My question is similar to this question, but the solution provided didn't tell whether increasing the sample size influences the prediction interval, so I would like to ask again.

The formulae for confidence interval: $$ \hat y \pm t_{\alpha/2, n-2} \sqrt{MSE} \sqrt{1/n + \frac{(x-\bar x)^2}{\sum (x_i - \bar x)^2}} $$

and prediction interval: $$ \hat y \pm t_{\alpha/2, n-2} \sqrt{MSE} \sqrt{1 + 1/n + \frac{(x-\bar x)^2}{\sum (x_i - \bar x)^2}} $$

If the sample size is increased, the standard error on the mean outcome given a new observation will decrease, then the confidence interval will become narrower. In my mind, at the same time, the prediction interval will also become narrower which is obvious from the fomular. However, my professor told me that the increasing sample size does not influence too much the prediction interval, so I am confused now. Could anybody give me some explanation?

  • $\begingroup$ I don't have the answer, but I can't help but notice (and point out) that there's a difference between not influencing the prediction interval, and not influencing the prediction interval too much. $\endgroup$
    – Ian_Fin
    Commented Aug 25, 2016 at 9:13
  • $\begingroup$ I think you should add the self-study tag. This is quite clearly "a routine question from a textbook, course, or test used for a class or self-study". $\endgroup$
    – DeltaIV
    Commented Aug 25, 2016 at 9:35
  • 2
    $\begingroup$ even ignoring the final term in both square roots (which would reduce the relative difference even further), compare $\sqrt{1}=1$ with $\sqrt{1+1/n}\approx 1+\frac{1}{2n}$. Now unless $n$ is tiny, those two tend to be very close and to only change a little if you change n. e.g. imagine you have $n=100$ and change it to $n=200$ ... by what percentage does $1+\frac{1}{2n}$ change? $\endgroup$
    – Glen_b
    Commented Aug 25, 2016 at 12:47

1 Answer 1


Confidence interval is an estimate of an interval in which mean of observations will fall when x=xi In its formula $$ 1/n + \frac{(x-\bar x)^2}{\sum (x_i - \bar x)^2} $$ Tends to 0.

Prediction interval is an estimate of an interval in which future individual observations will fall when x=xi In its formula $$ 1 + 1/n + \frac{(x-\bar x)^2}{\sum (x_i - \bar x)^2} $$ tends to 1

That means that the confidence interval for the mean of the outcomes at xi gets smaller as sample size grows. (as Central limit Theorem would suggest) which means that by increase of the sample size our estimate for the average (mean) outcome for xi gets better.

$$ \lim_{n->infinity}{CI = \hat y}$$

But the dispersion of the distribution of y|xi "the probability of an individual outcome" at xi, Doesn't change very much because central limit theorem is related to central tendencies not to individual behavior or outcomes. Therefore the prediction interval doesn't change very much. $$ \lim_{n->infinity}{PI = \hat y \pm t_{\alpha/2, n-2} \sqrt{MSE}}$$

Individual behavior remains uncertain no matter how much you increase your sample size ;)

  • $\begingroup$ when $n\to\infty$, $t_{\alpha/2,n-2}\to z_{\alpha/2}$, i.e., the t-distribution converges to the normal one for the number of degrees of freedom which goes to $\infty$. Also, assuming, as you implicitly did, that the last term in the square root goes to 0 when $n\to\infty$, requires that its denominator doesn't go to 0. Since $\bar{x}\to\mu$ by the LLN, this means requiring that $\max_i(x_i-\mu)^2$ doesn't go to 0 or goes to 0 sufficiently slow. $\endgroup$
    – DeltaIV
    Commented Aug 26, 2016 at 7:55

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