**Date:** Tue, 3 May 2005 11:14:30 +0200
**Reply-To:** Marta García-Granero
<biostatistics@terra.es>
**Sender:** "SPSSX(r) Discussion" <SPSSX-L@LISTSERV.UGA.EDU>
**From:** Marta García-Granero
<biostatistics@terra.es>
**Organization:** Asesoría Bioestadística
**Subject:** Re: Bland-Altman Correlations
**In-Reply-To:** <s2761d05.048@GWMAIN.luhs.org>
**Content-Type:** text/plain; charset=ISO-8859-1
Hi John,

You didn't send the message to the list, but to yourself with a hidden
copy to me (I guess). Be careful, I have a filter in my mail program
that deletes messages not addressed directly to me and blacklists the
sender. I was able to recover your message before the deletion and
added your email address to the friends list, to avoid future problems.

OK, now, the question

JN> I've been asked to look into how to compare two instruments
JN> using the Bland-Altman correlation, and have found nothing in the
JN> SPSS Help index, and little more on the internet. This is not a
JN> procedure with which I am familiar, having only just heard of it
JN> today.

JN> So, I'm hoping you might be able to clarify what this
JN> procedure does and how. From what little information I've been
JN> able to locate on the Internet, this procedure sounds as if it's a
JN> type of measurement of reliability or replicability (?). I was
JN> able to locate a couple references to primary texts (including the
JN> Bland Altman article) but would also appreciate a reference to
JN> some secondary documentation (perhaps a basic (bio)stats manual?)
JN> Finally, if SPSS does do this, can you refer me to the menu path?

The best source of documentation is Martin Blands page:

http://www-users.york.ac.uk/~mb55/meas//ba.htm
http://www-users.york.ac.uk/~mb55/meas/diffplot.htm
http://www-users.york.ac.uk/~mb55/meas/comfaq.htm
http://www-users.york.ac.uk/~mb55/talks/rcrtalk.htm

Once you've read all of them, you will have a clear picture of the
method.

SPSS doesn't perform Bland-Altman analysis (MedCalc and Analyse-it
do), but the basic steps are:

* For each data:
- Compute the difference between the two methods being compared
- Compute the average of the two methods

* Run descriptives to obtain the mean and SD of the differences (a
paired t test is also OK for that) and compute the limits of
agreement: Mean(diff)+/- 2*SD(diff). Discuss them (too wide?) as
the authors indicate in their Lancet paper.

* Compute (by hand) the Coefficient of
reproducibility:=100*SD(diff)/Mean(averages) (NOT mean of
differences, be careful).

* Plot difference (Y axis) vs average (X axis)
Edit the plot and add the three lines (Mean(diff), upper limit & lower
limit) as reference lines. Check for departures of the pattern as
described in Bland&Altman Lancet paper (bias, heterogeneity
proportional to the average...).

* If sample size is big enough, a histogram of the differences is also
useful.

This is the analysis using SPSS of the data presented by Bland&Altman
in their paper (we will ignore the second measure of each method,
because the use them to check for repeatability, a different
statistical problem, if you are interested, I also have some worked
examples using SPSS):

DATA LIST LIST/ subject wright1 wright2 mini1 mini2 (5 F8.0).
BEGIN DATA
1 494 490 512 525
2 395 397 430 415
3 516 512 520 508
4 434 401 428 444
5 476 470 500 500
6 557 611 600 625
7 413 415 364 460
8 442 431 380 390
9 650 638 658 642
10 433 429 445 432
11 417 420 432 420
12 656 633 626 605
13 267 275 260 227
14 478 492 477 467
15 178 165 259 268
16 423 372 350 370
17 427 421 451 443
END DATA.

VAR LABEL wright1'Wright peak flow meter #1'
/ wright2'Wright peak flow meter #2'
/mini1 'Mini Wright peak flow meter #1'
/mini2 'Mini Wright peak flow meter #2'.

COMPUTE dif1=wright1-mini1.
COMPUTE mean1=MEAN(wright1,mini1).
COMPUTE var1 = VARIANCE(wright1,mini1) .
EXECUTE .
GRAPH
/SCATTERPLOT(BIVAR)=mean1 WITH dif1.
GRAPH
/SCATTERPLOT(BIVAR)=mean1 WITH var1.
DESCRIPTIVES
VARIABLES=dif1 mean1
/STATISTICS=MEAN STDDEV .

* You will get:

N Mean Std.Dev.
dif1 17 -2.1176 38.76513
mean1 17 451.4118 113.07578

* CR = 100*38.765/451.4118 = 8.6%
* Lower: d - 2s = -2.1 - (2x38.8) = -79.7 l/min
* Upper: d + 2s = -2.1 + (2x38.8) = 75.5 l/min

* Now, edit the first graph and add the 3 reference lines (meanddiff, lower & upper).

--
Saludos,
Marta mailto:biostatistics@terra.es