Question

(b) the data seems to have a linear trend. so, yoko 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.

x	1	2	3	4	5	7	8
y	2	3	3	4	6	6	8

lin reg: ( y = ax + b )
( a approx 0.808 )
( b approx 1.109 )
( r^2 approx 0.929 )
( r approx 0.964 )

the linear equation that best fits the data is ( y = square ).

(c) use the equation from part (b) to help yoko predict the value of ( y ) when ( x = 6 ). write your answer to 3 decimal places.

( y = square )

Explanation:

Part (b)

Step1: Identify linear regression form

The linear regression model is given by \( y = ax + b \), where \( a \) is the slope and \( b \) is the y - intercept. From the given data, \( a\approx0.808 \) and \( b\approx1.109 \).

Step2: Write the linear equation

Substitute the values of \( a \) and \( b \) into the equation \( y = ax + b \). So the equation is \( y = 0.808x+1.109 \).

Step1: Use the linear equation

We have the linear equation \( y = 0.808x+1.109 \) from part (b). We need to find the value of \( y \) when \( x = 6 \).

Step2: Substitute \( x = 6 \) into the equation

Substitute \( x = 6 \) into \( y=0.808x + 1.109 \). So \( y=0.808\times6+1.109 \). First, calculate \( 0.808\times6=4.848 \). Then, add \( 1.109 \) to it: \( 4.848 + 1.109=5.957 \).

Answer:

\( y = 0.808x + 1.109 \)

Part (c)