Suppose that the data below are remission times (in weeks) of two treatment groups and that data follow an exponential distribution.

Group 1: (6,6,7,8,10,13,16,22,23,6+,9+,10+,11+,17+,19+,20+,25+,32+,32+,34+,35+)

Group 2: (1,1,2,2,3,4,4,5,5,7,8,8,9,11,11,12,12,15,17,22,23)

where $+$ mean censored data.

How I can use the likelihood ratio test in this case to test the hypothesis that the remission risk in Group 2 is 5 times the risk of Group 1?

Let $T\sim \exp(\alpha)$ then $$f(t)=\alpha\exp(-\alpha t)\qquad t>0$$

$$S(t)=P(T\geq t)=1-F_T(t)=\exp(-\alpha t)$$

then

$$\lambda(t)=\frac{f(t)}{S(t)}=\frac{\alpha\exp(-\alpha t)}{\exp(-\alpha t)}=\alpha$$

So I think if the $\lambda_1(t)=\alpha_1$ and $\lambda_2(t)=\alpha_2$ then the hypothesis is $$H_0:\lambda_2(t)=5\lambda_1(t)$$ in the particular case of exponential $$H_0:\alpha_2=5\alpha_1$$

What I think to do is find the likelihood estimate for two groups and compare

For Group 1

$$\alpha_1=\frac{\sum \delta_i}{\sum t_i}$$

where $\delta=1$ if are not censoring and $\delta=0$ if a particular time is censoring, and $t_i$ are the remission times.

I'm a litle lost. How the right way to test it?

This question is from an old exam, and need to be done at hand. I don't understand how I can use the likelihood test in this case, because if two groups follow a exponential distribution, the degrees of freedom in the likelihood test will be 0, what no make sense.