data set j: year vs. average gas price ($ per gallon): (2005, 2.30), (2007, 2.85), (2009, 2.50), (2011…

Explanation:

Step1: Recall linear - regression formula

The linear - regression equation is of the form $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. For a set of data points $(x_i,y_i)$ with $i = 1,\cdots,n$, the formulas for $m$ and $b$ are: [m=\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}}] [b=\frac{\sum_{i = 1}^{n}y_i - m\sum_{i = 1}^{n}x_i}{n}] Let $x$ be the year and $y$ be the gas - price. We have $n = 6$ data points: $(x_1,y_1)=(2005,2.30),(x_2,y_2)=(2007,2.85),(x_3,y_3)=(2009,2.50),(x_4,y_4)=(2011,3.40),(x_5,y_5)=(2013,3.60),(x_6,y_6)=(2015,2.80)$. First, calculate the necessary sums: (\sum_{i = 1}^{6}x_i=2005 + 2007+2009+2011+2013+2015=12060) (\sum_{i = 1}^{6}y_i=2.30 + 2.85+2.50+3.40+3.60+2.80 = 17.45) (\sum_{i = 1}^{6}x_i^{2}=2005^{2}+2007^{2}+2009^{2}+2011^{2}+2013^{2}+2015^{2}) (=4020025+4028049+4036081+4044121+4052169+4060225 = 24240670) (\sum_{i = 1}^{6}x_iy_i=2005\times2.30+2007\times2.85+2009\times2.50+2011\times3.40+2013\times3.60+2015\times2.80) (=4611.5+5729.95+5022.5+6837.4+7246.8+5642=35089.15)

Step2: Calculate the slope $m$

[m=\frac{6\times35089.15-12060\times17.45}{6\times24240670 - 12060^{2}}] [=\frac{210534.9-210447}{145444020 - 145443600}] [=\frac{87.9}{420}\approx0.2093]

Step3: Calculate the y - intercept $b$

[b=\frac{17.45-0.2093\times12060}{6}] [=\frac{17.45 - 2524.158}{6}] [=\frac{-2506.708}{6}\approx - 417.7847] The regression equation is $y = 0.2093x-417.7847$.

The formula for the correlation coefficient $r$ is: [r=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{\sqrt{\left[n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}\right]\left[n\sum_{i = 1}^{n}y_i^{2}-(\sum_{i = 1}^{n}y_i)^{2}\right]}}] First, calculate $\sum_{i = 1}^{6}y_i^{2}=2.30^{2}+2.85^{2}+2.50^{2}+3.40^{2}+3.60^{2}+2.80^{2}$ (=5.29+8.1225+6.25+11.56+12.96+7.84 = 51.0225) [r=\frac{6\times35089.15-12060\times17.45}{\sqrt{(6\times24240670 - 12060^{2})(6\times51.0225-17.45^{2})}}] [=\frac{87.9}{\sqrt{420\times(306.135 - 304.5025)}}] [=\frac{87.9}{\sqrt{420\times1.6325}}] [=\frac{87.9}{\sqrt{685.65}}] [=\frac{87.9}{26.185}\approx0.336]

Answer:

Question 10: $y = 0.2093x-417.7847$ Question 11: $r\approx0.336$