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.
Best Answer
If you would write the likelihood, then the answer would be easy. I use subscript $i$ for the first group and $j$ for the second group,
The likelihood for the model is $$ \prod_i \lambda_1 ^{\delta_i} \exp(-\lambda_1t_i) \prod_j \lambda_2^{\delta_j} \exp(-\lambda_2 t_j) $$ so the log-likelihood is $$ \ell = \log \lambda_1 \sum_i \delta_i - \lambda_1 \sum_i t_i + \log\lambda_2 \sum_j \delta_j - \lambda_2\sum_j t_j $$
Then you replace $\lambda_2 = 5 \lambda_1$ to obtain the (log)likelihood under the null $$ \ell_0 = \log \lambda_1 \sum_i \delta_i - \lambda_1 \sum_i t_i + (\log(5) + \log \lambda_1) \sum_j \delta_j - (5 + \lambda_1)\sum_j t_j $$
The difference in degrees of freedom is 1 since there is a difference of one parameter between the two models.
Then calculate the test statistic \begin{align*} 2(\ell - \ell_0) &= 2 * ((\log \lambda_2 - \log \lambda_1 - \log(5))\sum_j \delta_j + (-\lambda_2 + 5 + \lambda_1) \sum_j t_j ) \end{align*}
If you calculate $\lambda_1$ and $\lambda_2$, they are just the number of events / total time at risk (MLE of exponential distribution). You have the estimates $\lambda_1 = 9/361 = 0.024$ and $\lambda_2 = 21 / 182 = 0.11$ (maybe I made a mistake here, the most difficult part of the exercise).
Your test statistic is then $$ 2 * (21 * (-2.20 + 3.72 -1.6) + 182 * (-0.11 + 5 + 0.024)) = 1785$$ (again, calculations by hand might be wrong so you should check that).
Now if you vaguely know how the chi square distribution with 1 degree of freedom looks like, you'll have no trouble concluding that $1785 >> 1$ and that the null hypothesis is rejected.