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QuestionDetermine the least-squares line and correlation coefficient for points (−2,6),(2,1),(6,−3)(-2,6),(2,1),(6,-3). Plot the data and line.

Studdy Solution

STEP 1

Assumptions1. The given data points are (−,6),(,1),(6,−3)(-,6),(,1),(6,-3). We are looking for the line of least-squares fit, which minimizes the sum of the squares of the residuals (the distances of the points from the line)
3. We are also looking for the correlation coefficient, which measures the strength and direction of a linear relationship between two variables.

STEP 2

First, we need to calculate the means of the x-values and the y-values. The mean is calculated as the sum of the values divided by the number of values.
xˉ=1n∑i=1nxi\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_iyˉ=1n∑i=1nyi\bar{y} = \frac{1}{n}\sum_{i=1}^{n} y_i

STEP 3

Plug in the given values for the x-values and y-values to calculate the means.
xˉ=−2+2+63\bar{x} = \frac{-2 +2 +6}{3}yˉ=6+1−33\bar{y} = \frac{6 +1 -3}{3}

STEP 4

Calculate the means of the x-values and the y-values.
xˉ=63=2\bar{x} = \frac{6}{3} =2yˉ=43≈1.33\bar{y} = \frac{4}{3} \approx1.33

STEP 5

Next, we need to calculate the slope of the line of least-squares fit. The slope is calculated as the ratio of the sum of the products of the differences from the means for x and y to the sum of the squares of the differences from the mean for x.
m=∑i=1n(xi−xˉ)(yi−yˉ)∑i=1n(xi−xˉ)2m = \frac{\sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n} (x_i - \bar{x})^2}

STEP 6

Plug in the given values for the x-values, y-values, and the means to calculate the slope.
m=(−2−2)(6−1.33)+(2−2)(1−1.33)+(6−2)(−3−1.33)(−2−2)2+(2−2)2+(6−2)2m = \frac{(-2 -2)(6 -1.33) + (2 -2)(1 -1.33) + (6 -2)(-3 -1.33)}{(-2 -2)^2 + (2 -2)^2 + (6 -2)^2}

STEP 7

Calculate the slope of the line of least-squares fit.
m=−4(4.67)+0(−0.33)+4(−4.33)16+0+16=−1m = \frac{-4(4.67) +0(-0.33) +4(-4.33)}{16 +0 +16} = -1

STEP 8

Now that we have the slope, we can find the y-intercept of the line of least-squares fit. The y-intercept is calculated as the difference between the mean of the y-values and the product of the slope and the mean of the x-values.
b=yˉ−mxˉb = \bar{y} - m\bar{x}

STEP 9

Plug in the values for the slope and the means to calculate the y-intercept.
b=.33−(−)(2)b =.33 - (-)(2)

STEP 10

Calculate the y-intercept of the line of least-squares fit.
b=.33+2=3.33b =.33 +2 =3.33

STEP 11

Now that we have the slope and the y-intercept, we can write the equation of the line of least-squares fit.
y=mx+by = mx + b

STEP 12

Plug in the values for the slope and the y-intercept to write the equation of the line.
y=−x+.33y = -x +.33

STEP 13

Next, we need to calculate the correlation coefficient. The correlation coefficient is calculated as the slope of the line of least-squares fit times the standard deviation of the y-values divided by the standard deviation of the x-values.
r=msysxr = m\frac{s_y}{s_x}

STEP 14

First, we need to calculate the standard deviations of the x-values and the y-values. The standard deviation is calculated as the square root of the variance, which is the sum of the squares of the differences from the mean divided by the number of values.
sx=n∑i=n(xi−xˉ)2s_x = \sqrt{\frac{}{n}\sum_{i=}^{n} (x_i - \bar{x})^2}sy=n∑i=n(yi−yˉ)2s_y = \sqrt{\frac{}{n}\sum_{i=}^{n} (y_i - \bar{y})^2}

STEP 15

Plug in the given values for the x-values, y-values, and the means to calculate the standard deviations.
sx=(−2−2)2+(2−2)2+(−2)23s_x = \sqrt{\frac{(-2 -2)^2 + (2 -2)^2 + ( -2)^2}{3}}sy=(−.33)2+(−.33)2+(−3−.33)23s_y = \sqrt{\frac{( -.33)^2 + ( -.33)^2 + (-3 -.33)^2}{3}}

STEP 16

Calculate the standard deviations of the x-values and the y-values.
sx=16+0+163=10.67≈3.27s_x = \sqrt{\frac{16 +0 +16}{3}} = \sqrt{10.67} \approx3.27sy=22.09+0.11+18.783=13.66≈3.70s_y = \sqrt{\frac{22.09 +0.11 +18.78}{3}} = \sqrt{13.66} \approx3.70

STEP 17

Now that we have the standard deviations, we can calculate the correlation coefficient.
r=−3.703.27r = -\frac{3.70}{3.27}

STEP 18

Calculate the correlation coefficient.
r=−.13r = -.13The line of least-squares fit is y=−x+3.33y = -x +3.33 and the correlation coefficient is −.13-.13.To plot the data and graph the line, you would plot the given data points on a graph, then draw the line y=−x+3.33y = -x +3.33.

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