Question

q 10. this question is calculator active. use your calculator to find a linear model for this data 600 550 500 450 400 350 x 309 542 540 593 553 579 y clear all y = 909.105x - 0.821 y = -0.821x + 909.105 y = -0.52x + 579 y = -0.52x - 20.923

Explanation:

Step1: Recall linear model formula

A linear model is of the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y - intercept. We can use the method of least squares to find \( m \) and \( b \), or we can test the given options with the data points.

Step2: Test the first data point (\(x = 350,y = 579\))

• For option \(y=- 0.522x - 26.923\):

Substitute \(x = 350\) into the equation: \(y=-0.522\times350-26.923=-182.7 - 26.923=-209.623
eq579\)

• For option \(y=-0.52x + 579\):

Substitute \(x = 350\): \(y=-0.52\times350 + 579=-182+579 = 397
eq579\)

• For option \(y=-0.821x + 909.105\):

Substitute \(x = 350\): \(y=-0.821\times350+909.105=-287.35 + 909.105 = 621.755\approx622\) (close to 579? Let's check another point)
Substitute \(x = 600\): \(y=-0.821\times600+909.105=-492.6+909.105 = 416.505\). Wait, our \(y\) value at \(x = 600\) is 309. Not close. Wait, maybe I made a mistake. Wait, let's recalculate for \(x = 350\) in \(y=-0.821x + 909.105\): \(-0.821\times350=-287.35\), \(-287.35 + 909.105 = 621.755\), which is not equal to 579. Wait, let's check the fourth option \(y = 909.105x-0.821\): Substitute \(x = 350\), \(y=909.105\times350-0.821=318186.75 - 0.821=318185.929
eq579\)

Wait, maybe I should use the formula for slope \(m=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}\) and \(b=\frac{\sum y - m\sum x}{n}\)

First, list the data points:
\(n = 6\) (since there are 6 pairs of \((x,y)\): \((350,579),(400,553),(450,593),(500,540),(550,542),(600,309)\))

Calculate \(\sum x=350 + 400+450+500+550+600=350 + 400 = 750;750+450 = 1200;1200+500 = 1700;1700+550 = 2250;2250+600 = 2850\)

\(\sum y=579+553+593+540+542+309\)
\(579+553 = 1132;1132+593 = 1725;1725+540 = 2265;2265+542 = 2807;2807+309 = 3116\)

\(\sum xy=350\times579+400\times553+450\times593+500\times540+550\times542+600\times309\)
\(350\times579 = 202650\)
\(400\times553 = 221200\)
\(450\times593 = 266850\)
\(500\times540 = 270000\)
\(550\times542 = 298100\)
\(600\times309 = 185400\)
\(\sum xy=202650+221200+266850+270000+298100+185400\)
\(202650+221200 = 423850\)
\(423850+266850 = 690700\)
\(690700+270000 = 960700\)
\(960700+298100 = 1258800\)
\(1258800+185400 = 1444200\)

\(\sum x^{2}=350^{2}+400^{2}+450^{2}+500^{2}+550^{2}+600^{2}\)
\(350^{2}=122500\)
\(400^{2}=160000\)
\(450^{2}=202500\)
\(500^{2}=250000\)
\(550^{2}=302500\)
\(600^{2}=360000\)
\(\sum x^{2}=122500+160000+202500+250000+302500+360000\)
\(122500+160000 = 282500\)
\(282500+202500 = 485000\)
\(485000+250000 = 735000\)
\(735000+302500 = 1037500\)
\(1037500+360000 = 1397500\)

Now, calculate the slope \(m\):

\(m=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}\)

\(n = 6\), so:

\(n\sum xy=6\times1444200 = 8665200\)

\(\sum x\sum y=2850\times3116=2850\times3000+2850\times116 = 8550000+330600 = 8880600\)

\(n\sum x^{2}=6\times1397500 = 8385000\)

\((\sum x)^{2}=2850^{2}=8122500\)

\(m=\frac{8665200 - 8880600}{8385000-8122500}=\frac{-215400}{262500}\approx - 0.8206\approx - 0.821\)

Now calculate the y - intercept \(b\):

\(b=\frac{\sum y - m\sum x}{n}\)

\(\sum y=3116\), \(m\approx - 0.821\), \(\sum x = 2850\)

\(m\sum x=-0.821\times2850=-2340.85\)

\(\sum y - m\sum x=3116+2340.85 = 5456.85\)

\(b=\frac{5456.85}{6}\approx909.475\approx909.105\) (due to rounding differences)

So the linear model is \(y=-0.821x + 909.105\)

Answer:

\(y=-0.821x + 909.105\) (the option \(y=-0.821x + 909.105\))