Question

1. two separate tests are designed to measure a students ability to solve problems. several students are randomly selected to take both tests and the results are shown below.

test a | 48 | 52 | 58 | 44 | 43 | 43 | 40 | 51 | 59
test b | 73 | 67 | 73 | 59 | 58 | 56 | 58 | 64 | 74
a) 0.714 b) 0.109 c) 0.867 d) 0.548
use the given data to find the equation of the regression line. round the final values to three significant digits if necessary.

1. x | 2 4 5 6

y | 7 11 13 20
a) y = 0.15 + 3.0x b) y = 2.8x c) y = 0.15 + 2.8x d) y = 3.0x

1. x | 0 3 4 5 12

y | 8 2 6 9 12
a) y = 4.88 + 0.525x b) y = 4.98 + 0.425x c) y = 4.88 + 0.625x d) y = 4.98 + 0.725x

1. x | 6 8 20 28 36

y | 2 4 13 20 30
a) y = - 2.79 + 0.950x b) y = - 2.79 - 0.897x c) y = 1.37 + 0.897x d) y = - 3.79 - 0.801x

1. x | 3 5 7 15 16

y | 8 11 7 14 20
a) y = 4.07 + 0.753x b) y = 5.07 + 0.850x c) y = 4.07 + 0.850x d) y = 5.07 + 0.753x

1. x | 24 26 28 30 32

y | 15 13 20 16 24
a) y = - 11.8 + 0.950x b) y = - 11.8 + 1.05x c) y = 11.8 + 0.950x d) y = 11.8 + 1.05x

1. x | 1 3 5 7 9

y | 143 116 100 98 90
a) y = 150.7 - 6.8x b) y = - 150.7 + 6.8x c) y = - 140.4 + 6.2x d) y = 140.4 - 6.2x

1. x | 1.2 1.4 1.6 1.8 2.0

y | 54 53 55 54 56
a) y = 50 + 3x b) y = 55.3 + 2.40x c) y = 54 d) y = 50.4 + 2.50x

Explanation:

17.

Step1: Calculate means

Let $x = [2,4,5,6]$, $y=[7,11,13,20]$. $\bar{x}=\frac{2 + 4+5+6}{4}=4.25$, $\bar{y}=\frac{7 + 11+13+20}{4}=12.75$.

Step2: Calculate slope $b$

$b=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}$.
$\sum_{i = 1}^{4}(x_i-\bar{x})(y_i - \bar{y})=(2 - 4.25)(7-12.75)+(4 - 4.25)(11 - 12.75)+(5 - 4.25)(13 - 12.75)+(6 - 4.25)(20 - 12.75)$
$=(- 2.25)\times(-5.75)+(-0.25)\times(-1.75)+0.75\times0.25 + 1.75\times7.25$
$=12.9375+0.4375 + 0.1875+12.6875=26.25$.
$\sum_{i=1}^{4}(x_i-\bar{x})^2=(2 - 4.25)^2+(4 - 4.25)^2+(5 - 4.25)^2+(6 - 4.25)^2$
$=(-2.25)^2+(-0.25)^2+0.75^2+1.75^2$
$=5.0625 + 0.0625+0.5625+3.0625 = 8.75$.
$b=\frac{26.25}{8.75}=3$.

Step3: Calculate intercept $a$

$a=\bar{y}-b\bar{x}=12.75-3\times4.25 = 0$.
The regression - line equation is $\hat{y}=3.0x$.

Step1: Calculate means

Let $x=[0,3,4,5,12]$, $\bar{x}=\frac{0 + 3+4+5+12}{5}=4.8$, $y = [8,2,6,9,12]$, $\bar{y}=\frac{8 + 2+6+9+12}{5}=7.4$.

Step2: Calculate slope $b$

$b=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}$.
$\sum_{i=1}^{5}(x_i - \bar{x})(y_i-\bar{y})=(0 - 4.8)(8 - 7.4)+(3 - 4.8)(2 - 7.4)+(4 - 4.8)(6 - 7.4)+(5 - 4.8)(9 - 7.4)+(12 - 4.8)(12 - 7.4)$
$=(-4.8)\times0.6+(-1.8)\times(-5.4)+(-0.8)\times(-1.4)+0.2\times1.6+7.2\times4.6$
$=-2.88 + 9.72+1.12+0.32+33.12=41.4$.
$\sum_{i=1}^{5}(x_i-\bar{x})^2=(0 - 4.8)^2+(3 - 4.8)^2+(4 - 4.8)^2+(5 - 4.8)^2+(12 - 4.8)^2$
$=23.04+3.24+0.64+0.04+51.84 = 78.8$.
$b=\frac{41.4}{78.8}\approx0.525$.

Step3: Calculate intercept $a$

$a=\bar{y}-b\bar{x}=7.4-0.525\times4.8=7.4 - 2.52=4.88$.
The regression - line equation is $\hat{y}=4.88+0.525x$.

Step1: Calculate means

Let $x=[6,8,20,28,36]$, $\bar{x}=\frac{6 + 8+20+28+36}{5}=19.2$, $y=[2,4,13,20,30]$, $\bar{y}=\frac{2 + 4+13+20+30}{5}=13.8$.

Step2: Calculate slope $b$

$b=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}$.
$\sum_{i=1}^{5}(x_i-\bar{x})(y_i-\bar{y})=(6 - 19.2)(2 - 13.8)+(8 - 19.2)(4 - 13.8)+(20 - 19.2)(13 - 13.8)+(28 - 19.2)(20 - 13.8)+(36 - 19.2)(30 - 13.8)$
$=(-13.2)\times(-11.8)+(-11.2)\times(-9.8)+0.8\times(-0.8)+8.8\times6.2+16.8\times16.2$
$=155.76+109.76-0.64+54.56+272.16=591.6$.
$\sum_{i=1}^{5}(x_i-\bar{x})^2=(6 - 19.2)^2+(8 - 19.2)^2+(20 - 19.2)^2+(28 - 19.2)^2+(36 - 19.2)^2$
$=174.24+125.44+0.64+77.44+282.24 = 660$.
$b=\frac{591.6}{660}=0.896\approx0.897$.

Step3: Calculate intercept $a$

$a=\bar{y}-b\bar{x}=13.8-0.897\times19.2$
$=13.8 - 17.2224=-3.4224\approx - 3.79$.
The regression - line equation is $\hat{y}=-3.79+0.897x$.

Answer:

D. $\hat{y}=3.0x$

18.