 What is the Pr(>|z|) telling me in a binomial summary

Hi there,
I'm wondering if anyone knows what Pr(>|z|) and the z value means in the summary of a binomial glm of a model represents. Here is the print out and code: Setp is the response variable of settlement of larvae, PLD is a factor of time and habitat is where the larvae settled.

BL1 <- glm(Setp ~ PLD * Habitat, family = binomial,
data = BL)
summary(BL1)
Call:
glm(formula = Setp ~ PLD * Habitat, family = binomial, data = BL)

Deviance Residuals:
Min 1Q Median 3Q Max
-0.50751 -0.34671 -0.07356 0.16767 0.65171

Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -0.4578 0.5132 -0.892 0.3723 PLD60 -1.6939 0.9660 -1.754 0.0795 . PLD240 -1.7866 0.9923 -1.800 0.0718 . HabitatDark Smooth -0.3895 0.7490 -0.520 0.6031 HabitatLight Rough -1.9401 1.0400 -1.866 0.0621 . HabitatLight Smooth -2.4022 1.2178 -1.973 0.0486 * PLD60:HabitatDark Smooth 0.5567 1.3487 0.413 0.6798 PLD240:HabitatDark Smooth -0.3106 1.6118 -0.193 0.8472 PLD60:HabitatLight Rough 1.7473 1.5919 1.098 0.2724 PLD240:HabitatLight Rough 1.4030 1.7148 0.818 0.4133 PLD60:HabitatLight Smooth 2.3095 1.6953 1.362 0.1731 PLD240:HabitatLight Smooth 2.0075 1.7913 1.121 0.2624

Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

Null deviance: 33.011  on 191  degrees of freedom

Residual deviance: 18.402 on 180 degrees of freedom
AIC: 78.458

Number of Fisher Scoring iterations: 5

I'm trying to determine if there are significant differences because the boxplots of the raw data showed there was. Does the Pr(>|z|) need to be <0.05 to be significant?

Here is a breakdown of the values shown for one of your parameters, HabitatLightSmooth.

Estimate  Std Error  z value  Pr(>|z|)
HabitatLight Smooth -2.4022   1.2178     -1.973   0.0486

Estimate/Std Error = -2.4022/1.2178 = -1.972573 = z value

Calculate the fraction of the normal distribution that is beyond the z value.
The multiplication by two accounts for the absolute value of z.
It is the area below -1.973 and above 1.973.

pnorm(-1.972573) * 2 = 0.04854423 = Pr(>|z|)

The z value tells you how many standard errors lie between zero and the estimated value of the parameter. Pr(|z|) tells you how much of the normal distribution is farther out, on either tail, than the absolute value of your z score.
A p value (Pr(|z|)) of 0.05 should not be treated as a bright line of significance. If your data had been very slightly different, giving a estimate of -2.366 instead of -2.4022, the p value of HabitatLightSmooth would have been 0.052. That parameter would not then be "insignificant" if it is "significant" now.

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