[Rd] lm() gives different results to lm.ridge() and SPSS

[Viechtbauer Wolfgang (SP)]

Totally agree that standardizing the interaction term is nonsense. But in all fairness, SPSS doesn't do that. In fact, the 'REGRESSION' command in SPSS doesn't compute any interaction terms -- one has to first compute them 'by hand' and then add them to the model like any other variable. So somebody worked extra hard to standardize that interaction term in SPSS as well :/

Best,
Wolfgang

-----Original Message-----
From: R-devel [mailto:r-devel-bounces at r-project.org] On Behalf Of Fox, John
Sent: Friday, May 05, 2017 20:23
To: Nick Brown; peter dalgaard
Cc: r-devel at r-project.org
Subject: Re: [Rd] lm() gives different results to lm.ridge() and SPSS

Dear Nick,

On 2017-05-05, 9:40 AM, "R-devel on behalf of Nick Brown"
<r-devel-bounces at r-project.org on behalf of nick.brown at free.fr> wrote:

>>I conjecture that something in the vicinity of
>> res <- lm(DEPRESSION ~ scale(ZMEAN_PA) + scale(ZDIVERSITY_PA) +
>>scale(ZMEAN_PA * ZDIVERSITY_PA), data=dat)
>>summary(res) 
>> would reproduce the SPSS Beta values.
>
>Yes, that works. Thanks!

That you have to work hard in R to match the SPSS results isn’t such a bad
thing when you factor in the observation that standardizing the
interaction regressor, ZMEAN_PA * ZDIVERSITY_PA, separately from each of
its components, ZMEAN_PA and ZDIVERSITY_PA, is nonsense.

Best,
 John

-------------------------------------
John Fox, Professor
McMaster University
Hamilton, Ontario, Canada
Web: http://socserv.mcmaster.ca/jfox/

>----- Original Message -----
>
>From: "peter dalgaard" <pdalgd at gmail.com>
>To: "Viechtbauer Wolfgang (SP)"
><wolfgang.viechtbauer at maastrichtuniversity.nl>, "Nick Brown"
><nick.brown at free.fr>
>Cc: r-devel at r-project.org
>Sent: Friday, 5 May, 2017 3:33:29 PM
>Subject: Re: [Rd] lm() gives different results to lm.ridge() and SPSS
>
>Thanks, I was getting to try this, but got side tracked by actual work...
>
>Your analysis reproduces the SPSS unscaled estimates. It still remains to
>figure out how Nick got
>
>> 
>coefficients(lm(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1))
>
>(Intercept) ZMEAN_PA ZDIVERSITY_PA ZMEAN_PA:ZDIVERSITY_PA
>0.07342198 -0.39650356 -0.36569488 -0.09435788
>
>
>which does not match your output. I suspect that ZMEAN_PA and
>ZDIVERSITY_PA were scaled for this analysis (but the interaction term
>still obviously is not). I conjecture that something in the vicinity of
>
>res <- lm(DEPRESSION ~ scale(ZMEAN_PA) + scale(ZDIVERSITY_PA) +
>scale(ZMEAN_PA * ZDIVERSITY_PA), data=dat)
>summary(res) 
>
>would reproduce the SPSS Beta values.
>
>> On 5 May 2017, at 14:43 , Viechtbauer Wolfgang (SP)
>><wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:
>> 
>> I had no problems running regression models in SPSS and R that yielded
>>the same results for these data.
>> 
>> The difference you are observing is from fitting different models. In
>>R, you fitted: 
>> 
>> res <- lm(DEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=dat)
>> summary(res) 
>> 
>> The interaction term is the product of ZMEAN_PA and ZDIVERSITY_PA. This
>>is not a standardized variable itself and not the same as "ZINTER_PA_C"
>>in the png you showed, which is not a variable in the dataset, but can
>>be created with: 
>> 
>> dat$ZINTER_PA_C <- with(dat, scale(ZMEAN_PA * ZDIVERSITY_PA))
>> 
>> If you want the same results as in SPSS, then you need to fit:
>> 
>> res <- lm(DEPRESSION ~ ZMEAN_PA + ZDIVERSITY_PA + ZINTER_PA_C,
>>data=dat) 
>> summary(res) 
>> 
>> This yields: 
>> 
>> Coefficients: 
>> Estimate Std. Error t value Pr(>|t|)
>> (Intercept) 6.41041 0.01722 372.21 <2e-16 ***
>> ZMEAN_PA -1.62726 0.04200 -38.74 <2e-16 ***
>> ZDIVERSITY_PA -1.50082 0.07447 -20.15 <2e-16 ***
>> ZINTER_PA_C -0.58955 0.05288 -11.15 <2e-16 ***
>> --- 
>> Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>> 
>> Exactly the same as in the png.
>> 
>> Peter already mentioned this as a possible reason for the discrepancy:
>>https://stat.ethz.ch/pipermail/r-devel/2017-May/074191.html ("Is it
>>perhaps the case that x1 and x2 have already been scaled to have
>>standard deviation 1? In that case, x1*x2 won't be.")
>> 
>> Best, 
>> Wolfgang