§15.3--estimation and prediction - furman...
TRANSCRIPT
![Page 1: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/1.jpg)
§15.3–Estimation and Prediction
Tom Lewis
Fall Term 2009
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 1 / 10
![Page 2: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/2.jpg)
Outline
1 Estimating the conditional mean of the response variable
2 Estimating the observed value of the response variable
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 2 / 10
![Page 3: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/3.jpg)
Estimating the conditional mean of the response variable
The problem
We will assume that y = β1x + β0 + ε satisfies the assumptions of theregression model and that the standard deviation of ε is σ. Given a fixedvalue xp of the explanatory variable, what is our best estimate for thevalue of β1xp + β0?
The basic idea
Given a sample {x1, x2, . . . , xn} of explanatory variables, we canconstruct the regression equation y = b1x + b0. The value
yp = b1xp + b0
should be a close approximation to β1xp + β0. This value is called theconditional mean of the response variable (at x = xp).
To understand how good an approximation yp gives, we need tounderstand its variability.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 3 / 10
![Page 4: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/4.jpg)
Estimating the conditional mean of the response variable
The problem
We will assume that y = β1x + β0 + ε satisfies the assumptions of theregression model and that the standard deviation of ε is σ. Given a fixedvalue xp of the explanatory variable, what is our best estimate for thevalue of β1xp + β0?
The basic idea
Given a sample {x1, x2, . . . , xn} of explanatory variables, we canconstruct the regression equation y = b1x + b0. The value
yp = b1xp + b0
should be a close approximation to β1xp + β0. This value is called theconditional mean of the response variable (at x = xp).
To understand how good an approximation yp gives, we need tounderstand its variability.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 3 / 10
![Page 5: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/5.jpg)
Estimating the conditional mean of the response variable
The problem
We will assume that y = β1x + β0 + ε satisfies the assumptions of theregression model and that the standard deviation of ε is σ. Given a fixedvalue xp of the explanatory variable, what is our best estimate for thevalue of β1xp + β0?
The basic idea
Given a sample {x1, x2, . . . , xn} of explanatory variables, we canconstruct the regression equation y = b1x + b0. The value
yp = b1xp + b0
should be a close approximation to β1xp + β0. This value is called theconditional mean of the response variable (at x = xp).
To understand how good an approximation yp gives, we need tounderstand its variability.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 3 / 10
![Page 6: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/6.jpg)
Estimating the conditional mean of the response variable
The problem
We will assume that y = β1x + β0 + ε satisfies the assumptions of theregression model and that the standard deviation of ε is σ. Given a fixedvalue xp of the explanatory variable, what is our best estimate for thevalue of β1xp + β0?
The basic idea
Given a sample {x1, x2, . . . , xn} of explanatory variables, we canconstruct the regression equation y = b1x + b0. The value
yp = b1xp + b0
should be a close approximation to β1xp + β0. This value is called theconditional mean of the response variable (at x = xp).
To understand how good an approximation yp gives, we need tounderstand its variability.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 3 / 10
![Page 7: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/7.jpg)
Estimating the conditional mean of the response variable
Theorem
The random variable yp is normally distributed.
The mean of yp is β1xp + β0
The standard deviation of yp is
σ
√1
n+
(xp − x)2
Sxx
In other words
z =yp − (β1xp + β0)
σ√
1n +
(xp−x)2
Sxx
is a standard normal random variable.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 4 / 10
![Page 8: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/8.jpg)
Estimating the conditional mean of the response variable
Theorem
The random variable yp is normally distributed.
The mean of yp is β1xp + β0
The standard deviation of yp is
σ
√1
n+
(xp − x)2
Sxx
In other words
z =yp − (β1xp + β0)
σ√
1n +
(xp−x)2
Sxx
is a standard normal random variable.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 4 / 10
![Page 9: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/9.jpg)
Estimating the conditional mean of the response variable
Theorem
The random variable yp is normally distributed.
The mean of yp is β1xp + β0
The standard deviation of yp is
σ
√1
n+
(xp − x)2
Sxx
In other words
z =yp − (β1xp + β0)
σ√
1n +
(xp−x)2
Sxx
is a standard normal random variable.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 4 / 10
![Page 10: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/10.jpg)
Estimating the conditional mean of the response variable
Theorem
The random variable yp is normally distributed.
The mean of yp is β1xp + β0
The standard deviation of yp is
σ
√1
n+
(xp − x)2
Sxx
In other words
z =yp − (β1xp + β0)
σ√
1n +
(xp−x)2
Sxx
is a standard normal random variable.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 4 / 10
![Page 11: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/11.jpg)
Estimating the conditional mean of the response variable
Theorem
The random variable yp is normally distributed.
The mean of yp is β1xp + β0
The standard deviation of yp is
σ
√1
n+
(xp − x)2
Sxx
In other words
z =yp − (β1xp + β0)
σ√
1n +
(xp−x)2
Sxx
is a standard normal random variable.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 4 / 10
![Page 12: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/12.jpg)
Estimating the conditional mean of the response variable
Theorem
The random variable
t =yp − (β1xp + β0)
se
√1n +
(xp−x)2
Sxx
has a t-distribution with n − 2 degrees of freedom.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 5 / 10
![Page 13: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/13.jpg)
Estimating the conditional mean of the response variable
Problem
The data in the set LinModSec3.txt was created according to the modely = −2x + 30 + ε, where ε is a standard normal random variable. Here aresome summary statistics:
Sxx Syy Sxy b1 b0
82.5 338.536 -164.6 -1.9952 30.093
therefore
SST SSR SSE
338.536 328.4019394 10.13406061
Create a 95% confidence interval for the conditional mean of the responsevariable at xp = 8.5.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 6 / 10
![Page 14: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/14.jpg)
Estimating the observed value of the response variable
The problem
We will assume that y = β1x + β0 + ε satisfies the assumptions of theregression model and that the standard deviation of ε is σ. Given a fixedvalue xp of the explanatory variable, what is our best estimate for thevalue of yp, the observed value of the response variable?
The basic idea
Given a sample {x1, x2, . . . , xn} of explanatory variables, we canconstruct the regression equation y = b1x + b0. The value
yp = b1xp + b0
should be a close approximation to yp.
To understand how good an approximation yp gives, we need tounderstand the variability of yp − yp.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 7 / 10
![Page 15: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/15.jpg)
Estimating the observed value of the response variable
The problem
We will assume that y = β1x + β0 + ε satisfies the assumptions of theregression model and that the standard deviation of ε is σ. Given a fixedvalue xp of the explanatory variable, what is our best estimate for thevalue of yp, the observed value of the response variable?
The basic idea
Given a sample {x1, x2, . . . , xn} of explanatory variables, we canconstruct the regression equation y = b1x + b0. The value
yp = b1xp + b0
should be a close approximation to yp.
To understand how good an approximation yp gives, we need tounderstand the variability of yp − yp.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 7 / 10
![Page 16: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/16.jpg)
Estimating the observed value of the response variable
The problem
We will assume that y = β1x + β0 + ε satisfies the assumptions of theregression model and that the standard deviation of ε is σ. Given a fixedvalue xp of the explanatory variable, what is our best estimate for thevalue of yp, the observed value of the response variable?
The basic idea
Given a sample {x1, x2, . . . , xn} of explanatory variables, we canconstruct the regression equation y = b1x + b0. The value
yp = b1xp + b0
should be a close approximation to yp.
To understand how good an approximation yp gives, we need tounderstand the variability of yp − yp.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 7 / 10
![Page 17: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/17.jpg)
Estimating the observed value of the response variable
The problem
We will assume that y = β1x + β0 + ε satisfies the assumptions of theregression model and that the standard deviation of ε is σ. Given a fixedvalue xp of the explanatory variable, what is our best estimate for thevalue of yp, the observed value of the response variable?
The basic idea
Given a sample {x1, x2, . . . , xn} of explanatory variables, we canconstruct the regression equation y = b1x + b0. The value
yp = b1xp + b0
should be a close approximation to yp.
To understand how good an approximation yp gives, we need tounderstand the variability of yp − yp.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 7 / 10
![Page 18: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/18.jpg)
Estimating the observed value of the response variable
Theorem
The random variable yp − yp is normally distributed.
The mean of yp − yp is 0.
The standard deviation of yp − yp is
σ
√1 +
1
n+
(xp − x)2
Sxx
In other words
z =yp − yp
σ√
1 + 1n +
(xp−x)2
Sxx
is a standard normal random variable.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 8 / 10
![Page 19: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/19.jpg)
Estimating the observed value of the response variable
Theorem
The random variable yp − yp is normally distributed.
The mean of yp − yp is 0.
The standard deviation of yp − yp is
σ
√1 +
1
n+
(xp − x)2
Sxx
In other words
z =yp − yp
σ√
1 + 1n +
(xp−x)2
Sxx
is a standard normal random variable.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 8 / 10
![Page 20: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/20.jpg)
Estimating the observed value of the response variable
Theorem
The random variable yp − yp is normally distributed.
The mean of yp − yp is 0.
The standard deviation of yp − yp is
σ
√1 +
1
n+
(xp − x)2
Sxx
In other words
z =yp − yp
σ√
1 + 1n +
(xp−x)2
Sxx
is a standard normal random variable.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 8 / 10
![Page 21: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/21.jpg)
Estimating the observed value of the response variable
Theorem
The random variable yp − yp is normally distributed.
The mean of yp − yp is 0.
The standard deviation of yp − yp is
σ
√1 +
1
n+
(xp − x)2
Sxx
In other words
z =yp − yp
σ√
1 + 1n +
(xp−x)2
Sxx
is a standard normal random variable.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 8 / 10
![Page 22: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/22.jpg)
Estimating the observed value of the response variable
Theorem
The random variable yp − yp is normally distributed.
The mean of yp − yp is 0.
The standard deviation of yp − yp is
σ
√1 +
1
n+
(xp − x)2
Sxx
In other words
z =yp − yp
σ√
1 + 1n +
(xp−x)2
Sxx
is a standard normal random variable.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 8 / 10
![Page 23: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/23.jpg)
Estimating the observed value of the response variable
Theorem
The random variable
t =yp − yp
se
√1 + 1
n +(xp−x)2
Sxx
has a t-distribution with n − 2 degrees of freedom.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 9 / 10
![Page 24: §15.3--Estimation and Prediction - Furman Universitymath.furman.edu/~tlewis/math241/weiss/chap15/sec3.pdfTom Lewis 15.3–Estimation and Prediction Fall Term 2009 2 / 10 Estimating](https://reader030.vdocuments.us/reader030/viewer/2022040917/5e9220addee68a30f56acb63/html5/thumbnails/24.jpg)
Estimating the observed value of the response variable
Problem
The data in the set LinModSec3.txt was created according to the modely = −2x + 30 + ε, where ε is a standard normal random variable. Here aresome summary statistics:
Sxx Syy Sxy b1 b0
82.5 338.536 -164.6 -1.9952 30.093
therefore
SST SSR SSE
338.536 328.4019394 10.13406061
Create a 95% confidence interval for the observed value of the responsevariable at xp = 8.5.
Tom Lewis () §15.3–Estimation and Prediction Fall Term 2009 10 / 10