Language used to interpret a quasibinominal logit GLM for APA

Hey, I'm attempting to discuss my GLM results in an APA format, however, I am struggling with the language used. This is because from my understanding the size of the intercept number does not increase or decrease the size of the effect? Therefore, it is hard to discuss.

Here is my code.

glm(formula = counter_prop ~ appear_selected * label_selected, 
    family = quasibinomial(link = "logit"), data = subset(aggDataOut, 
        identityResponse == 1))

Deviance Residuals: 
     Min        1Q    Median        3Q       Max  
-0.85593  -0.13662  -0.01543   0.09408   1.00935  

                                      Estimate Std. Error t value Pr(>|t|)    
(Intercept)                           -0.87439    0.03792 -23.060  < 2e-16 ***
presume_selectedvegan                  -0.79385    0.06049 -13.123  < 2e-16 ***
label_selectedNL                       0.25689    0.05201   4.939 1.03e-06 ***
appear_selectedvegan:label_selectedNL  0.09171    0.08161   1.124    0.262    
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for quasibinomial family taken to be 0.04270381)

    Null deviance: 42.151  on 582  degrees of freedom
Residual deviance: 25.051  on 579  degrees of freedom

Number of Fisher Scoring iterations: 4

Here how I have tried to interpret the results ;

This is a general linear model investigates the impact on food choice from two characterstics of the food, whether the food is presumed to be vegan or not and If the food has a vegan label or not, but only for the identity response “1 – not vegan".

This model finds a significant negative effect for food choice that is presumed to be vegan (est = -0.79385, Std. Error = 0.06049, P < 0.05). There is also a positive significant effect for food without a vegan label (est = 0.25689, Std.Error = 0.05201, P < 0.05 ), however, this effect is smaller? (not sure how to word this).

Also, because I am conducting 6 GLM for this analysis, one for each identity response, does this mean that I have to multiply the confidence value (normally 0.05) by 6? If so how do I do this in R? :slight_smile:

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