Sampling Distributions Questions and Answers

Answer: A

All three statements are true. In probability theory an outcome is the result of a random experiment. Statistical inference aims to estimate the uncertainty, or the sample to sample variation, in what is learnt from a sample. A sampling error is the difference between the true population values (parameters) and the measured sample values (statistics). The other options mark at least one of these as false, which is wrong. An outcome such as getting a 4 on one roll of a die is a result of a random experiment, which is the idea in statement i.

Answer: D

\(E\left(\frac{Y}{X}\right)\) is equal to \(\frac{5}{7}\).

For a chi-squared variable with \(k\) degrees of freedom, \(E(\chi_{k}^{2}) = k\) and, for \(k \gt 2\), \(E\left(\frac{1}{\chi_{k}^{2}}\right) = \frac{1}{k - 2}\). By independence:

\[ \begin{aligned} E\left(\tfrac{X}{Y}\right) &= \tfrac{m}{n - 2} = 3, \quad m + n = 14 \\ 14 - n &= 3n - 6 \Rightarrow n = 5,\ m = 9 \\ E\left(\tfrac{Y}{X}\right) &= \tfrac{n}{m - 2} = \tfrac{5}{7} \end{aligned} \]

Taking \(E(Y/X)\) as the reciprocal \(\frac{1}{3}\) is the common mistake: the expectation of a reciprocal is not the reciprocal of the expectation, and \(\frac{1}{3}\) is not offered.

Answer: C

The variance is \(\lambda\).

A Poisson(\(\lambda\)) variable has mean \(\lambda\) and variance \(\lambda\). \(W_{n}\) is the sample mean of independent, identically distributed variables with finite variance, so by the central limit theorem

\[ \begin{aligned} \sqrt{n}(W_{n} - \lambda) &\to N(0, \sigma^{2}) \\ &= N(0, \lambda) \end{aligned} \]

The scaling by \(\sqrt{n}\) is exactly what keeps the variance at \(\sigma^{2}\) rather than shrinking it to \(\sigma^{2}/n\). \(\sqrt{\lambda}\) is the standard deviation, not the variance; \(\lambda^{2}\) squares the variance again; and 1 is the variance only after dividing by \(\sqrt{\lambda}\) as well.

Answer: C

The maximum likelihood estimator is \(1/(\overline{X} - 1)\).

The density is an exponential shifted to start at 1. For a sample with every \(x_{i} \ge 1\) the log-likelihood is

\[ \begin{aligned} \ell(\theta) &= n\ln\theta - \theta\sum_{i=1}^{n}(x_{i} - 1) \\ \ell'(\theta) &= \frac{n}{\theta} - \sum_{i=1}^{n}(x_{i} - 1) = 0 \\ \hat{\theta} &= \frac{n}{\sum x_{i} - n} = \frac{1}{\overline{X} - 1} \end{aligned} \]

\(\ell''(\theta) = -n/\theta^{2} \lt 0\), so this is a maximum. \(1/\overline{X}\) is the estimator for the unshifted exponential, which ignores the shift by 1, and \((1/\overline{X}) - 1\) applies the shift in the wrong place. \(\overline{X}\) estimates a mean, not a rate.