Interpreting split_ci output

[BrauseT]

Hi!

I am calculating split half reliabilities using the splifthalfr-package. For consistency reasons I would like to report regular 95% confidence intervals, i.e. the parameter estimate and the corresponding upper and lower limits. In another discussion, I read that split_ci delegates the calculations to the function bcajack in the bcaboot-package. After reading https://rdrr.io/cran/bcaboot/man/bcajack.html I figured that the standard limits are reported in the third column (std) of the $lims-table, whereas the parameter estimate can be found in the first row (est) and first column (theta) of the $stats-table (which is equal to the 0.5 value in the third column of the $lims-table). Is that correct?

Using the example output from the website above, this would lead to the following parameter estimate and 95% confidence interval:
r = .51, 95% CI [.44, .57].

Example output (taken from here):

$lims

            bca      jacksd       std   pct
0.025 0.4349782 0.003100685 0.4421256 0.006
…
0.5   0.4989099 0.002384237 0.5065603 0.311
…
0.975 0.5615867 0.001996201 0.5709950 0.928

$stats

        theta       sdboot         z0            a     sdjack
est 0.5065603 0.0328754548 -0.2455895 -0.001290911 0.01864547
jsd 0.0000000 0.0007288154  0.0483231  0.000000000 0.00000000

[tayllana]

Yes, you are correct in your interpretation of the output from the bcaboot package. Here's how you can interpret it:

• The parameter estimate (theta) is indeed found in the first row (est) and first column (theta) of the $stats table. In your example, the parameter estimate (theta) is 0.5065603.
• The standard limits (std) are indeed reported in the third column (std) of the $lims table. These are essentially the confidence interval bounds. In your example, the lower limit would be 0.4421256 and the upper limit would be 0.5065603.

So, using the output you provided:

• The parameter estimate (r) is approximately 0.51 (rounded from 0.5065603).
• The 95% confidence interval for this estimate would be approximately [0.44, 0.57] (rounded from [0.4421256, 0.5065603]).

Your understanding seems accurate based on the information you've provided.

[tpronk]

Thanks for helping out @tayllana!

[BrauseT]

Thanks a lot @tayllana and @tpronk! :-)