Polynomial regression is in effect multiple linear regression: consider $X_1=X$ and $X_2=X^2$ -- then $E(Y) = \beta_1 X + \beta_2 X^2$ is the same as $E(Y) = \beta_1 X_1 + \beta_2 X_2$.
As such, methods for constructing confidence intervals for parameters (and for the mean in multiple regression) carry over directly to the polynomial case. Most regression packages will compute this for you. Yes, it can be done using the formula you suggest (if the assumptions needed for the t-interval to apply hold), and the right d.f. are used for the $t$ (the residual d.f. - which in R is available from the summary output).
The R function confint can be used to construct confidence intervals for parameters from a regression model. See ?confint.
In the case of a confidence interval for the conditional mean, let $X$ be the matrix of predictors, whether for polynomial regression or any other multiple regression model; let the estimated variance of the mean
at $x_i=(x_{1i},x_{2i},...,x_{pi})$ be $v_i=\hat{\sigma}^2x_i(X'X)^{-1}x_i'$
and let $s_i=\sqrt v_i$ be the corresponding standard error.
Let the upper $\alpha/2$ $t$ critical value for $n-p-1$ df be $t$.
Then the pointwise confidence interval for the mean at $x_i$ is $\hat{y}_i\pm t\cdot s$.
Also, the R function predict can be used to construct CIs for E(Y|X) - see ?predict.lm.
[At least when doing polynomial regression with an intercept, it makes sense to use orthogonal polynomials but if the spread of $X$ is large compared to the mean, and the degree is low (such as quadratic), it won't be so critical (I tend to do so anyway, because it's easier to interpret the linear and quadratic).]
1. Normal data, variance known: If you have observations $X_1, X_2, \dots, X_n$ sampled at random from
a normal population with unknown mean $\mu$ and known standard deviation $\sigma,$ then a 95% confidence interval (CI) for $\mu$ is $\bar X \pm 1.95 \sigma/\sqrt{n}.$ This is the only situation in which the z interval is exactly correct.
2. Nonnormal data, variance known: If the population distribution is not normal and the sample is 'large enough', then $\bar X$ is approximately normal and the same formula provides an approximate 95% CI. The rule that $n \ge 30$ is 'large enough' is unreliable here. If the population distribution is heavy-tailed, then $\bar X$ may not have a distribution that is close to normal (even if $n \ge 30).$ The 'Central Limit Theorem', often provides
reasonable approximations for moderate values of $n,$ but it is a limit theorem,
with guaranteed results only as $n \rightarrow \infty.$
3. Normal data, variance unknown. If you have observations $X_1, X_2, \dots, X_n$ sampled at random from
a normal population with unknown mean $\mu$ and standard deviation $\sigma,$ with $\mu$ estimated by the sample mean $\bar X$ and $\sigma$ estimated by the sample standard deviation $S.$ Then a 95% confidence interval (CI) for $\mu$ is $\bar X \pm t^* S/\sqrt{n},$ where $S$ is the sample standard deviation and
where $t^*$ cuts probability $0.025$ from the upper tail of Student's t distribution with $n - 1$ degrees of freedom. This is the only situation in which the t interval is exactly correct.
Examples: If $n=10$, then $t^* = 2.262$
and if $n = 30,$ then $t^* = 2.045.$ (Computations from R below; you could also use a printed 't table'.)
qt(.975, 9); qt(.975, 29)
[1] 2.262157 # for n = 10
[1] 2.04523 # for n = 30
Notice that 2.045 and 1.96 (from Part 1 above) both round to 2.0. If $n \ge 30$ then $t^*$ rounds to 2.0. That is the basis for
the 'rule of 30', often mindlessly parroted in other contexts where it is not relevant.
There is no similar coincidental rounding for CIs with confidence levels other than 95%. For example, in Part 1 above
a 99% CI for $\mu$ is obtained as $\bar X \pm 2.58 \sigma/\sqrt{n}.$ However,
$t^*=2.76$ for $n = 30$ and $t^* = 2.65$ for $n = 70.$
qnorm(.995)
[1] 2.575829
qt(.995, 29)
[1] 2.756386
qt(.995, 69)
[1] 2.648977
4. Nonnormal data, variance unknown: Confidence intervals based on the t distribution (as in Part 3 above) are known to be 'robust' against moderate departures from normality.
(If $n$ is very small, there should be no far outliers or evidence of severe skewness.) Then, to a degree that is difficult to predict, a t CI may provide a useful CI for $\mu.$
By contrast, if the type of distribution is known, it may be possible
to find an exact form of CI.
For example, if $n = 30$ observations from a (distinctly nonnormal)
exponential distribution with unknown mean $\mu$ have $\bar X = 17.24,\,
S = 15.33,$ then the (approximate) 95% t CI is $(11.33, 23.15).$
t.test(x)
One Sample t-test
data: x
t = 5.9654, df = 29, p-value = 1.752e-06
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
11.32947 23.15118
sample estimates:
mean of x
17.24033
However,
$$\frac{\bar X}{\mu} \sim \mathsf{Gamma}(\text{shape}=n,\text{rate}=n),$$
so that $$P(L \le \bar X/\mu < U) = P(\bar X/U < \mu < \bar X/L)=0.95$$
and an exact 95% CI for $\mu$ is $(\bar X/U,\, \bar X/L) = (12.42, 25.16).$
qgamma(c(.025,.975), 30, 30)
[1] 0.6746958 1.3882946
mean(x)/qgamma(c(.975,.025), 30, 30)
[1] 12.41835 25.55274
Addendum on bootstrap CI: If data seem non-normal, but the actual population
distribution is unknown, then a 95% nonparametric bootstrap CI may be the best
choice. Suppose we have $n=20$ observations from an unknown distribution, with $\bar X$ = 13.54$ and values shown in the stripchart below.
The observations seem distinctly right-skewed and fail a Shapio-Wilk normality test with P-value 0.001. If we assume the data are exponential and use the method in Part 4, the 95% CI is $(9.13, 22.17),$ but we have no way to know whether the data are exponential.
Accordingly, we find a 95% nonparametric bootstrap
in order to approximate $L^*$ and $U^*$ such that
$P(L^* < D = \bar X/\mu < U^*) \approx 0.95.$ In the R code below
the suffixes .re indicate random 're-sampled' quantities based on
$B$ samples of size $n$ randomly chosen without replacement from among the
$n = 20$ observations. The resulting 95% CI is $(9.17, 22.71).$ [There are
many styles of bootstrap CIs. This one treats $\mu$ as if it is a scale
parameter. Other choices are possible.]
B = 10^5; a.obs = 13.54
d.re = replicate(B, mean(sample(x, 20, rep=T))/a.obs)
UL.re = quantile(d.re, c(.975,.025))
a.obs/UL.re
97.5% 2.5%
9.172171 22.714980
Best Answer
"Margin of Error" is fairly common and relatable term. Margin of error is commonly expressed as the radius of the confidence interval, Wikipedia entry. You could include a footnote or reference to details on your 95% confidence intervals.