Interpretation of coefficients in logistic regression output

Question

I am doing logistic regression in R on a binary dependent variable with only one independent variable. I found the odd ratio as 0.99 for an outcomes. This can be shown in following. Odds ratio is defined as, $ratio_{odds}(H) = \frac{P(X=H)}{1-P(X=H)}$. As given earlier $ratio_{odds} (H) = 0.99$ which implies that $P(X=H) = 0.497$ which is close to 50% probability. This implies that the probability for having a H cases or non H cases 50% under the given condition of independent variable. This does not seem realistic from the data as only ~20% are found as H cases. Please give clarifications and proper explanations of this kind of cases in logistic regression.

I am hereby adding the results of my model output:

M1 <- glm(H~X, data=data, family=binomial())
summary(M1)

Call:
glm(formula = H ~ X, family = binomial(), data = data)

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-1.8563   0.6310   0.6790   0.7039   0.7608  

Coefficients:
                Estimate      Std. Error      z value     Pr(>|z|)    
(Intercept)    1.6416666      0.2290133      7.168      7.59e-13 ***
   X          -0.0014039      0.0009466     -1.483      0.138    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 1101.1  on 1070  degrees of freedom
Residual deviance: 1098.9  on 1069  degrees of freedom
  (667 observations deleted due to missingness)
AIC: 1102.9

Number of Fisher Scoring iterations: 4


exp(cbind(OR=coef(M1), confint(M1)))
Waiting for profiling to be done...
                                      OR           2.5 %       97.5 %
(Intercept)                    5.1637680       3.3204509     8.155564
     X                         0.9985971       0.9967357     1.000445

I have 1738 total dataset, of which H is a dependent binomial variable. There are 19.95% fall in (H=0) category and remaining are in (H=1) category. Further this binomial dependent variable compare with the covariate X whose minimum value is 82.23, mean value is 223.8 and maximum value is 391.6. The 667 missing values correspond to the covariate X i.e 667 data for X is missing in the dataset out of 1738 data.

Answer 1

Summary

The question misinterprets the coefficients.

The software output shows that the log odds of the response don't depend appreciably on $X$, because its coefficient is small and not significant ($p=0.138$). Therefore the proportion of positive results in the data, equal to $100 - 19.95\% \approx 80\%$, ought to have a log odds close to the intercept of $1.64$. Indeed,

$$\log\left(\frac{80\%}{20\%}\right) = \log(4) \approx 1.4$$

is only about one standard error ($0.22$) away from the intercept. Everything looks consistent.

---

Detailed analysis

This generalized linear model supposes that the log odds of the response $H$ being $1$ when the independent variable $X$ has a particular value $x$ is some linear function of $x$,

$$\text{Log odds}(H=1\,|\,X=x) = \beta_0 + \beta_1 x.\tag{1}$$

The glm command in R estimated these unknown coefficients with values $$\hat\beta_0 = 1.641666\pm 0.2290133$$ and $$\hat\beta_1 = -0.0014039\pm 0.0009466.$$

The dataset contains a large number $n$ of observations with various values of $x$, written $x_i$ for $i=1, 2, \ldots, n$, which range from $82.3$ to $391.6$ and average $\bar x = 223.8$. Formula $(1)$ enables us to compute the estimated probabilities of each outcome, $\Pr(H=1\,|\,X=x_i)$. If the model is any good, the average of those probabilities ought to be close to the average of the outcomes.

Since the odds are, by definition, the ratio of a probability to its complement, we can use simple algebra to find the estimated probabilities in terms of the log odds

$$\widehat\Pr(H=1\,|\,X=x) = 1 - \frac{1}{1 + \exp\left(\hat\beta_0 + \hat\beta_1 x\right)}.$$

As a nonlinear function of $x$, that's difficult to average. However, provided $\beta_1 x$ is small (much less than $1$ in size) and $1+\exp(\hat\beta_0)$ is not small (it exceeds $6$ in this case), we can safely use a linear approximation

$$\frac{1}{1 + \exp\left(\hat\beta_0 + \hat\beta_1 x\right)} = \frac{1}{1 + \exp(\hat\beta_0)}\left(1 - \hat\beta_1 x \frac{\exp{\hat\beta_0}}{1 + \exp(\hat\beta_0)}\right) + O\left(\hat\beta_1 x\right)^2.$$

Since the $x_i$ never exceed $391.6$, $|\hat\beta_1 x_i|$ never exceeds $391.6\times 0.0014039 \approx 0.55$, so we're ok. Consequently, the average of the outcomes may be approximated as

$$\eqalign{ \frac{1}{n}\sum_{i=1}^n \widehat\Pr(H=1\,|\,X=x) &\approx \frac{1}{n}\sum_{i=1}^n \left(1 - \frac{1}{1 + \exp(\hat\beta_0)}\left(1 - \hat\beta_1 x_i \frac{\exp{\hat\beta_0}}{1 + \exp(\hat\beta_0)}\right)\right)\\ &= 0.162238 + 0.000190814 \bar{x} \\ &= 20.4943\%. }$$

Although that's not exactly equal to the $19.95\%$ observed in the data, it is more than close enough, because $\hat\beta_1$ has a relatively large standard error. For example, if $\beta_1$ were increased by only $0.3$ of its standard error to $-0.0011271$, then the previous calculation would produce $19.95\%$ exactly.