Question

the gas mileage of an automobile first increases and then decreases as the speed increases. suppose this relationship is very regular, as shown by the data in the table on speed (miles per hour) and mileage (miles per gallon). to access the data, click the link for your preferred software format. csv excel (xls) excel (xlsx) jmp mac - text minitab14 - 18 minitab18+ pc - text r spss ti crunchit! make a scatterplot of mileage versus speed. show that the correlation between speed and mileage is r = 0. find the seven standardized values for each variable. finally, find the values for z_xz_y for each observation. give your answers rounded to four decimal places. z_xz_y(20,21)= z_xz_y(30,26)= z_xz_y(40,29)= z_xz_y(50,30)= z_xz_y(60,29)= z_xz_y(70,26)= z_xz_y(80,21)=

Explanation:

Step1: Calculate mean of speed ($\bar{x}$)

$\bar{x}=\frac{20 + 30+40+50+60+70+80}{7}=\frac{350}{7} = 50$

Step2: Calculate mean of mileage ($\bar{y}$)

$\bar{y}=\frac{21 + 26+29+30+29+26+21}{7}=\frac{182}{7}=26$

Step3: Calculate standard - deviation of speed ($s_x$)

$s_x=\sqrt{\frac{(20 - 50)^2+(30 - 50)^2+(40 - 50)^2+(50 - 50)^2+(60 - 50)^2+(70 - 50)^2+(80 - 50)^2}{7-1}}$
$=\sqrt{\frac{900 + 400+100+0+100+400+900}{6}}=\sqrt{\frac{2800}{6}}\approx21.6025$

Step4: Calculate standard - deviation of mileage ($s_y$)

$s_y=\sqrt{\frac{(21 - 26)^2+(26 - 26)^2+(29 - 26)^2+(30 - 26)^2+(29 - 26)^2+(26 - 26)^2+(21 - 26)^2}{7 - 1}}$
$=\sqrt{\frac{25+0 + 9+16+9+0+25}{6}}=\sqrt{\frac{84}{6}}=\sqrt{14}\approx3.7417$

Step5: Calculate $z_{x - y}$ for each pair

For $(20,21)$

$z_{x(20)}=\frac{20 - 50}{21.6025}\approx - 1.3887$
$z_{y(21)}=\frac{21 - 26}{3.7417}\approx - 1.3363$
$z_{x - y(20,21)}=z_{x(20)}\times z_{y(21)}\approx(-1.3887)\times(-1.3363)\approx1.8557$

For $(30,26)$

$z_{x(30)}=\frac{30 - 50}{21.6025}\approx - 0.9258$
$z_{y(26)}=\frac{26 - 26}{3.7417}=0$
$z_{x - y(30,26)}=z_{x(30)}\times z_{y(26)} = 0$

For $(40,29)$

$z_{x(40)}=\frac{40 - 50}{21.6025}\approx - 0.4629$
$z_{y(29)}=\frac{29 - 26}{3.7417}\approx0.8018$
$z_{x - y(40,29)}=z_{x(40)}\times z_{y(29)}\approx(-0.4629)\times0.8018\approx - 0.3712$

For $(50,30)$

$z_{x(50)}=\frac{50 - 50}{21.6025}=0$
$z_{y(30)}=\frac{30 - 26}{3.7417}\approx1.0690$
$z_{x - y(50,30)}=z_{x(50)}\times z_{y(30)} = 0$

For $(60,29)$

$z_{x(60)}=\frac{60 - 50}{21.6025}\approx0.4629$
$z_{y(29)}=\frac{29 - 26}{3.7417}\approx0.8018$
$z_{x - y(60,29)}=z_{x(60)}\times z_{y(29)}\approx0.4629\times0.8018\approx0.3712$

For $(70,26)$

$z_{x(70)}=\frac{70 - 50}{21.6025}\approx0.9258$
$z_{y(26)}=\frac{26 - 26}{3.7417}=0$
$z_{x - y(70,26)}=z_{x(70)}\times z_{y(26)} = 0$

For $(80,21)$

$z_{x(80)}=\frac{80 - 50}{21.6025}\approx1.3887$
$z_{y(21)}=\frac{21 - 26}{3.7417}\approx - 1.3363$
$z_{x - y(80,21)}=z_{x(80)}\times z_{y(21)}\approx1.3887\times(-1.3363)\approx - 1.8557$

Answer:

$z_{x - y(20,21)}\approx1.8557$
$z_{x - y(30,26)} = 0$
$z_{x - y(40,29)}\approx - 0.3712$
$z_{x - y(50,30)} = 0$
$z_{x - y(60,29)}\approx0.3712$
$z_{x - y(70,26)} = 0$
$z_{x - y(80,21)}\approx - 1.8557$