# What's the right interpretation for parameter estimates in loglinear modelling?

I'm doing a loglinear analysis of the following data.

Male is coded as 1, Female as 2. Senior workers are coded as 1, middle level as 2, and shopfloor as 3. A is coded as 1 and is the most positive appraisal. B is coded as 2 and is intermediate, while C is coded as 3 and represents the worst possible appraisal.

I ran backwards elimination in SPSS.

On the basis of that I decided to remove from the model the three way interaction and also sex*appraise.

I then generated the following parameter estimates.

I'm unsure of how to interpret these parameter estimates, and the texts I usually rely on don't seem to interpret this table at all.

Some questions:

1. What do the estimates mean? How should I interpret the results?
2. Some parameters are set to zero because they're redundant. Why did they get set to zero, and what does it mean for this parameter to be zero? What does it mean to say that they are redundant?
3. Can I say things like: "Position3*Appraise1 is -1.06, Position3*Appraise2 is +0.273, and Position3*Appraise3 is 0. Therefore people in Position 3 hardly ever get Appraise1, they very often get Appraise2, and they're intermediately likely to get Appraise3." I often hear interpretations of this sort in casual conversation, but I feel that perhaps they are not entirely accurate.

1. Looking at note C in the last table, it looks like the model is missing main effects and only includes interaction terms which is unusual and usually a bad idea. But we can still recreate the full equation equation:

\begin{align} \log \mu_{ijk} &= \lambda + \lambda_{ij}^{PS} + \lambda_{ik}^{PA} \\ \mu_{ijk} &= e^{\lambda + \lambda_{ij}^{PS} + \lambda_{ik}^{PA}} \\ \mu_{ijk} &= e^{\lambda}e^{\lambda_{ij}^{PS}}e^{\lambda_{ik}^{PA}} \end{align}

The parameters are a multiplicative effect on the estimated count (reference). For example, if $\rm{position}=1$ and $\rm{sex}=1$ and $\rm{appraisal}=1$ then the estimated count is:

$$\mu_{ijk} = e^{4.742}e^{-1.691}e^{0.439} = 32.786$$

2. The parameters are zero since they are the reference level for the model. The Constant parameter is used for the count for $\rm{position}=3$, $\rm{sex}=2$, and $\rm{appraise}=3$ which is $e^{4.742} e^0 e^0=114.66$. Notice if you change any of the 3 variables, at least one of the other two parameters will be non-zero.

3. After going through point 1 and 2, you can see that this interpretation is not quite correct. The model gives us counts and not "likely". So instead, we can say for people in $\rm{position}=3$, more are $\rm{appraise}=2$ than $\rm{appraise}=3$. And for people in $\rm{position}=3$, less are $\rm{appraise}=1$ than $\rm{appraise}=3$. To answer your comment about accuracy, you can use the 95% confidence intervals for the parameters and compute count confidence intervals. There are also other methods (possibly better) for determining statistical significance.

If you're wondering about "likely", one question that people tend to ask would be: Are men in $\rm{position}=3$ more likely to be $\rm{appraise}=1$ than women in $\rm{position}=3$? To answer this, we can look at odds ratios and look at the standard errors there.

The odds of a man in $\rm{position}=3$ being $\rm{appraise}=1$ vs otherwise is $\frac{62}{257 + 198}=0.13$ to 1

The odds of a woman in $\rm{position}=3$ being $\rm{appraise}=1$ vs otherwise is $\frac{45}{149+ 111}=0.17$ to 1

The odds ratio is then $\frac{62 * (149+111)}{45 * (257+198)} = \frac{0.13}{0.17}=0.78$

The standard error is defined in log odds ratio and is approximately

\begin{align} SE &= \sqrt{\frac{1}{n_{11}} + \frac{1}{n_{10}} + \frac{1}{n_{01}}+ \frac{1}{n_{00}}} \\ &= \sqrt{\frac{1}{62} + \frac{1}{257+198} + \frac{1}{45} + \frac{1}{149+111}} \\ &= 0.21 \end{align}

To combine the two, we need to compute the log odds ratio, adjust for the standard error, then convert out of log scale. This gives a 95% confidence interval of

$$\text{Lower Bound} = e^{\log{(Odds Ratio)} - 1.96 SE} = e^{\log{0.78} - 1.96 * 0.21} = 0.52$$ $$\text{Upper Bound} = e^{\log{(Odds Ratio)} + 1.96 SE} = e^{\log{0.78} + 1.96 * 0.21} = 1.19$$

Since we're at least 95% significant, we can say that men have 0.78 times (lower) odds than womens' odds.

Finally, take a look at Agresti's Categorical Data Analysis. Chapter 9,10 [3rd ed.] contains lots of information about interpretation and statistical significance for log linear models. It also has a better model selection technique using likelihood ratios rather than looking at a parameter's statistical significance.

• Thanks for your response; it has clarified things very well. However, are you able to tell me why is it generally a bad idea to exclude the main effects? Can you think of any circumstances in which it would make sense to exclude them? – user1205901 - Reinstate Monica Apr 25 '15 at 23:16
• It is general practice. stats.stackexchange.com/questions/11009/…. But for Log Linear Models specifically, Agresti has a few paragraphs about this (section 9.1.5. 3rd ed.). "A reason for including lower-order terms is that, otherwise, the statistical significance and the interpretation of a higher-order term depends on how variables are coded. This is undesirable, and with hierarchical models the same results no matter how variables are coded" – Eric Farng Apr 26 '15 at 10:26