section 12.2 linear regression hawkes learning systems math courseware specialists copyright © 2008...
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Section 12.2
Linear Regression
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Copyright © 2008 by Hawkes Learning
Systems/Quant Systems, Inc.
All rights reserved.
• If r is statistically significant, then a regression line can be used to make predictions regarding the data.
• Regression Line – the line from which the average variation from the data is the smallest. Also known as the line of best fit. There is only one regression line for each data set.
Other commonly used regression line formulas:
Definitions:
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12.2 Linear Regression
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b0 y-interceptb1 slope
Slope:
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12.2 Linear Regression
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y-Intercept:
When calculating the slope, round your answers to three decimal places.
When calculating the y-intercept, round your answers to three decimal places.
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TI-84 Plus Instructions:
1. Press STAT, then EDIT
2. Type the x-variable values into L1
3. Type the y-variable values into L2
4. Press STAT, then CALC
5. Choose 4: Linreg(ax+b)
6. Press ENTER
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12.2 Linear Regression
a b1 slopeb b0 y-intercept
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Determine the regression line:The local school board wants to evaluate the relationship between class size and performance on the state achievement test. They decide to collect data from various schools in the county. A portion of the data collected is listed below. Each pair represents the class size and corresponding average score on the achievement test for 8 schools.
Solution:
First decide which variable should be the x variable and which variable should be y.Class size is the independent variable, x, and average score is the dependent variable, y.
We can use a calculator or the equations to find the regression line.
Class Size 15 17 18 20 21 24 26 29
Average Score 85.3 86.2 85.0 82.7 81.9 78.8 75.3 72.1
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12.2 Linear Regression
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Solution (continued):
Using the equations for slope and y-intercept:n 8, ∑x170, ∑y 647.3, ∑xy13588.7, ∑x 23772
Now place the slope and y-intercept into the regression line equation:
–1.043
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12.2 Linear Regression
103.085
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Predictions:
A prediction should not be made with a regression model if:
1. The correlation is not statistically significant.
2. You are using a value outside of the range of the sample data.
3. The population is different than that of the sample data.
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12.2 Linear Regression
Calculate the prediction:
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a. 16
b. 19
c. 25
d. 45
Since 45 is outside the range of the original data of 15 to 29, we cannot predict the test score.
Use the equation of the regression line,
to predict what the average achievement test score will be for the following class size:
Regression, Inference, and Model Building
12.2 Linear Regression
86.397
83.268
77.010
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Putting it all together:The table below gives the average monthly temperature and corresponding monthly precipitation totals for one year in Key West, Florida.
a. Create a scatter plot.b. Calculate the correlation coefficient, r.c. Verify that the correlation coefficient is statistically significant at the 0.05
level of significance.d. Calculate the equation of the line of best fit.e. Calculate and interpret the coefficient of determination, r 2.f. If appropriate, make a prediction for the monthly precipitation for a
month in which the average temperature is 80 degrees.g. If appropriate, make a prediction for the monthly precipitation in Destin,
FL for a month in which the average temperature is 83 degrees.
Temperatures (in °F) and Precipitation (in inches) in Key West, FL
Average Temperature
75 76 79 82 85 88 89 90 88 85 81 77
Inches in Precipitation
2.22 1.51 1.86 2.06 3.48 4.57 3.27 5.40 5.45 4.34 2.64 2.14
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12.2 Linear Regression
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Solution (continued):
a. Create a scatter plot.
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12.2 Linear Regression
The scatter plot shows a positive trend in data.
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Solution (continued):
b. Calculate the correlation coefficient, r.
Regression, Inference, and Model Building
12.2 Linear Regression
n 12, ∑x995, ∑y38.94, ∑xy3299.23, ∑x282815, ∑y2147.847
0.859
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Solution (continued):
c. Verify that the correlation coefficient is statistically significant at the 0.05 level of significance.
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12.2 Linear Regression
n 12, r 0.859, 0.05
r
Statistically significant if |r | > r .
Therefore, r is statistically significant at the 0.05 level.
0.576
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Solution (continued):
n 12, ∑x995, ∑y38.94, ∑xy3299.23, ∑x282815
Now place the slope and y-intercept into the regression line equation:
0.225
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12.2 Linear Regression
–15.424
d. Calculate the equation of the line of best fit.
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r 0.859
r 2
So about 73.8% of the variation in precipitation can be attributed to the linear relationship between temperature and precipitation. The remaining 26.2% of the variation is from unknown sources.
0.738
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12.2 Linear Regression
Solution (continued):
e. Calculate and interpret the coefficient of determination, r 2.
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Thus a reasonable estimate for the precipitation for a month in which the average temperature is 80 degrees is about 2.58 inches.
Regression, Inference, and Model Building
12.2 Linear Regression
Solution (continued):
f. If appropriate, make a prediction for the monthly precipitation for a month in which the average temperature is 80 degrees.
2.576
g. If appropriate, make a prediction for the monthly precipitation in Destin, FL for a month in which the average temperature is 83 degrees.
The data was collected in Key West, not Destin, FL. Therefore, it is not appropriate to use the linear regression line to make predictions regarding the precipitation in Destin.