Question

816 the following table represents a sample of sale prices, in thousands of dollars, and number of new homes available at that price in 2017.

sale price, p (in thousands of dollars)	160	180	200	220	240	260	280

state the linear regression function, f(p), that estimates the number of new homes available at a specific sale price, p. round all values to the nearest hundredth. state the correlation coefficient of the data to the nearest hundredth. explain what this means in the context of the problem.

Explanation:

Step1: Calculate mean of p and f(p)

$p = [160,180,200,220,240,260,280]$, $\bar{p} = \frac{160+180+200+220+240+260+280}{7} = 220$
$f(p) = [126,103,82,75,82,40,20]$, $\bar{f(p)} = \frac{126+103+82+75+82+40+20}{7} \approx 75.43$

Step2: Compute slope (m)

$m = \frac{\sum (p_i - \bar{p})(f(p_i)-\bar{f(p)})}{\sum (p_i - \bar{p})^2}$
Numerator: $(160-220)(126-75.43)+(180-220)(103-75.43)+(200-220)(82-75.43)+(220-220)(75-75.43)+(240-220)(82-75.43)+(260-220)(40-75.43)+(280-220)(20-75.43) = -12240$
Denominator: $(160-220)^2+(180-220)^2+(200-220)^2+(220-220)^2+(240-220)^2+(260-220)^2+(280-220)^2 = 11200$
$m = \frac{-12240}{11200} \approx -1.09$

Step3: Find y-intercept (b)

$b = \bar{f(p)} - m\bar{p} = 75.43 - (-1.09)(220) \approx 315.23$

Step4: Calculate correlation coefficient (r)

$r = \frac{\sum (p_i - \bar{p})(f(p_i)-\bar{f(p)})}{\sqrt{\sum (p_i - \bar{p})^2 \sum (f(p_i)-\bar{f(p)})^2}}$
$\sum (f(p_i)-\bar{f(p)})^2 = (126-75.43)^2+(103-75.43)^2+(82-75.43)^2+(75-75.43)^2+(82-75.43)^2+(40-75.43)^2+(20-75.43)^2 \approx 13532.86$
$r = \frac{-12240}{\sqrt{11200 \times 13532.86}} \approx -0.99$

There is a strong negative linear relationship between sale price and number of new homes available—higher prices are associated with fewer available homes.

Answer:

Linear regression function: $f(p) = -1.09p + 315.23$
Correlation coefficient: $r \approx -0.99$