Question

create a linear model for the data in the table.
x | 5 | 8 | 11 | 14 | 17 | 20
y | 7 | 16 | 21 | 28 | 38 | 43
write a linear model for the data in the table.
y = □x + (□)
(type integers or decimals rounded to three decimal places as needed.)

Explanation:

Step1: Calculate the slope \( m \)

The formula for the slope \( m \) of a linear model using two points \((x_1, y_1)\) and \((x_2, y_2)\) is \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Let's take the first two points \((5, 7)\) and \((8, 16)\).
\( m=\frac{16 - 7}{8 - 5}=\frac{9}{3} = 3 \). We can verify with another pair, say \((8, 16)\) and \((11, 21)\): \( m=\frac{21 - 16}{11 - 8}=\frac{5}{3}\approx1.666 \)? Wait, no, maybe I made a mistake. Wait, let's use the formula for the slope of the line of best fit. The formula for the slope \( m \) is \( m=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}} \), where \( n = 6 \) (number of data points).

First, calculate \( \sum x \), \( \sum y \), \( \sum xy \), \( \sum x^{2} \):

\( \sum x=5 + 8+11 + 14+17+20=75 \)

\( \sum y=7 + 16+21+28+38+43 = 153 \)

\( \sum xy=(5\times7)+(8\times16)+(11\times21)+(14\times28)+(17\times38)+(20\times43)=35 + 128+231+392+646+860 = 2292 \)

\( \sum x^{2}=5^{2}+8^{2}+11^{2}+14^{2}+17^{2}+20^{2}=25 + 64+121+196+289+400 = 1095 \)

Now, \( n = 6 \)

\( m=\frac{6\times2292-75\times153}{6\times1095-(75)^{2}}=\frac{13752 - 11475}{6570 - 5625}=\frac{2277}{945}\approx2.410 \)

Step2: Calculate the y - intercept \( b \)

The formula for the y - intercept \( b \) is \( b=\frac{\sum y - m\sum x}{n} \)

\( b=\frac{153-2.410\times75}{6}=\frac{153 - 180.75}{6}=\frac{- 27.75}{6}\approx - 4.625 \)

Wait, let's recalculate the slope more accurately.

\( 6\times2292=13752 \), \( 75\times153 = 11475 \), \( 13752-11475 = 2277 \)

\( 6\times1095 = 6570 \), \( 75^{2}=5625 \), \( 6570 - 5625=945 \)

\( m=\frac{2277}{945}=\frac{2277\div9}{945\div9}=\frac{253}{105}\approx2.410 \) (rounded to three decimal places)

\( \sum y=153 \), \( m\sum x=2.4095238\times75\approx180.714 \)

\( b=\frac{153 - 180.714}{6}=\frac{- 27.714}{6}\approx - 4.619 \) (Wait, maybe my initial point - slope was wrong because the data may not be perfectly linear, so we need to use the least - squares method.)

Wait, let's use two points again. Let's take \((5,7)\) and \((20,43)\). The slope \( m=\frac{43 - 7}{20 - 5}=\frac{36}{15}=2.4 \)

Then using the point - slope form \( y - y_1=m(x - x_1) \), with \((x_1,y_1)=(5,7)\)

\( y - 7=2.4(x - 5) \)

\( y=2.4x-12 + 7 \)

\( y = 2.4x-5 \)

Wait, let's check with \( x = 8 \): \( y=2.4\times8-5=19.2 - 5 = 14.2 \), but the actual \( y \) is 16. Hmm, not good.

Wait, let's use the least - squares formula correctly.

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

\( n = 6 \), \( \sum x=75 \), \( \sum y = 153 \), \( \sum xy=2292 \), \( \sum x^{2}=1095 \)

\( m=\frac{6\times2292-75\times153}{6\times1095 - 75^{2}}=\frac{13752-11475}{6570 - 5625}=\frac{2277}{945}=2.41 \) (rounded to two decimal places)

\( b=\frac{\sum y - m\sum x}{n}=\frac{153-2.41\times75}{6}=\frac{153 - 180.75}{6}=\frac{-27.75}{6}=-4.625 \)

Now check with \( x = 5 \): \( y=2.41\times5-4.625 = 12.05 - 4.625 = 7.425 \), close to 7.

\( x = 8 \): \( y=2.41\times8-4.625=19.28 - 4.625 = 14.655 \), actual \( y = 16 \), still some error.

Wait, maybe I miscalculated \( \sum xy \). Let's recalculate \( \sum xy \):

\( 5\times7 = 35 \)

\( 8\times16=128 \)

\( 11\times21 = 231 \)

\( 14\times28=392 \)

\( 17\times38 = 646 \)

\( 20\times43 = 860 \)

\( 35+128 = 163 \), \( 163+231 = 394 \), \( 394+392 = 786 \), \( 786+646 = 1432 \), \( 1432+860 = 2292 \). That's correct.

\( \sum x = 5 + 8+11+14+17+20=75 \), correct.

\( \sum y=7 + 16+21+28+38+43 = 153 \), correct.

\( \sum x^{2}=25 + 64+121+196+289+400 = 1095 \), correct.

So \( m=\frac{6\times2292-75\times153}{6\times1095 - 75^{2}}=\frac{13752 - 11475}{6570 - 5625}=\frac{2277}{945}=2.41 \) (exact fraction is \( \frac{253}{105}\approx2.410 \))

\( b=\frac{153-2.4095238\times75}{6}=\frac{153 - 180.7142857}{6}=\frac{-27.7142857}{6}\approx - 4.619 \)

So the linear model is \( y = 2.410x-4.619 \) (rounded to three decimal places)

Wait, let's check with \( x = 11 \): \( y=2.410\times11-4.619=26.51 - 4.619 = 21.891 \), actual \( y = 21 \), close.

\( x = 14 \): \( y=2.410\times14-4.619=33.74 - 4.619 = 29.121 \), actual \( y = 28 \), close.

\( x = 17 \): \( y=2.410\times17-4.619=40.97 - 4.619 = 36.351 \), actual \( y = 38 \), a bit off, but it's a linear model.

Alternatively, if we use the two - point formula with \((5,7)\) and \((17,38)\):

\( m=\frac{38 - 7}{17 - 5}=\frac{31}{12}\approx2.583 \), then \( y - 7=\frac{31}{12}(x - 5) \), \( y=\frac{31}{12}x-\frac{155}{12}+7=\frac{31}{12}x-\frac{155 - 84}{12}=\frac{31}{12}x-\frac{71}{12}\approx2.583x - 5.917 \). But this is less accurate than the least - squares method.

So using the least - squares method:

\( m\approx2.410 \), \( b\approx - 4.619 \)

Answer:

\( y=\boxed{2.410}x+(\boxed{-4.619}) \)