title: Calculating length of 95%-CI using dplyr
body: <p>Last time I asked how it was possible to calculate the average score per measurement occasion (week) for a variable (procras) that has been measured repeatedly for multiple respondents. So my (simplified) dataset in long format looks for example like the following (here two students, and 5 time points, no grouping variable):</p>
  
  <pre><code>studentID  week   procras
     1        0     1.4
     1        6     1.2
     1        16    1.6
     1        28    NA
     1        40    3.8
     2        0     1.4
     2        6     1.8
     2        16    2.0
     2        28    2.5
     2        40    2.8
  </code></pre>
  
  <p>Using dplyr I would get the average score per measurement occasion</p>
  
  <pre><code>mean_data &lt;- group_by(DataRlong, week)%&gt;% summarise(procras = mean(procras, na.rm = TRUE))
  </code></pre>
  
  <p>Looking like this e.g.:</p>
  
  <pre><code>Source: local data frame [5 x 2]
          occ  procras
        (dbl)    (dbl)
      1     0 1.993141
      2     6 2.124020
      3    16 2.251548
      4    28 2.469658
      5    40 2.617903
  </code></pre>
  
  <p>With ggplot2 I could now plot the average change over time, and by easily adjusting the group_data() of dplyr I could also get means per sub groups (for instance, the average score per occasion for men and women).
  Now I would like to add a column to the mean_data table which includes the length for the 95%-CIs around the average score per occasion.</p>
  
  <p><a href="http://www.cookbook-r.com/Graphs/Plotting_means_and_error_bars_(ggplot2)/" rel="noreferrer">http://www.cookbook-r.com/Graphs/Plotting_means_and_error_bars_(ggplot2)/</a> explains how to get and plot CIs, but this approach seems to become problematic as soon as I wanted to do this for any subgroup, right? So is there a way to let dplyr also include the CI (based on group size, ect.) automatically in the mean_data?
  After that it should be fairly easy to plot the new values as CIs into the graphs I hope.
  Thank you.</p>