Question

(b) the data seems to have a linear trend. so, leila wants to find a linear equation that could approximately model the data. select \compute\ to generate the values used for a linear regression model. write the values in your answer to 3 decimal places.
the linear equation that best fits the data is ( y = square ).
(c) use the equation from part (b) to help leila predict the value of ( y ) when ( x = 9 ). write your answer to 3 decimal places.
( y = square )

Explanation:

Step1: Calculate necessary sums

First, compute sums for regression:
$n=7$
$\sum x = 1+2+3+5+6+7+8 = 32$
$\sum y = 1+3+2+6+7+7+9 = 35$
$\sum xy = (1×1)+(2×3)+(3×2)+(5×6)+(6×7)+(7×7)+(8×9) = 1+6+6+30+42+49+72 = 206$
$\sum x^2 = 1^2+2^2+3^2+5^2+6^2+7^2+8^2 = 1+4+9+25+36+49+64 = 188$
$\sum y^2 = 1^2+3^2+2^2+6^2+7^2+7^2+9^2 = 1+9+4+36+49+49+81 = 229$

Step2: Calculate slope $a$

Use regression slope formula:
$$a = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$
$$a = \frac{7×206 - 32×35}{7×188 - 32^2} = \frac{1442 - 1120}{1316 - 1024} = \frac{322}{292} ≈ 1.103$$

Step3: Calculate intercept $b$

Use regression intercept formula:
$$b = \frac{\sum y - a\sum x}{n}$$
$$b = \frac{35 - 1.103×32}{7} = \frac{35 - 35.296}{7} = \frac{-0.296}{7} ≈ -0.042$$

Step4: Calculate correlation $r$

Use correlation coefficient formula:
$$r = \frac{n\sum xy - \sum x \sum y}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}}$$
$$n\sum y^2 - (\sum y)^2 = 7×229 - 35^2 = 1603 - 1225 = 378$$
$$r = \frac{322}{\sqrt{292×378}} = \frac{322}{\sqrt{110376}} ≈ \frac{322}{332.23} ≈ 0.969$$

Step5: Calculate $r^2$

Square the correlation coefficient:
$$r^2 = (0.969)^2 ≈ 0.939$$

Step6: Predict $y$ at $x=9$

Use the linear equation $y=ax+b$:
$$y = 1.103×9 + (-0.042) = 9.927 - 0.042 = 9.885$$

Answer:

Part (b)

$a=1.103$, $b=-0.042$, $r^2=0.939$, $r=0.969$
The linear equation that best fits the data is $y = 1.103x - 0.042$

Part (c)

$y=9.885$