If I plot a symmetrical errorbar (let's say SEM = 0.5), should it show 0.5 on each side, or 0.25 (half) on each side?
Solved – Should error bars be halved
data visualizationstandard error
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You are totally correct in your assumption that error bars representing the standard error of the mean are totally inappropriate for within-subject designs. However, the question of overlapping error bars and significance is yet another topic, to which I will come back at the end of this commented reference list.
There is rich literature from Psychology on within-subject confidence intervals or error bars which do exactly what you want. The reference work is clearly:
Loftus, G. R., & Masson, M. E. J. (1994). Using confidence intervals in within-subject designs. Psychonomic Bulletin & Review, 1(4), 476–490. doi:10.3758/BF03210951
However, their problem is that they use the same error term for all levels of a within-subject factor. This does not seem to be a huge problem for your case (2 levels). But there are more modern approaches solving this problem. Most notably:
Franz, V., & Loftus, G. (2012). Standard errors and confidence intervals in within-subjects designs: Generalizing Loftus and Masson (1994) and avoiding the biases of alternative accounts. Psychonomic Bulletin & Review, 1–10. doi:10.3758/s13423-012-0230-1
Baguley, T. (2011). Calculating and graphing within-subject confidence intervals for ANOVA. Behavior Research Methods. doi:10.3758/s13428-011-0123-7 [can be found here]
Further references can be found in the latter two papers (which I think are both worth a read).
How do researchers interpret CIs? Bad according to the following paper:
Belia, S., Fidler, F., Williams, J., & Cumming, G. (2005). Researchers Misunderstand Confidence Intervals and Standard Error Bars. Psychological Methods, 10(4), 389–396. doi:10.1037/1082-989X.10.4.389
How should we interpret overlapping and non-overlapping CIs?
Cumming, G., & Finch, S. (2005). Inference by Eye: Confidence Intervals and How to Read Pictures of Data. American Psychologist, 60(2), 170–180. doi:10.1037/0003-066X.60.2.170
One final thought (although this is not relevant to your case): If you have a split-plot design (i.e., within- and between-subject factors) in one plot, you can forget about error bars all together. I would (humbly) recommend my raw.means.plot function in the R package plotrix.
The standard error is the standard deviation of an estimator; the SEM therefore arises when you are using the sample mean as an estimator of the true underlying population mean. In this case, the estimated standard error will generally be much smaller than the sample standard deviation of the original data points, since the mean estimator is less variable than the data itself.
To see how this works more specifically, let $X_1,...,X_n \sim \text{IID Dist}$ be your observable sample values and let $\bar{X} = \sum_{i=1}^n X_i / n$ be the resulting sample mean, which is taken to be an estimator of the underlying population mean $\mu = \mathbb{E}(X_i)$. If we let $\sigma^2 = \mathbb{V}(X_i)$ be the underlying population variance then the true standard error of the sample mean is:
$$\begin{equation} \begin{aligned} \text{se} \equiv \text{se}(\bar{X}) \equiv \mathbb{S}(\bar{X}) &= \sqrt{\mathbb{V}(\bar{X})} \\[6pt] &= \sqrt{\mathbb{V} \Big( \frac{1}{n} \sum_{i=1}^n X_i \Big)} \\[6pt] &= \sqrt{\frac{1}{n^2} \sum_{i=1}^n \mathbb{V} (X_i)} \\[6pt] &= \sqrt{\frac{1}{n^2} \sum_{i=1}^n \sigma^2} \\[6pt] &= \sqrt{\frac{n \sigma^2}{n^2} } \\[6pt] &= \sqrt{\frac{\sigma^2}{n} } \\[6pt] &= \frac{\sigma^2}{\sqrt{n}}. \\[6pt] \end{aligned} \end{equation}$$
Substituting the unknown paraeter $\sigma$ with the observable sample standard deviation $s$ yields the estimated standard error:
$$\widehat{\text{se}} = \frac{s^2}{\sqrt{n}}.$$
The estimated standard error is not an estimate of the dispersion of the underlying data; it is an estimate of the dispersion of the estimator in your problem, which is the sample mean in this case. Since the sample mean averages over all the observed values, it is much less variable than those initial values. Specifically, we can see from the above result that the estimated standard error of the mean is equal to the sample standard deviation of the underlying data, divided by $\sqrt{n}$. Now, obviously as $n$ gets bigger, the SEM is going to be substantially less than the sample standard deviation of the underlying data.
Once you have calculated the estimated SEM, it is usual to use this to give a confidence interval for the true underlying population mean $\mu$ at some specified confidence level $1-\alpha$. This can be done using the standard interval formula for a population mean:
$$\text{CI}_\mu(1-\alpha) = \Big[ \bar{X} \pm t_{n-1,\alpha/2} \cdot \widehat{se} \Big] = \Big[ \bar{X} \pm \frac{t_{n-1,\alpha/2}}{\sqrt{n}} \cdot s \Big].$$
Contrary to the goal stated in your question, it is never a good idea to report the interval $\bar{X} \pm \widehat{se}$; this is just a confidence interval using the strange requirement that $t_{n-1,\alpha/2}=1$, which is likely to be misleading to your reader. Instead, you should choose a sensible confidence level $1-\alpha$, and give a proper confidence interval, reporting your confidence level to your reader.
Application to your data: It appears from your analysis that you are seeking to aggregate your data, ignoring the time value covariates, and therefore analysing them as a single IID sample. This is not necessarily the best way to analyse the data, but I will proceed this way in order to use your method, to focus on the aspects of the SEM in your question. On this basis, you have $n=30$ and $s = 0.7722$ (which I calculated from the thirty values in your table). The estimated standard error of the mean should then be $\widehat{\text{se}} = 0.7722/\sqrt{30} = 0.1410$. It is unclear to me how you got the contrary value reported in your question.
In any case, you can see that the estimated standard error $\widehat{\text{se}} = 0.1410$ is substantially lower than the sample standard deviation $s = 0.7722$. As noted above, this is not surprising, since the former is the estimated standard deviation of a sample mean, and the sample mean is less variable due to averaging across multiple data points. Taking $\alpha=0.05$ we obtain $t_{n-1,\alpha/2} = t_{29,0.025} = 2.0452$, so the resulting $95$% confidence interval for the true population mean is:
$$\text{CI}_\mu(0.95) = \Big[ 7.7920 \pm 2.0452 \cdot 0.1410 \Big] = \Big[ 7.7920 \pm 0.2884 \Big] = \Big[ 7.5038, 8.0804 \Big].$$
As noted, this analysis ignores the time data, and simply treats all the values as a single IID sample, so it is important to remember that this confidence interval is contingent on that treatment of the data (which seems to be what you are after). This is not the best form of analysis; a better approach would be to use the time covariate in a regression model.
Best Answer
It should ideally show twice on each side, representing an approximate 95% confidence interval for the true value of the parameter. This makes it easy to see for which possible values you have enough evidence to claim are not candidates for the true parameter value. You 100% should not display halved error bars, as this is never done and is misleading. Display the full error bar on each side is fine, but doesn't really convey much information, and many readers will simply double its length in their head to approximate a 95% confidence interval.