Lesson 5: Auxillary Data and Regression Estimation
Lesson 5: Auxillary Data and Regression EstimationOverview
This lesson discusses when and how to use regression estimation. An example for using regression is given. Then we compare the regression estimate with simply using the sample mean, not taking advantage of the auxiliary information. To illustrate that one has to choose the right model, we use the ratio estimate for the example even though the condition for using the ratio estimate was not satisfied. And, not surprisingly, ratio estimate performs poorly since it is not the appropriate model for that data set.
Lesson 5: Ch. 8.1 of Sampling by Steven Thompson, 3rd edition
 know why and when to use regression estimates,
 know how to check the condition to see whether one can use the regression estimate,
 compute the regression estimate and its estimated variance,
 compute confidence interval based on regression estimate,
 see that the regression estimate does perform better than the expansion estimate when auxiliary data is useful, and
 see that the regression estimate does perform better than the ratio estimate when the condition for using the ratio estimate is not satisfied
5.1  Linear Regression Estimator
5.1  Linear Regression EstimatorLooking at the data, how will we find things that will work, or which model should we use? These are key questions. The variance for the estimators will be an important indicator.
The Idea Behind Regression Estimation
When the auxiliary variable x is linearly related to y but does not pass through the origin, a linear regression estimator would be appropriate. This does not mean that the regression estimate cannot be used when the intercept is close to zero. The two estimates, regression and ratio may be quite close in such cases and you can choose the one you want to use.
In addition, if multiple auxiliary variables have a linear relationship with y, multiple regression estimates may be appropriate.
To estimate the mean and total of yvalues, denoted as \(\mu\) and \(\tau\), one can use the linear relationship between y and known xvalues.
Let's start with a simple example:
\(\hat{y}=a+bx\) , which is our basic regression equation.
Then,
\(b=\dfrac{\sum\limits_{i=1}^n(x_i\bar{x})(y_i\bar{y})}{\sum\limits_{i=1}^n(x_i\bar{x})^2}\) and
\(a=\bar{y}b\bar{x}\)
Then to estimate the mean for y, substitute as follows:
\(x=\mu_x,\quad a=\bar{y}b\bar{x},\text{then}\)
\(\hat{\mu}_L=(\bar{y}b\bar{x})+b\mu_x\)
\(\hat{\mu}_L=\bar{y}+b(\mu_x\bar{x}),\quad \hat{\mu}_L=a+b\mu_x\)
Note that even though \(\hat{\mu}_L\) is not unbiased under simple random sampling, it is roughly so (asymptotically unbiased) for large samples.
Thus, the Mean Square Error of the estimate, denoted as MSE (\(\hat{\mu}_L\)) is not the same as Var(\(\hat{\mu}_L\))due to the bias but can be roughly estimated by the following when the sample size is large:
\begin{align}
\hat{V}ar(\hat{\mu}_L) &=\dfrac{Nn}{N \times n}\cdot \dfrac{\sum\limits_{i=1}^n(y_iabx_i)^2}{n2}\\
&= \dfrac{Nn}{N \times n}\cdot MSE\\
\end{align}
where MSE is the MSE of the linear regression model of y on x.
Therefore, an approximate (1\(\alpha\))100% CI for \(\mu\) is:
\(\hat{\mu}_L \pm t_{n2,\alpha/2}\sqrt{\hat{V}ar(\hat{\mu}_L)}\)
It follows that:
\(\hat{\tau}_L=N\cdot \hat{\mu}_L=N\bar{y}+b(\tau_xN\bar{x})\)
\begin{align}
\hat{V}ar(\hat{\tau}_L) &= N^2 \hat{V}ar(\hat{\mu}_L) \\
&= \dfrac{N \times (Nn)}{n} \cdot MSE\\
\end{align}
And, an approximate (1\(\alpha\))100% CI for \(\tau\) is:
\(\hat{\tau}_L \pm t_{n2,\alpha/2}\sqrt{\hat{V}ar(\hat{\tau}_L)}\)
Example 51: Average First Year Calculus Scores
Reference: p. 205 of Scheaffer, Mendenhall and Ott
The institutional researcher of a college wants to estimate the average first year Calculus score of the first year students. Since the students take the Calculus class from different instructors, it is expensive to find out their Calculus scores. The researcher only takes a simple random sample of 10 students and find out their first year Calculus scores. The researcher has a record of the college mathematics achievement test that the 486 first year students took prior to entering the college. And the average achievement test score for the 486 students was 52. The scatterplot of the 10 samples with both scores are given below. The researcher would like to use these information to help estimate the average first year calculus score of these 486 students.
The scatter plot shows that there is a strong positive linear relationship.
\(\hat{\mu}_L=\bar{y}+b(\mu_x\bar{x})=a+b\mu_x\)
Student 
Achievement test score X

Calculus score Y

1

39

65

2

43

78

3

21

52

4

64

82

5

57

92

6

47

89

7

28

73

8

75

98

9

34

56

10

52

75

Minitab output
Regression Analysis
The regression equation is
Y = 40.8 + 0.766 X
Analysis of Variance
Source  DF  SS  MS  F  P 

Regression  1  1450.0  1450.0  19.14  0.002 
Resid. Error  8  606.0  75.8  
Total  9  2056.0 
Coefficients
Predictor  Coef  StDev  T  P 

Constant  40.784  8.507  4.79  0.001 
X  0.7656  0.1750  4.38  0.002 
S = 8.704 RSq = 70.5% RSq(adj) = 66.8%
Try it!
\begin{align}
\hat{\mu}_L &= 40.8+0.766 \times 52\\
&= 80.63\\
\end{align}
The minitab output provides us with pvalues for the constant and the coefficient of X. We can see that both terms are significant. (ratio estimate is not appropriate since the constant term is nonzero).
Now we can compute the variance.
Try it!
\begin{align}
\hat{V}ar(\hat{\mu}_L) &=\dfrac{Nn}{N \times n}\cdot MSE\\
&= \dfrac{48610}{486 \times 10} \times 75.8\\
&= 7.42\\
\end{align}
Try it!
\(\hat{\mu}_L \pm t_{n2}\sqrt{\hat{V}ar(\hat{\mu}_L)}, \quad df=8\)
\begin{array}{lcl}
& = & 80.63 \pm 2.306 \times \sqrt{7.42} \\
& = & 80.63 \pm 6.28
\end{array}
5.2  Comparison of Estimators
5.2  Comparison of EstimatorsCompare the regression estimate to the estimate \(\bar{y}\)
To compare the regression estimate to the estimate \(\bar{y}\), (which does not use auxiliary result of x), we see that:
\(\hat{V}ar(\bar{y})=\dfrac{Nn}{N}\cdot \dfrac{s^2}{n}\)
\(s^2\) for y values is: \((15.11)^2\)
Try it!

What is the \(Var(\bar{y})\)?
\begin{align}
\hat{V}ar(\bar{y}) &= \dfrac{48610}{486 \times 10} \cdot 228.31 \\
&= 22.36\\
\end{align} 
Next, what is an approximate 95% CI for μ ?
\(\bar{y} \pm t_{n1}\sqrt{\hat{V}ar(\bar{y})}\)
\begin{array}{lcl}
& = & 76 \pm 2.262 \times \sqrt{22.36} \\
& = & 76 \pm 10.70
\end{array}
Recall that the 95% confidence interval using regression estimate is 80.63 \(\pm\) 6.28; a much shorter confidence interval.
This regression estimate is more precise than \(\bar{y}\).
Additionally, we have another estimator that we can look at.
Compare \(\hat{\mu}_L\) to the ratio estimator \(\hat{\mu}_r\)
Next, Minitab was used to find out the mean and standard deviation for X and Y.
Variable  N  Mean  StDev  SE Mean 

X  10  46.00  16.58  5.24 
Y  10  76.00  15.11  4.78 
The ratio estimate is inappropriate for this example. However, just to show a counter example, we can compute the variance of the ratio estimate using the following Minitab print out and compare this to the regression estimate.
X  Y  Y rX 

39  65  0.572 
43  78  6.964 
21  52  17.308 
64  82  23.728 
57  92  2.164 
47  89  11.356 
28  73  26.744 
75  98  25.900 
34  56  0.168 
52  75  10.904 
Sum of squares (uncorrected) of Y rX = 2550.03
For the Calculus Scores example we should not use the ratio estimator \(\hat{\mu}_r\) because the pvalue for the constant term is 0.001. This implies that it does not go through the origin and for this reason the ratio estimate is not appropriate. But for the purposes of a counter example we will work it out here anyway:
\(\hat{\mu}_r=r\mu_x=\dfrac{\bar{y}}{\bar{x}}\cdot \mu_x=\dfrac{76}{46}\cdot 52=85.91\)
Next, we need to figure out the variance and for this we need the MSE while using ratio estimate. From the Minitab output we have the SS / n1, therefore, the
\(s^2_r=\dfrac{1}{101} \sum\limits_{i=1}^{10} (y_irx_i)^2=283.33\) (this is huge!)
Now we can compute the variance:
Try it!
\begin{align}
\hat{V}ar(\hat{\mu}_r) &=\dfrac{Nn}{N}\cdot \dfrac{s^2_r}{n}\\
&= \dfrac{48610}{486}\cdot \dfrac{283.33}{10}=27.75\\
\end{align}
Now we can compute a 95% confidence interval for \(\mu\).
Try it!
\(\hat{\mu}_r \pm t_{n1}\sqrt{\hat{V}ar(\hat{\mu}_r)}\)
\begin{array}{lcl}
& = & 85.91 \pm 2.262 \times \sqrt{27.75} \\
& = & 85.91 \pm 11.92
\end{array}
We can see that the ratio estimate is even worse than \(\bar{y}\) when it is used in an inappropriate situation.
The width of the interval is larger than the one for the regression estimate.
The moral to this story here is, "Use the right model!"