7.3.4. Assuming the observations are failure times, are the failure rates (or Mean Times To Failure) for two distributions the same?

7. Product and Process Comparisons
7.3. Comparisons based on data from two processes

7.3.4. Assuming the observations are failure times, are the failure rates (or Mean Times To Failure) for two distributions the same?

Comparing two exponential distributions is to compare the means or hazard rates

The comparison of two (or more) life distributions is a common objective when performing statistical analyses of lifetime data. Here we look at the one-parameter exponential distribution case.

In this case, comparing two exponential distributions is equivalent to comparing their means (or the reciprocal of their means, known as their hazard rates).

Type II Censored data

Definition of Type II censored data

Definition: Type II censored data occur when a life test is terminated exactly when a pre-specified number of failures have occurred. The remaining units have not yet failed. If $n$ units were on test, and the pre-specified number of failures is $r$ (where $r$ is less than or equal to $n$), then the test ends at $t_r$, the time of the $r$-th failure.

Two exponential samples oredered by time

Suppose we have Type II censored data from two exponential distributions with means $\theta_1$ and $\theta_2$. We have two samples from these distributions, of sizes $n_1$ on test with $r_1$ failures and $n_2$ on test with $r_2$ failures, respectively. The observations are time to failure and are therefore ordered by time. $$ t_{1(1)} \, < \, \ldots \, < t_{1(n)} \,\,\,\,\, (r_1 \le n_1) $$ $$ t_{2(1)} \, < \, \ldots \, < t_{2(n)} \,\,\,\,\, (r_2 \le n_2) $$

Test of equality of $\theta_1$ and $\theta_2$ and confidence interval for $\theta_1 / \theta_2$

Letting $$ T_i = \sum_{j=1}^{r_i} t_{i(j)} + (n_i - r_i) t_{i(r_i)} \,\,\,\,\, i = 1, \, 2 \, .$$

Then $$ \frac{2T_1}{\theta_1} \approx \chi_{2r_1}^2 \,\,\,\,\, \mbox{ and } \,\,\,\,\, \frac{2T_2}{\theta_2} \approx \chi_{2r_2}^2 \, , $$ with $T_1$ and $T_2$ independent. Thus $$ U = \frac{2T_1 / (2 r_1 \theta_1)}{2T_2 / (2 r_2 \, \theta_2)} = \frac{\hat{\theta}_1 \theta_2}{\hat{\theta}_2 \, \theta_1} \, , $$ where $$ \hat{\theta}_1 = \frac{T_1}{r_1} \,\,\,\,\, \mbox{ and } \,\,\,\,\, \hat{\theta}_2 = \frac{T_2}{r_2} \, . $$ $U$ has an $F$ distribution with $(2 r_1, \, 2 r_2)$ degrees of freedom. Tests of equality of $\theta_1$ and $\theta_2$ can be performed using tables of the $F$ distribution or computer programs. Confidence intervals for $\theta_1 / \theta_2$, which is the ratio of the means or the hazard rates for the two distributions, are also readily obtained.

Numerical example

A numerical application will illustrate the concepts outlined above.

For this example, $$ H_0: \,\, \theta_1 / \theta_2 = 1 $$ $$ H_a: \,\, \theta_1 / \theta_2 \ne 1 $$ Two samples of size 10 from exponential distributions were put on life test. The first sample was censored after 7 failures and the second sample was censored after 5 failures. The times to failure were:

Sample 1:	125	189	210	356	468	550	610
Sample 2:	170	234	280	350	467

So $r_1$ = 7, $r_2$ = 5, and $t_{1, \, r_1}$ = 610, $t_{2, \, r_2}$ = 467.

Then $T_1$ = 4338 and $T_2$ = 3836.

The estimator for $\theta_1$ is 4338 / 7 = 619.71 and the estimator for $\theta_2$ is 3836 / 5 = 767.20.

The ratio of the estimators = $U$ = 619.71 / 767.20 = 0.808.

If the means are the same, the ratio of the estimators, $U$, follows an $F$ distribution with $(2 r_1, \, 2 r_2)$ degrees of freedom. The $P(F < 0.808) = 0.348$. The associated p-value is 2(0.348) = 0.696. Based on this $p$-value, we find no evidence to reject the null hypothesis (that the true but unknown ratio = 1). Note that this is a two-sided test, and we would reject the null hyposthesis if the $p$-value is either too small (i.e., less or equal to 0.025) or too large (i.e., greater than or equal to 0.975) for a 95 % significance level test.

We can also put a 95 % confidence interval around the ratio of the two means. Since the 0.025 and 0.975 quantiles of $F_{14, \, 10}$ are 0.3178 and 3.5504, respectively, we have $$ P(U/3.5504 < \theta_1 / \theta_2 < U/0.3178) = 0.95 $$ and (0.228, 2.542) is a 95 % confidence interval for the ratio of the unknown means. The value of 1 is within this range, which is another way of showing that we cannot reject the null hypothesis at the 95 % significance level.