Statistics and Econometrics Questions and Answers

Q: Logit and Probit models are used to analyze data that are :

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.

Q: If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a normally distributed population with mean \(\mu\) and variance \(\sigma^{2}\), then the following follows a standard normal distribution :

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.

Q: Suppose \(Y_{i} = \alpha + \beta X_{i} + e_{i}\) is a linear regression model. Goldfeld-Quandt test is a test of :

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.

Q: The following is the formula for Pearson's coefficient of correlation :

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.

Q: If \(Y_{1} = \alpha + \beta X_{i} + e_{i}\) is a linear regression model and there is serial correlation among the error terms, then :

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.

Q: Suppose, in a regression model, \(Y_{i} = \alpha + \beta X_{i} + e_{i}\); covariance between X and e is not zero. Then :

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.

Q: A random walk time series is :

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.

Q: Rejecting the null hypothesis when it is true is referred to as :

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.

21.

For accepting a null hypothesis, the computed value of a test statistic should be ............................ than the critical value.

A.Greater
B.Less
C.Superior
D.Significant

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

Discuss in Forum

22.

Logit and Probit models are used to analyze data that are :

A.Binary
B.Continuous
C.Missing
D.Used to analyze business cycles exclusively

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

Discuss in Forum

23.

If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a normally distributed population with mean \(\mu\) and variance \(\sigma^{2}\), then the following follows a standard normal distribution :

A.\(\dfrac{\bar{X} - u}{\sigma^{2}}\)
B.\(\dfrac{(\bar{X} - u)^{2}}{\sigma^{2}}\)
C.\(\dfrac{\sigma^{2}}{(\bar{X} - u)^{2}}\)
D.\(\dfrac{\bar{X} - u}{\sigma/\sqrt{n}}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

Discuss in Forum

24.

Suppose \(Y_{i} = \alpha + \beta X_{i} + e_{i}\) is a linear regression model. Goldfeld-Quandt test is a test of :

A.Autocorrelation
B.Data mistakes
C.Heteroscedasticity
D.Correlation between X and Y

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

Discuss in Forum

25.

The following is the formula for Pearson's coefficient of correlation :

A.\(\dfrac{\text{Covariance(X, Y)}}{\sigma_{X}\sigma_{Y}}\)
B.\(\dfrac{\text{Covariance(X, Y)}}{\sigma_{X}^{2}\sigma_{Y}^{2}}\)
C.\(\dfrac{\sigma_{X}^{2}}{\sigma_{Y}^{2}}\) Covariance(X, Y)
D.\(\sigma_{X}^{2} \times\) Covariance(X, Y) \(\times \sigma_{Y}^{2}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

Discuss in Forum

26.

If \(Y_{1} = \alpha + \beta X_{i} + e_{i}\) is a linear regression model and there is serial correlation among the error terms, then :

A.OLS estimators are unbiased and efficient
B.OLS estimators are unbiased but not efficient
C.OLS estimators are wrong
D.OLS estimators cannot be calculated

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

Discuss in Forum

27.

Probability of \((A \cup B) = P(A \cup B) =\)

A.P(A) + P(B)
B.\(P(A) + P(B) - P(A \cap B)\)
C.\(P(A) + P(B) + P(A \cup B)\)
D.P(A)/P(B)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

Discuss in Forum

28.

Suppose, in a regression model, \(Y_{i} = \alpha + \beta X_{i} + e_{i}\); covariance between X and e is not zero. Then :

A.OLs estimates are biased and inconsistent
B.OLs estimates are biased but consistent
C.OLs estimates are unbiased but inconsistent
D.OLs estimates cannot be calculated

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

Discuss in Forum

29.

A random walk time series is :

A.Stationary
B.Impossible to analyze
C.Non-stationary
D.Not useful in real world

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

Discuss in Forum

30.

Rejecting the null hypothesis when it is true is referred to as :

A.Type II error
B.Type 0 error
C.Type I error
D.Type III error

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

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