Cox regression. Find 95% confidence interval for comparison of two groups

I am working with pbc dataset in R and would like to build 95% confidence interval for the comparison of two groups:

1. 60-year-old males on DPCA with bilirubin = 1 mg/dL
2. 40-year-old females on placebo with bilirubin = 0.5 mg/dL

In order to to do this I built Cox regression:

cox.adj = coxph(Surv(time, status) ~ trt + age.cat + bili + sex, data = data)
coef exp(coef) se(coef)      z Pr(>|z|)
trt1                 0.10349   1.10904  0.18644  0.555  0.57882
age.cat 2. [42, 50) -0.01111   0.98895  0.30553 -0.036  0.97099
age.cat 3. [50, 57)  0.52052   1.68291  0.28611  1.819  0.06887 .
age.cat 4. >= 57     0.87540   2.39983  0.28495  3.072  0.00213 **
bili                 0.14951   1.16127  0.01343 11.136  < 2e-16 ***
sexf                -0.50976   0.60064  0.24207 -2.106  0.03522 *
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

exp(coef) exp(-coef) lower .95 upper .95
trt1                   1.1090     0.9017    0.7696    1.5982
age.cat 2. [42, 50)    0.9890     1.0112    0.5434    1.7999
age.cat 3. [50, 57)    1.6829     0.5942    0.9606    2.9485
age.cat 4. >= 57       2.3998     0.4167    1.3729    4.1950
bili                   1.1613     0.8611    1.1311    1.1922
sexf                   0.6006     1.6649    0.3737    0.9653

data_test1
age.cat sex trt bili
1  4. >= 57   m   1  1.0
2    1. <42   f   0  0.5

prediction = predict(cox.adj, newdata = data_test1, se = T)
prediction
$fit 1 2 0.7007238 -0.8626783$se.fit
1         2
0.2556641 0.2074737

Hazard ratio:

exp(0.7007238) / exp(-0.8626783) = 4.78

So, 60-year-old males on DPCA with bilirubin = 1.0 mg/dL have 4.78 times the hazard of mortality when compared to 40-year-old females on placebo with bilirubin = 0.5 mg/dL.

The question is how to calculate 95% cofidence interval for this comparison? My approach is:

exp(0.7007238 - 2 * 0.2556641) / exp(-0.8626783 - 2 * 0.2074737)
exp(0.7007238 + 2 * 0.2556641) / exp(-0.8626783 + 2 * 0.2074737)

which gives me [4.336299, 5.258169] but I have information that 95% confidence interval probably should be about [2.3, 10]. So, my question is - are my calculations correct or not? If not, how should I calculate 95% confidence interval for this comparison?

Here are slides which I am trying to replicate in R:

If $$x_1$$ and $$x_2$$ are column vectors of predictor values that you want to compare, take the vector difference between them, $$X = x_2 -x_1$$. If $$\hat \beta$$ is the column vector of coefficient estimates, then the point estimate of the difference in log hazard ratio between the 2 situations is the inner product $$X^T \hat \beta$$. If $$V$$ is the variance-covariance matrix among the coefficient estimates, the variance of the estimated difference is the product $$X^T V X$$. Use that to get the 95% CI for the log hazard ratio difference, assuming a normal distribution. Finally you exponentiate to get the hazard ratio CI between the 2 situations.