Free CAS MAS-I (Modern Actuarial Statistics I) Statistics Practice Questions

For an exponential distribution with rate parameter $\lambda$, the variance is $1/\lambda^2$. Given $\text{Var}(X) = 1/9$, we have $1/\lambda^2 = 1/9$, so $\lambda = 3$ and the mean is $1/\lambda = 1/3$. The median of an exponential distribution is $\frac{\ln 2}{\lambda} = \frac{\ln 2}{3} = \frac{0.6931}{3} \approx 0.2310$. This falls in $[0.20, 0.25)$.

Choice A is incorrect because 0.2310 is not less than 0.15.
Choice B is incorrect because 0.2310 is not less than 0.20.
Choice
Choice E is correct because $0.20 \leq 0.2310 < 0.25$.
Choice C is incorrect because 0.2310 is less than 0.25.
Choice D is incorrect because 0.2310 is less than 0.30.

(A) is correct.

The Nelson-Dalen estimator is:
$$\hat{H}(3) = \frac{d_1}{n_1} + \frac{d_2}{n_2} + \frac{d_3}{n_3} = \frac{8}{80} + \frac{10}{68} + \frac{7}{52}$$
$$= 0.10000 + 0.14706 + 0.13462 = 0.38168 \approx 0.382$$

Choice E is incorrect because 0.217 omits the third term, computing only through time 2 and understating it.
Choice B is incorrect because 0.282 uses $d_j/(n_j + d_j)$ instead of $d_j/n_j$.
Choice D is incorrect because 0.452 results from using reduced risk sets as if censoring preceded deaths.
Choice C is incorrect because 0.547 is $1 - \hat{S}(3)$ from the Kaplan-Meier, not the cumulative hazard.

Choice A is correct.

The joint density is:
$$\prod_{i=1}^n f(x_i \mid \theta) = \prod_{i=1}^n \theta x_i^{\theta-1} = \theta^n \left(\prod_{i=1}^n x_i\right)^{\theta-1}$$

Rewriting using $T = \prod x_i$:
$$= \theta^n \exp\left[(\theta - 1) \sum_{i=1}^n \ln x_i\right] = \theta^n T^{\theta-1}$$

This factors as $g(T, \theta) = \theta^n T^{\theta-1}$ and $h(\mathbf{x}) = 1$. Since the $\theta$-dependent part depends on the data only through $T = \prod X_i$, by the factorization theorem $\prod X_i$ is sufficient for $\theta$. Equivalently, $\sum \ln X_i$ is sufficient since it is a one-to-one transformation of $\prod X_i$.

Choice E is incorrect because $\sum X_i$ does not appear in the factored likelihood; the joint density involves $\prod x_i$ raised to a power, not $\sum x_i$.
Choice C is incorrect because $\sum X_i^\theta$ depends on the parameter $\theta$ inside the statistic, which does not yield a valid sufficient statistic.
Choice D is incorrect because the support $(0,1)$ does not depend on $\theta$, so the maximum order statistic carries no special role; it does not capture the $\prod x_i$ structure.
Choice B is incorrect because $\sum X_i^2$ does not appear in the factorization and is not sufficient for this power-family distribution.