Question

1. a candle is measured as it burns. the height of the candle at various times as it burns is recorded in the table below:

time (hours)	2	4	7	15
height (inches)	5.75	5.25	4.5	2.5

a. write a linear regression model to represent this data. round to the nearest hundredth when necessary.
b. predict the height of the candle after 10 hours.
c. predict the time it takes for the candle to burn out.

Explanation:

Step1: Calculate the means of x and y

Let \(x\) be the time (hours) and \(y\) be the height (inches).
\(x = [2,4,7,15]\), \(\bar{x}=\frac{2 + 4+7+15}{4}=\frac{28}{4}=7\)
\(y=[5.75,5.25,4.5,2.5]\), \(\bar{y}=\frac{5.75 + 5.25+4.5+2.5}{4}=\frac{18}{4} = 4.5\)

Step2: Calculate the slope \(m\)

\[

$$\begin{align*} m&=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}}\\ (x_1-\bar{x})(y_1 - \bar{y})&=(2 - 7)(5.75-4.5)=(-5)\times1.25=- 6.25\\ (x_2-\bar{x})(y_2 - \bar{y})&=(4 - 7)(5.25-4.5)=(-3)\times0.75=-2.25\\ (x_3-\bar{x})(y_3 - \bar{y})&=(7 - 7)(4.5-4.5)=0\times0 = 0\\ (x_4-\bar{x})(y_4 - \bar{y})&=(15 - 7)(2.5-4.5)=8\times(-2)=-16\\ \sum_{i = 1}^{4}(x_{i}-\bar{x})(y_{i}-\bar{y})&=-6.25-2.25 + 0-16=-24.5\\ (x_1-\bar{x})^2&=(2 - 7)^2=25\\ (x_2-\bar{x})^2&=(4 - 7)^2 = 9\\ (x_3-\bar{x})^2&=(7 - 7)^2=0\\ (x_4-\bar{x})^2&=(15 - 7)^2=64\\ \sum_{i=1}^{4}(x_{i}-\bar{x})^{2}&=25 + 9+0+64=98\\ m&=\frac{-24.5}{98}=-0.25 \end{align*}$$

\]

Step3: Calculate the y - intercept \(b\)

Using the formula \(y=mx + b\) and substituting \(\bar{x}\) and \(\bar{y}\) and \(m\):
\(4.5=-0.25\times7 + b\)
\(4.5=-1.75 + b\)
\(b=4.5 + 1.75=6.25\)
The linear - regression model is \(y=-0.25x + 6.25\)

Step4: Predict the height at \(x = 10\)

Substitute \(x = 10\) into \(y=-0.25x + 6.25\)
\(y=-0.25\times10+6.25=-2.5 + 6.25 = 3.75\) inches

Step5: Predict the time when \(y = 0\)

Set \(y = 0\) in \(y=-0.25x + 6.25\)
\(0=-0.25x+6.25\)
\(0.25x=6.25\)
\(x=\frac{6.25}{0.25}=25\) hours

Answer:

A. \(y=-0.25x + 6.25\)
B. 3.75 inches
C. 25 hours