# Calculating the mean using regression data

I have some data to analyze where $y$ is dependent of $x$ - a linear regression was used.

It's a question from an exam, so I think it should be solvable. The regression was used to estimate the mean miles per gallon (response) from the amount of miles driven (predictor).

I have the following statistics available:

• Correlation coefficient (0.117)
• Standard deviation (0.482)
• Number of observations (101)

An ANOVA of this regression yields (Regression and residuals, respectively):

• df: 1, 99
• SS: 0.319, 22.96
• MS: 0.319, 0.232
• F-value: 1.374, critical F-value: 0.244

The regression itself (Intercept and Slope, respectively):

• Coefficients: 6.51, -0.00024
• Standard deviations: 0.186, 0.0002
• t-Values: 34.90, -1.17
• p-Values: 1.93E-57, 0.2439

Also, the "upper and lower 95% and 99%" are given for the above regression (although I'm not sure what that means).

Now, I am asked to calculate the mean $y$ for several values $x$, that's relatively easy, I just use the coefficients. So for example, I can calculate the mean miles per gallon for 500 miles driven.

Part where I'm stuck: I need to calculate the 99% confidence interval for the mean of $y$.. Obviously, this is what the example is all about - the introduction states that the mileage of a car should be estimated.

My question: How can I find out the mean of $y$ using the data provided above? (And, subsequently, the 99% confidence interval, although I seem to have the standard deviation, so that shouldn't be the problem)

• With great difficulty. The problem is that almost all the information you have is location indpendent. The only location information you have is the data about the estimated intercept. You know that the point $(\bar{x}, \bar{y})$ lies on the regression line, so knowing the mean of $x$ should give you the mean of $y$. Mar 25, 2011 at 10:25
• @Henry Your point seems valid. However the question is from an exam, so I doubt that it's not possible to find an answer. In this case, how do I get the mean of $x$? Mar 25, 2011 at 11:43
• Perhaps, having found what you call "the mean $y$ for several $x$", you are now being asked for the confidence intervals around those estimated values of $y$. Might that be possible? Mar 25, 2011 at 11:50
• @Henry I don't think so. I updated my question to reflect a practical example (with numbers). The thing is, it says that the regression is used to estimate the mean. Mar 25, 2011 at 12:11
• @slhck These data suffice to compute the range of the desired confidence interval, but they do not determine the mean of y. To see this, consider what would change if you were to add a constant $c$ to all the $x_i$ and simultaneously add $-0.00024 c$ to all the $y_i$: the answer is, nothing. Consequently you cannot estimate the mean of $y$. Thus, if the original question is answerable at all, understanding the vague "upper and lower 95% and 99%" is crucial. I would guess these would be confidence limits for the mean of $x$. (See the replies for why I'm wrong. :-)
– whuber
Mar 25, 2011 at 13:23

Contrary to @whuber's claim, the mean of x and y are contained in the information given.

Okay, so you have the line equation

$$y_i=\alpha +x_i\beta + e_i$$

estimates $\hat{\beta}=r\frac{s_y}{s_x}$ and $\hat{\alpha}=\overline{y}-\hat{\beta}\overline{x}$.

where $r$ is the correlation. The question doesn't state whether the standard deviation (0.482) is for $s_y$ or $s_x$ (the MLE standard deviation, with divisor $n$). Either way, you can work out the either from the info given. for their ratio must satisfy:

$$\frac{\hat{\beta}}{r}=\frac{s_y}{s_x}$$

The slope can't be negative if the correlation is positive, so I have assumed that you have done something incorrectly (for you have correlation of 0.117, and slope of -0.00024; this is impossible). This will affect the numbers, but not the general method. So I will assume the standard deviations are both known, but not write in the specific values. The same goes for the rest of the actual numbers.

Now the variance of $\hat{\beta}$ is given by:

$$var(\hat{\beta})=s_e^2(X^TX)^{-1}_{22}=\frac{s_e^2 (X^TX)_{11}}{|X^TX|}$$

Note that $(X^TX)_{11}=n$ and $s_e^2$ is the "mean square error". The variance of $\alpha$ is given by:

$$var(\hat{\alpha})=s_e^2(X^TX)^{-1}_{11}=\frac{s_e^2 (X^TX)_{22}}{|X^TX|}$$

Now $(X^TX)_{22}=\sum_i x_i^2 = n(s_x^2+n\overline{x}^2)$

And dividing these two variances gives:

$$\frac{var(\hat{\alpha})}{var(\hat{\beta})}=\frac{(X^TX)_{22}}{(X^TX)_{11}}=\frac{n(s_x^2+n\overline{x}^2)}{n}=s_x^2+n\overline{x}^2$$

Now all quantities in the equation are known, except for the mean $\overline{x}$. So we can re-arrange this equation and solve for the mean:

$$\overline{x}=\pm\sqrt{\frac{\frac{var(\hat{\alpha})}{var(\hat{\beta})}-s_x^2}{n}}$$

But we know from the start that $x_i>0$ - you can't drive "negative miles". So only the positive square root is to be taken. The rest is straight-forward CI stuff. The estimate of the mean $\hat{\overline{y}}$ is given by:

$$\hat{\overline{y}}=\hat{\alpha}+\hat{\beta}\overline{x}=\hat{\alpha}+\hat{\beta}\sqrt{\frac{\frac{var(\hat{\alpha})}{var(\hat{\beta})}-s_x^2}{n}}=\overline{y}$$

And the variance is given by:

$$var(\hat{\overline{y}})=var(\hat{\alpha})+\overline{x}^2 var(\hat{\beta})+2\overline{x}cov(\hat{\alpha},\hat{\beta})$$

Now the covariance is equal to: $$cov(\hat{\alpha},\hat{\beta})=s_e^2(X^TX)^{-1}_{21}=-\frac{s_e^2 (X^TX)_{21}}{|X^TX|}=-\frac{s_e^2 n\overline{x}}{ns_x^2}=-\frac{s_e^2 \overline{x}}{s_x^2}$$

And so the variance is given by:

$$var(\hat{\overline{y}})=var(\hat{\alpha})+\overline{x}^2 var(\hat{\beta})-2\frac{s_e^2 \overline{x}^2}{s_x^2}=var(\hat{\alpha})+\frac{\frac{var(\hat{\alpha})}{var(\hat{\beta})}-s_x^2}{n}\left(var(\hat{\beta})-2\frac{s_e^2}{s_x^2}\right)$$

So you construct your $100(1-P)$% confidence interval by choosing $T_{1-P/2}^{(n-2)}$ as the $P/2$ quantile of standard T distribution with $n-1$ degrees of freedom (which effectively equal to the standard normal, as $n-1=100$), and you have:

$$CI=\overline{y}\pm T_{1-P/2}^{(n-2)}\sqrt{var(\hat{\overline{y}})}$$

And all quantities are calculable, given the information.

• +1 Very good! I didn't think that the CI on the intercept would reveal the mean, but of course it does: the CI depends on the squared mean, as you point out. Once we are reminded of that, the rest is algebra.
– whuber
Mar 25, 2011 at 16:06
• Nice, thank you very much. Concerning the positive correlation and the negative slope, I double checked, it's what the data says. Given the above numbers, is it even valid to say that there's a dependency between $x$ and $y$? Mar 25, 2011 at 17:11
• @slhck - given the numbers, it is valid to say an error has been made in either calculating the slope or in calculating the correlation, or both (unless the correlation is $r^2=0.117$ and not $r$). But putting this aside, given the absolute value of the slope, the t-stat is just over 1 - not very strong evidence of a slope. Correlation does not imply dependence in a physical sense, only a logical one. So $x$ and $y$ may be physically dependent (but on a small scale), but given the data, it is pretty safe to assume logical independence. Mar 25, 2011 at 22:24