Question

temperatures: following are average temperatures, in degrees fahrenheit, during the months of january and july for 6 u.s. cities. use the ti - 84 plus calculator as needed.

city	january temperature	july temperature
st. louis	29.6	80.2
denver	29.2	73.4
chicago	22.0	73.3
houston	51.8	83.6
washington, d.c.	34.9	79.2

send data to excel

part: 0 / 4

part 1 of 4

(a) compute the least - squares regression line for predicting july temperature from january temperature.
the equation for the least - squares regression line is $hat{y}=square$. round the slope and y - intercept to four decimal places as needed.

Explanation:

Step1: Define variables

Let $x$ be the January temperature and $y$ be the July temperature. We have $n = 6$ data - points: $(x_1,y_1)=(32.1,76.5),(x_2,y_2)=(29.6,80.2),(x_3,y_3)=(29.2,73.4),(x_4,y_4)=(22.0,73.3),(x_5,y_5)=(51.8,83.6),(x_6,y_6)=(34.9,79.2)$.

Step2: Calculate sums

$\sum_{i = 1}^{n}x_i=32.1 + 29.6+29.2+22.0+51.8+34.9 = 199.6$
$\sum_{i = 1}^{n}y_i=76.5 + 80.2+73.4+73.3+83.6+79.2 = 466.2$
$\sum_{i = 1}^{n}x_i^2=32.1^2+29.6^2+29.2^2+22.0^2+51.8^2+34.9^2$
$=1030.41+876.16+852.64+484+2683.24+1218.01 = 7144.46$
$\sum_{i = 1}^{n}x_iy_i=32.1\times76.5+29.6\times80.2+29.2\times73.4+22.0\times73.3+51.8\times83.6+34.9\times79.2$
$=2455.65+2373.92+2143.28+1602.6+4330.48+2764.08 = 15670.01$

Step3: Calculate slope $b_1$

$b_1=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2}$
$=\frac{6\times15670.01 - 199.6\times466.2}{6\times7144.46-199.6^2}$
$=\frac{94020.06-93053.52}{42866.76 - 39840.16}$
$=\frac{966.54}{3026.6}=0.3194$

Step4: Calculate y - intercept $b_0$

$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{199.6}{6}\approx33.2667$
$\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{466.2}{6}=77.7$
$b_0=\bar{y}-b_1\bar{x}=77.7-0.3194\times33.2667$
$=77.7 - 10.6254=67.0746$

Answer:

$\hat{y}=67.0746 + 0.3194x$