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 counterexample, we can compute the variance of the ratio estimate using the following Minitab printout 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 
The sum of squares (uncorrected) of Y rX = 2550.03
Note!
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 counterexample, 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 a 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 of this story here is, "Use the right model!"