Statistics and Econometrics Questions and Answers

Answer: B

To accept the null hypothesis, the computed statistic should be Less than the critical value.

In a significance test, the critical value marks the boundary of the rejection region at the chosen level of significance. If the computed value of the test statistic (in absolute terms, for a two-tailed test) is less than the critical value, it falls in the acceptance region and the null hypothesis is not rejected. If it is greater, the null is rejected.

'Greater' is the condition for rejection, while 'superior' and 'significant' are not comparisons of the two values.

Answer: A

Logit and probit models are used for binary data.

They model a dependent variable that takes two values, such as 1 if a household owns a house and 0 if not. The probability of the outcome is modelled through a cumulative distribution function, the logistic for logit and the standard normal for probit, which keeps predicted probabilities between 0 and 1, unlike the linear probability model.

Continuous dependent variables call for ordinary regression, missing data for imputation methods, and these models are not tied to business cycles.

Answer: D

The standard normal statistic is \(\dfrac{\bar{X} - u}{\sigma/\sqrt{n}}\) (the paper writes \(u\) for the mean \(\mu\)).

For a random sample from \(N(\mu, \sigma^{2})\), the sample mean is \(N(\mu, \sigma^{2}/n)\). Subtracting its mean and dividing by its standard deviation \(\sigma/\sqrt{n}\) gives a standard normal variable.

Dividing by \(\sigma^{2}\) instead of the standard error gives the wrong scale. The squared forms are non-negative, so they cannot be normal; \((\bar{X} - \mu)^{2}/(\sigma^{2}/n)\) would be chi-square with one degree of freedom, and the reciprocal form has no standard distribution here.

Answer: C

The Goldfeld-Quandt test is a test of heteroscedasticity.

The observations are ordered by the variable thought to drive the error variance, a block of central observations is dropped, and separate regressions are fitted to the two ends. The ratio of their residual sums of squares, each divided by its degrees of freedom, follows an F distribution if the error variance is constant. A large ratio indicates heteroscedasticity.

Autocorrelation is tested with the Durbin-Watson or Breusch-Godfrey tests, and the correlation between X and Y is measured by the correlation coefficient, not by this test.

Answer: A

Pearson's coefficient is \(\dfrac{\text{Covariance(X, Y)}}{\sigma_{X}\sigma_{Y}}\).

Dividing the covariance by the product of the two standard deviations removes the units of measurement and gives a number between -1 and 1. Equivalently,

\[ \begin{aligned} r &= \frac{\sum(x - \bar{x})(y - \bar{y})}{\sqrt{D}}, \\ D &= \sum(x - \bar{x})^{2}\sum(y - \bar{y})^{2} \end{aligned} \]

Dividing by the product of the variances gives a quantity that depends on the units, and the forms that multiply the covariance by variances are not correlation coefficients at all.

Answer: B

With serial correlation, OLS estimators are unbiased but not efficient.

Unbiasedness of OLS needs only that the errors have zero mean given the regressors; it does not depend on the errors being uncorrelated. But the Gauss-Markov theorem, which makes OLS the best (minimum variance) linear unbiased estimator, assumes uncorrelated errors. With autocorrelation, generalised least squares has smaller variance, so OLS is no longer efficient, and the usual OLS standard errors are biased.

The estimates can still be calculated, and they are not 'wrong' in the sense of being biased.

Answer: B

\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), the addition rule of probability.

When the probabilities of A and of B are added, the outcomes in both events, \(A \cap B\), are counted twice. Subtracting \(P(A \cap B)\) once corrects the double count.

\(P(A) + P(B)\) alone is right only for mutually exclusive events, where \(P(A \cap B) = 0\). Adding \(P(A \cup B)\) again makes the formula circular and too large, and \(P(A)/P(B)\) has nothing to do with a union; a ratio of this kind appears in conditional probability.

Answer: A

The OLS estimates are biased and inconsistent.

OLS requires the regressor to be uncorrelated with the error term. If \(\mathrm{Cov}(X, e) \neq 0\), the regressor is endogenous, and the estimator of the slope converges to

\[ \begin{aligned} \hat{\beta} &\to \beta \\ &+ \frac{\mathrm{Cov}(X, e)}{\mathrm{Var}(X)} \end{aligned} \]

which differs from \(\beta\). The error does not vanish as the sample grows, so the estimator is inconsistent as well as biased. Instrumental variables are the usual remedy.

Bias in small samples with consistency in large ones is the case of a lagged dependent variable with serially uncorrelated errors, not of a correlated regressor. The estimates can still be computed; they are simply wrong on average.

Answer: C

A random walk is non-stationary.

A random walk is \(Y_{t} = Y_{t-1} + e_{t}\). Each shock is permanently added in, so \(Y_{t} = Y_{0} + e_{1} + \cdots + e_{t}\) and its variance \(t\sigma^{2}\) grows with time. A series whose variance changes over time is not stationary; this one has a unit root.

It is not impossible to analyse: taking first differences gives \(\Delta Y_{t} = e_{t}\), which is stationary, and unit-root tests such as Dickey-Fuller are built for it. It is also very useful in practice, as the standard model of stock prices and exchange rates.

Answer: C

Rejecting a true null hypothesis is a Type I error.

A Type I error is a false rejection: the null hypothesis is true but the test rejects it. Its probability is the significance level \(\alpha\) chosen for the test, such as 5 per cent.

A Type II error is the opposite mistake, accepting (failing to reject) a null hypothesis that is false; its probability is \(\beta\), and \(1 - \beta\) is the power of the test. 'Type 0' is not a standard term, and 'Type III' is an informal name for answering the wrong question, not part of the classical framework.