© 2010 Pearson Prentice Hall. All rights reserved

Least Squares Regression Models

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The least-squares regression model is given by

where• yi is the value of the response variable for the

ith individual• 0 and 1 are the parameters to be estimated

based on sample data• xi is the value of the explanatory variable for the

ith individual• i is a random error term with mean 0 and

variance , the error terms are independent and normally distributed.

• i=1,…,n, where n is the sample size (number of ordered pairs in the data set)

y i 1x i 0 i

i

2 2

Formulas for the slope and intercept estimates.

ii xbby 10ˆ For the estimated regression equation given by the formula:

The slope b1 is calculated by:

n

xx

n

yxxy

b 2

2

1)(

)(

)()()(

And the intercept b0 can be found with :

)(10 xbyb

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The standard error of the estimate, se, is found using the formula

se y i ˆ y i 2n 2

residuals2n 2

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Parallel Example 2: Compute the Standard Error

Compute the standard error of the estimate for thedrilling data which is presented on the next slide.

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Depth at Which Time to DrillDrilling Begins, x 5 Feet, y

(in feet) (in minutes)35 5.8850 5.9975 6.7495 6.1120 7.47130 6.93145 6.42155 7.97160 7.92175 7.62185 6.89190 7.9

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Solution

Step 1: Using technology (i.e. Minitab), we find the least squares regression line to be

Step 2, 3: The predicted values as well as the residuals for the 12 observations are given in the table on the next slide

ˆ y 0.0116x 5.5273

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Depth, x Time, y35 5.88 5.9333 -0.0533 0.002850 5.99 6.1073 -0.1173 0.013875 6.74 6.3973 0.3427 0.117495 6.1 6.6293 -0.5293 0.2802120 7.47 6.9193 0.5507 0.3033130 6.93 7.0353 -0.1053 0.0111145 6.42 7.2093 -0.7893 0.6230155 7.97 7.3253 0.6447 0.4156160 7.92 7.3833 0.5367 0.2880175 7.62 7.5573 0.0627 0.0039185 6.89 7.6733 -0.7833 0.6136190 7.9 7.7313 0.1687 0.0285

ˆ y

y ˆ y

y ˆ y 2

residuals 2 2.7012

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SolutionStep 4: We find the sum of the squared

residuals by summing the last column of the table:

Step 5: The standard error of the estimate is then given by:

residuals2 2.7012

se residuals2

n 2

2.7012

100.5197

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CAUTION!

Be sure to divide by n-2 when computing the standard error of the estimate.

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Parallel Example 4: Compute the Standard Error

Verify that the residuals from the drilling example are normally distributed.

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Conclusion: We have insufficient evidence at the 5% level of significance to support the claim that the residual errors from this model are not normally distributed.

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Hypothesis Test Regarding the Slope Coefficient, 1

To test whether two quantitative variables are linearlyrelated, we use the following steps provided that

1. the sample is obtained using random sampling.2. the residuals are normally distributed with

constant error variance.

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Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:

Step 2: Select a level of significance, , depending on the seriousness of making a Type I error.

Two-tailed Left-Tailed Right-Tailed

H0: 1 = 0 H0: 1 = 0 H0: 1 = 0

H1: 1 0 H1: 1 < 0 H1: 1 > 0

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Step 3: Compute the test statistic

which follows Student’s t-distribution with n-2 degrees of freedom. Remember, when computing the test statistic, we assume the null hypothesis to be true. So, we assume that 1=0.

t0 b1 1

sb1

b1

sb1

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Step 4: Use Table VI to estimate the P-value using n-2 degrees of freedom.

P-Value Approach

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P-Value Approach

Two-Tailed

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P-Value Approach

Left-Tailed

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P-Value Approach

Right-Tailed

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Step 5: If the P-value < , reject the null hypothesis.

P-Value Approach

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Step 6: State the conclusion.

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CAUTION!

Before testing H0: 1 = 0, be sure to draw a residual plot to verify that a linear model is appropriate.

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Parallel Example 5: Testing for a Linear Relation

Test the claim that there is a linear relation between drill depth and drill time at the = 0.05 level of significance using the drilling data.

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Solution

Verify the requirements:• We assume that the experiment was randomized so

that the data can be assumed to represent a random sample.

• In Parallel Example 4 we confirmed that the residuals were normally distributed by constructing a normal probability plot.

• To verify the requirement of constant error variance, we plot the residuals against the explanatory variable, drill depth.

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There is no discernable pattern.

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Solution

Step 1: We want to determine whether a linear relation exists between drill depth and drill time without regard to the sign of the slope. This is a two-tailed test with

H0: 1 = 0 versus H1: 1 0

Step 2: The level of significance is = 0.05.

Step 3: Using technology, we obtained an estimate of 1 in Parallel Example 2, b1=0.0116. To determine the standard deviation of b1, we compute .

The calculations are on the next slide.

x i x 2

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Depth, x35 -91.25 8326.562550 -76.25 5814.062575 -51.25 2626.562595 -31.25 976.5625120 -6.25 39.0625130 3.75 14.0625145 18.75 351.5625155 28.75 826.5625160 33.75 1139.0625175 48.75 2376.5625185 58.75 3451.5625190 63.75 4064.0625

x i x

x i x 2

xi x 2 30006 .25

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Solution

Step 3, cont’d: We have

The test statistic is

sb1

se

x i x 2

0.5197

30006.250.0030

t0 b1

sb1

0.0116

0.0033.867

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Solution: P-Value Approach

Step 4: Since this is a two-tailed test, the P-value is the sum of the area under the t-distribution with 12-2=10 degrees of freedom to the left of -t0 = -3.867 and to the right of t0 = 3.867. Using Table VI we find that with 10 degrees of freedom, the value 3.867 is between 3.581 and 4.144 corresponding to right-tail areas of 0.0025 and 0.001, respectively. Thus, the P-value is between 0.002 and 0.005.

Step 5: Since the P-value is less than the level of significance, 0.05, we reject the null hypothesis.

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Solution

Step 6: There is sufficient evidence at the = 0.05 level of significance to conclude that a linear relation exists between drill depth and drill time.

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Confidence Intervals for the Slope of the Regression Line

A (1- )100% confidence interval for the slope of the true regression line, 1, is given by the following formulas:

Lower bound:

Upper bound:

Here, t/2 is computed using n-2 degrees of freedom.

b1 t 2 se

x i x 2b1 t 2 sb1

b1 t 2 se

x i x 2b1 t 2 sb1

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Note: The confidence interval formula for 1 can be computed only if the data are randomly obtained, the residuals are normally distributed, and there is constant error variance.

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Parallel Example 7: Constructing a Confidence Interval for the Slope of the True Regression Line

Construct a 95% confidence interval for the slope of the least-squares regression line for the drilling example.

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Solution

The requirements for the usage of the confidence interval formula were verified in previous examples.We also determined• b1 = 0.0116• in previous examples.

sb10.0030

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Solution

Since t0.025=2.228 for 10 degrees of freedom, we have

Lower bound = 0.0116-2.2280.003=0.0049Upper bound = 0.0116+2.2280.003=0.0183.

We are 95% confident that the mean increase in the time it takes to drill 5 feet for each additional foot of depth at which the drilling begins is between 0.005 and 0.018 minutes.

The Coefficient of Determination

The Coefficient of Determination is the proportion of the variability in the response variable that can be attributed to the least squares regression model.

How to calculate R2

Using the sum of squares technique:

2

22

)(

)(1

yy

residualsR

But for the SLR models we can simplify the calculation slightly and , where r is the correlation between the response and predictor variables.

22 )(rR

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Parallel Example 8: Calculating the Coefficient of Determination

Using technology for our drilling example we can calculate the correlation between the response and predictor to be 0.772822. Using the simplified calculation for the coefficient of determination that means:

5973.0)772822.0( 22 R

Interpretation: Our model using the depth at which drilling begins as a predictor is able to explain 59.73% of the natural variability in the time it takes to drill 5 feet.