We now use a maximum likelihood test to compare two random samples from exponential distributions.

Example

Clinical records give the survival time in months from diagnosis of 30 sufferers from a certain disease as

9.73
5.56
4.28
4.87
1.55
6.20
1.08
7.17
28.65
6.10
16.16
9.92
2.40
6.19
7.67
1.11
4.66
4.35
7.31
3.28
13.38
3.08
0.41
4.33
2.16
4.49
0.75
4.45
10.29
0.90

In a clinical trial of a new drug treatment, 21 sufferers had survival times of

22.07
12.47
6.42
8.15
0.64
20.04
17.49
2.22
3.00
28.09
3.94
8.59
4.26
32.82
8.32
2.12
18.53
9.95
4.25
3.70
5.82

Assuming that survival times are exponentially distributed, perform a likelihood ratio test for whether the death rate is different for those getting the new drug.

The assumption of exponential distributions means that

\[ \{X_{C,1}, \dots, X_{C,30}\} \;\;\sim\;\; \ExponDistn(\lambda_C) \\ \{X_{D,1}, \dots, X_{D,21}\} \;\;\sim\;\; \ExponDistn(\lambda_D) \]

The hypotheses of that we want to test are:

The log-likelihood for the big model is

\[ \begin{align} \ell(\lambda_C, \lambda_D) \;\;&=\;\; \left(30 \log \lambda_C - \left(\sum_{i}{x_{C,i}}\right) \lambda_C \right) + \left(21 \log \lambda_D - \left(\sum_{i}{x_{D,i}}\right) \lambda_D \right) \\ &=\;\; 30 \log \lambda_C - 182.48 \lambda_C + 21 \log \lambda_D - 222.87 \lambda_D \end{align} \]

which simplifies for the small model to be

\[ \ell(\lambda) \;\;=\;\; 51 \log \lambda - 405.35 \lambda \]

Big model, \(\mathcal{M}_B\)

For the big model, the maximum likelihood estimates of the two unknown parameters are

\[ \hat{\lambda}_C \;\;=\;\; \frac{n_C}{\sum{x_{C,i}}} \;\;=\;\; \frac{30}{182.48} \;\;=\;\; 0.1644 \\ \hat{\lambda}_D \;\;=\;\; \frac{n_D}{\sum{x_{D,i}}} \;\;=\;\; \frac{21}{222.87} \;\;=\;\; 0.0942 \]

The maximum possible value for the log-likelihood is found by replacing \(\lambda_C\) and \(\lambda_D\) in the log-likelihood with these values,

\[ \ell(\mathcal{M}_B) \;\;=\;\; \ell(\hat{\lambda}_C,\hat{\lambda}_D) \;\;=\;\; -154.77 \]

Small model, \(\mathcal{M}_S\)

When the two exponential parameters are equal, their maximum likelihood estimate is

\[ \hat{\lambda} \;\;=\;\; \frac{n}{\sum{x_i}} \;\;=\;\; \frac{51}{405.35} \;\;=\;\; 0.1258 \]

giving log-likelihood

\[ \ell(\mathcal{M}_S) \;\;=\;\; \ell(\hat{\lambda}) \;\;=\;\; -156.72 \]

Likelihood ratio test

The test statistic is

\[ \chi^2 \;\;=\;\; 2\left(\ell(\mathcal{M}_B) - \ell(\mathcal{M}_S)\right) \;\;=\;\; 3.906 \]

Since there is one more unknown parameter in the big model, this should be compared to the \(\ChiSqrDistn(1 \text{ df})\) distribution. Its upper tail probability above 3.906 is 0.048 and this is the p-value for the test.

Since the p-value is just under 5%, we should conclude that there is only moderately strong evidence that the survival rate is different for those getting the new drug treatment.