Question

815 stephen collected data from a travel website. the data included a hotels distance from times square in manhattan and the cost of a room for one weekend night in august. a table containing these data appears below.

distance from times square (city blocks) (x)	0	0	1	1	3	4	7	11	14	19

write the linear regression equation for this data set. round all values to the nearest hundredth. state the correlation coefficient for this data set, to the nearest hundredth. explain what the sign of the correlation coefficient suggests in the context of the problem.

Explanation:

Step1: List data pairs

x: [0,0,1,1,3,4,7,11,14,19]; y: [293,263,244,224,185,170,219,153,136,111]

Step2: Calculate sums

$\sum x=0+0+1+1+3+4+7+11+14+19=60$
$\sum y=293+263+244+224+185+170+219+153+136+111=2008$
$\sum xy=0×293+0×263+1×244+1×224+3×185+4×170+7×219+11×153+14×136+19×111=0+0+244+224+555+680+1533+1683+1904+2109=8932$
$\sum x^2=0^2+0^2+1^2+1^2+3^2+4^2+7^2+11^2+14^2+19^2=0+0+1+1+9+16+49+121+196+361=754$
$n=10$

Step3: Compute slope (b)

$b=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=\frac{10×8932-60×2008}{10×754-60^2}=\frac{89320-120480}{7540-3600}=\frac{-31160}{3940}≈-7.91$

Step4: Compute intercept (a)

$\bar{x}=\frac{60}{10}=6$, $\bar{y}=\frac{2008}{10}=200.8$
$a=\bar{y}-b\bar{x}=200.8-(-7.91)×6=200.8+47.46=248.26$

Step5: Linear regression equation

$\hat{y}=a+bx$

Step6: Calculate correlation coefficient (r)

$\sum y^2=293^2+263^2+244^2+224^2+185^2+170^2+219^2+153^2+136^2+111^2=85849+69169+59536+50176+34225+28900+47961+23409+18496+12321=420042$
$r=\frac{n\sum xy-\sum x\sum y}{\sqrt{[n\sum x^2-(\sum x)^2][n\sum y^2-(\sum y)^2]}}=\frac{-31160}{\sqrt{3940×(10×420042-2008^2)}}=\frac{-31160}{\sqrt{3940×(4200420-4032064)}}=\frac{-31160}{\sqrt{3940×168356}}≈\frac{-31160}{\sqrt{663322640}}≈\frac{-31160}{25755.05}≈-0.98$

Answer:

Linear regression equation: $\hat{y}=248.26-7.91x$
Correlation coefficient: $-0.98$
The negative sign suggests a strong negative linear relationship: as distance from Times Square increases, room cost tends to decrease.