Exam MAS-I · practice
Exam MAS-I — Statistics & Simulation Practice Flashcards
Thirty exam-realistic multiple-choice problems on CAS Exam MAS-I statistics and simulation — point estimation and estimator properties, maximum likelihood with Fisher-information variance, bias and MSE, confidence intervals for means and proportions, hypothesis tests with z/t/chi-square statistics and power, and inverse-transform simulation with Monte Carlo and bootstrap reasoning — each with a fully worked solution.
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- Bias, variance & MSEAn estimator $\hat\theta$ of $\theta$ satisfies $E[\hat\theta]=\theta+\dfrac{3}{n}$ and $\text{Var}(\hat\theta)=\dfrac{2\theta^{2}}{n}$. For $n=6$ and $\theta=4$, calculate the mean squared error of $\hat\theta$. (A) $0.25$ (B) $5.33$ (C) $5.58$ (D) $5.83$ (E) $32.25$**Answer: (C).** Bias $=E[\hat\theta]-\theta=\dfrac{3}{n}=\dfrac{3}{6}=0.5$, so $\text{bias}^{2}=0.25$. Variance $=\dfrac{2\theta^{2}}{n}=\dfrac{2(4)^{2}}{6}=\dfrac{32}{6}\approx 5.3333$. $\text{MSE}=\text{Var}+\text{bias}^{2}=5.3333+0.25=5.5833\approx 5.58$. Reporting only the variance $5.33$ (forgetting the bias-squared term) gives distractor (B); reporting only $\text{bias}^{2}=0.25$ gives (A); adding the bias itself rather than its square gives $5.83$ (D).
- Maximum likelihoodExponential lifetimes with mean $\theta$ and density $f(x)=\dfrac{1}{\theta}e^{-x/\theta}$ are observed as $3,8,5,12,7,1$. Calculate the maximum-likelihood estimate $\hat\theta$. (A) $5.0$ (B) $5.6$ (C) $6.0$ (D) $7.2$ (E) $36.0$**Answer: (C).** The exponential MLE for the mean is the sample mean, $\hat\theta=\bar X$. $\sum x_{i}=3+8+5+12+7+1=36$, with $n=6$. $\hat\theta=\dfrac{36}{6}=6.0$. Reporting the total $\sum x_{i}=36$ (forgetting to divide by $n$) gives distractor (E); dividing by $n-1=5$ instead of $n$ gives $7.2$ (D).
- Confidence intervalsA sample of $n=100$ losses has $\bar x=850$ and known population standard deviation $\sigma=200$. Calculate the upper endpoint of the $95\%$ confidence interval for the mean loss. Use $z_{0.025}=1.96$. (A) $810.80$ (B) $850.00$ (C) $872.90$ (D) $889.20$ (E) $928.00$**Answer: (D).** Standard error $=\dfrac{\sigma}{\sqrt n}=\dfrac{200}{\sqrt{100}}=\dfrac{200}{10}=20$. Margin $=z_{0.025}\cdot 20=1.96(20)=39.2$. Upper endpoint $=\bar x+\text{margin}=850+39.2=889.2$. The lower endpoint $850-39.2=810.8$ is distractor (A); using the one-sided $z=1.645$ gives $850+32.9=882.9$, near (C).
- Maximum likelihoodClaim counts $x_{1},\dots,x_{n}$ are i.i.d. Poisson with rate $\lambda$, whose Fisher information for one observation is $I(\lambda)=\dfrac{1}{\lambda}$. A sample of $n=50$ has $\bar x=4$. Calculate the estimated asymptotic standard deviation of the MLE $\hat\lambda$. (A) $0.080$ (B) $0.283$ (C) $0.400$ (D) $0.566$ (E) $2.000$**Answer: (B).** The MLE is $\hat\lambda=\bar X=4$. Its asymptotic variance is $\dfrac{1}{nI(\lambda)}=\dfrac{\lambda}{n}$, estimated at $\hat\lambda$: $\widehat{\text{Var}}(\hat\lambda)=\dfrac{\hat\lambda}{n}=\dfrac{4}{50}=0.08$. Standard deviation $=\sqrt{0.08}\approx 0.2828$. Reporting the variance $0.08$ as the standard deviation is distractor (A) — a classic variance-for-SD slip. Using $\dfrac{\lambda^{2}}{n}=0.32$ then taking the root gives $0.566$ (D).
- Hypothesis testingTest $H_0:\mu=200$ versus $H_1:\mu\neq 200$ at $\alpha=0.05$ with $n=64$, $\bar x=191$, and known $\sigma=40$. Determine the test statistic and the conclusion. Use $z_{0.025}=1.96$. (A) $z=-1.80$; fail to reject $H_0$ (B) $z=-1.80$; reject $H_0$ (C) $z=-1.44$; fail to reject $H_0$ (D) $z=-1.44$; reject $H_0$ (E) $z=-9.00$; reject $H_0$**Answer: (A).** Standard error $=\dfrac{\sigma}{\sqrt n}=\dfrac{40}{\sqrt{64}}=\dfrac{40}{8}=5$. $z=\dfrac{\bar x-\mu_0}{\sigma/\sqrt n}=\dfrac{191-200}{5}=\dfrac{-9}{5}=-1.80$. Two-tailed critical value is $z_{0.025}=1.96$. Since $|-1.80|=1.80<1.96$, **fail to reject $H_0$**. Treating $1.80>1.645$ (a one-tailed cutoff) and rejecting gives distractor (B); not taking $\sqrt n$ in the standard error mis-scales $z$.
- Monte Carlo & bootstrapA continuous loss has cdf $F(x)=1-e^{-x/500}$ for $x>0$. Using the inverse-transform method with the uniform draw $U=0.30$, calculate the simulated loss $X=F^{-1}(U)$. Use $\ln 0.70\approx -0.3567$. (A) $150.00$ (B) $178.34$ (C) $356.67$ (D) $601.99$ (E) $834.03$**Answer: (B).** Set $F(x)=U$: $1-e^{-x/500}=0.30\Rightarrow e^{-x/500}=0.70\Rightarrow -\dfrac{x}{500}=\ln 0.70$. $\ln 0.70\approx -0.356675$, so $x=-500(-0.356675)=178.34$. Using $X=-500\ln(U)=-500\ln 0.30\approx 601.99$ — inverting with $U$ rather than $1-U$ — gives distractor (D). The correct form here is $X=-500\ln(1-U)$.
- Point estimationTwo unbiased estimators of $\mu$ have variances $\text{Var}(\hat\mu_{1})=\dfrac{\sigma^{2}}{n}$ and $\text{Var}(\hat\mu_{2})=\dfrac{3\sigma^{2}}{2n}$. Calculate the relative efficiency of $\hat\mu_{1}$ relative to $\hat\mu_{2}$, defined as $\dfrac{\text{Var}(\hat\mu_{2})}{\text{Var}(\hat\mu_{1})}$. (A) $0.50$ (B) $0.67$ (C) $1.00$ (D) $1.50$ (E) $2.00$**Answer: (D).** The relative efficiency of $\hat\mu_{1}$ relative to $\hat\mu_{2}$ is $\dfrac{\text{Var}(\hat\mu_{2})}{\text{Var}(\hat\mu_{1})}$ (the more efficient estimator on top makes the ratio exceed $1$). $\dfrac{\text{Var}(\hat\mu_{2})}{\text{Var}(\hat\mu_{1})}=\dfrac{3\sigma^{2}/(2n)}{\sigma^{2}/n}=\dfrac{3}{2}=1.50$. So $\hat\mu_{2}$ has $1.5$ times the variance of $\hat\mu_{1}$; $\hat\mu_{1}$ is the more efficient one. Inverting the ratio gives $0.67$ (distractor B).
- Bias, variance & MSEA sample of size $n$ from a distribution with variance $\sigma^{2}$ uses $T=\dfrac{1}{n}\sum_{i=1}^{n}(X_{i}-\bar X)^{2}$ to estimate $\sigma^{2}$. If $\sigma^{2}=90$ and $n=10$, calculate the bias of $T$. (A) $-9.0$ (B) $-1.0$ (C) $0.0$ (D) $9.0$ (E) $81.0$**Answer: (A).** The expected sum of squared deviations about $\bar X$ is $(n-1)\sigma^{2}$, so $E[T]=\dfrac{(n-1)\sigma^{2}}{n}=\left(1-\dfrac{1}{n}\right)\sigma^{2}$. Bias $=E[T]-\sigma^{2}=-\dfrac{\sigma^{2}}{n}=-\dfrac{90}{10}=-9.0$. The divisor-$n$ estimator underestimates $\sigma^{2}$, hence a negative bias. Assuming $T$ is unbiased (the $n-1$ version) gives $0$ (distractor C).