\bar{x})^2=(134 - 83.25)^2=(50.75)^2=2575.5625$.
$\sum_{i=1}^{4}(x_i-\bar{x})^2=370.5625+3875.5625+945.5625+2575.5625 = 7767.25$.
$m=\frac{194.5}{7767.25}\approx0.025$.

Step3: Calculate the y - intercept (b)

Using the formula $b=\bar{y}-m\bar{x}$, we substitute $m = 0.025$, $\bar{x}=83.25$ and $\bar{y}=1.5$.
$b=1.5-0.025\times83.25=1.5 - 2.08125=- 0.58125\approx - 0.58$.

Answer:

$y = 0.025x-0.58$