Incorrect! In the framework where there are a total of k "regressors" including a constant (i.e. k parameters to estimate) and T observations are used, t-test statistics will follow a t-distribution with T-k degrees of freedom. In this case, T=27 and k=3. The fact that the observations are quarterly is irrelevant. We would thus be looking in the t-tables in the degrees of freedom=24 row and the 2.5% row (so that 2.5% is in each tail for a 5% 2-sided test). The critical value would be 2.06. Answer b would be incorrectly obtained if you had mistakenly looked at the 5% column, which would have been the right one if a 2-sided test at 10% significance level or a 5% one-sided test were being used. You would have obtained answer d if you had forgotten to take the number of parameters estimated away from the number of observations in the degrees of freedom calculation. 1.64 is the 5% one-sided normal critical value.

Incorrect! u-hat will have the same dimensions as u - that is, T ( 1. u-hat transposed will therefore have dimensions 1 ( T, and u-hat transposed multiplied by u-hat will have dimensions 1 ( T ( T ( 1, which is 1 ( 1. This can also be seen as correct since u-hat transposed u-hat is the residual sum of squares written in a matrix form. This quantity will always be a scalar, and so it must be 1 ( 1.

Incorrect! This question requires you to do the work of both of the previous questions. First, you need to calculate the intercept coefficient, which will be obtained by multiplying the first row of the inverse(X(X) matrix by the (X(y) column - i.e.

(-0.3 ( 0.5) + (0.2 ( 0.4) + (-0.1 ( 0.2) = -0.09.

The standard error will be given by the square root of (0.022 ( 0.3), which is 0.0816.

The test statistic will be calculated as the estimate divided by its standard error (since the value of beta tested under the null hypothesis is zero). Thus the test statistic is

-0.09 / 0.0816 = -1.10.

Incorrect! It would be wrong to remove the intercept, since it should always be retained in a regression model, even if it is not significant in the regression model. The reason is that excluding the intercept can lead to biased and inconsistent estimates on all of the slope parameters. Also, any forecasts made from a regression where there is no intercept term could also be biased. There would be no need to recomputed the test statistic, since it is quite plausible that the intercept coefficient is not significant. The regression line would only go through the origin if the intercept estimate were exactly zero. In practice, that is extremely unlikely to be the case.

Incorrect! The answer is no. If we use a 5% one-sided test with a single time series sample of data, we would expect 5% of the managers to outperform the market in a statistically significant way 5% of the time by chance alone. In other words, if we took 100 completely random trading rules and applied them to the data, we would expect on average 5% of the rules to beat the market statistically. Since, in this case, only 3 fund managers were able to beat the market in a statistical sense, this does not show evidence for stock market inefficiency, so b is correct. Note that if we had many separate sets of time series observations, and we found that the same 3 fund managers were significantly outperforming the market every time, this would be evidence against the EMH.

Incorrect! The way to do this would be as follows. First, rearrange the restriction so that it is written with one of the betas as the subject of the expression - for example, beta2 = 1-beta3. Next, impose this on the regression model by substituting in for beta2 into the equation. The equation would then be

y = beta1 + (1-beta3) x2 + beta3 x3 + beta4 x4 + u

Then, expanding the parentheses would give

y = beta1 + x2 - beta3 x2 + beta3 x3 + beta4 x4 + u

Then expand the parentheses and rearrange the expression, gathering any variables without parameters attached on the left-hand side, and gathering together all terms in each of the betas.

y - x2 = beta1 + beta3 (x3 - x2) + beta4 x4 + u

Finally, in order to be able to estimate the model, a substitution would be made to make the last equation appear as a standard linear regression model, e.g. P = (y - x2) and Q = (x3 - x2), so that the estimated regression would be

P = beta1 + beta3 Q + beta4 x4 + u

One obvious question to ask would be what would have happened if beta3 had been made the subject of the restriction instead of beta2 (i.e. if we write beta3 = 1-beta2). If this had been the case, once the restriction had been imposed, the residual sum of squares would be identical to that which would result if beta2 is substituted out.

Incorrect! In order to be tested using an F-test, restrictions must be linear (in the parameters), and clearly (iii) fits this description. It is important to also recall that single hypotheses such as (i) can also be tested using an F-test as well as a t-test, so that (iii) can be tested using an F-test. (iv) is obviously a non-linear function of the parameters so cannot be tested using an F-test, and similarly (ii) involves a squared term so that this cannot be tested either. Therefore, the correct answer is b.

Incorrect! The null hypothesis in question 9 involves m = 1 restriction (although the single restriction involves two parameters), and we are told that the number of observations, T, is 1,000. k, the number of "regressors" in the unrestricted regression including a constant, or the number of parameters estimated in the unrestricted regression, which is 4. Therefore, the appropriate critical value to look up would be an F(1,996), which would be found in the column corresponding to the first column and the 996th row of the F-table. Since there is no 996th row, the most appropriate value to use would be in the "infinity" row (i.e. the last row) of the table. The appropriate critical value (assuming a significance level of 5%) would be 3.84. You would have incorrectly got 3.00 if you had used m = 2 restrictions, while wrongly obtaining a critical value of 253 or 254 would arise from getting the rows and the columns mixed up (e.g. looking up an F(996,1) by mistake).

Incorrect! Recalling the result that the restricted residual sum of squares is always the larger of the two, and that the formula for the test statistic is

test-statistic = ((RRSS - URSS) / URSS) ( (T-k)/m

The result is

test-statistic = ((30.2 - 28.1) / 28.1) ( (996)/1

= 74.43

So the correct answer is c. A result of 37.2 would have arisen if you had thought there were two restrictions (where there is in fact only one) and a negative F-statistic would have wrongly arisen if the two RSS had been mixed up.

Incorrect! Goodness of fit statistics show how well the estimated model fits the data - that is - how well the sample regression function fits the data. It would be nice to know how well the estimated model describes the true relationship between the variables embodied in the population regression function, but this is never possible since in practice we will never have access to the whole population of data. Thus the population regression function is never known, and hence all of the answers incorporating the PRF are wrong!

Incorrect! All of (i) to (iii) are true. If the value of R-squared is exactly zero, this can only occur if the residual sum of squares is equal to the total sum of squares so that the regression has not explained any of the variability of y about its mean value, and the explained sum of squares is zero. In other words, the regression line has not explained any of the variability of y about its mean value, so (iii) is correct. If this were the case, all of the slope coefficient estimates must be zero ((i) is correct), although the intercept estimate may or may not be zero, so (iv) is false. Hence if we plotted y separately against each of the explanatory variables, the resulting line would be horizontal in each case ((ii) is true).

Incorrect! It is true that model 2 must have an R-squared at least as high as that of model 1 since the former contains an additional variable, and R-squared can never fall when an extra variable is added. The value of the adjusted R-squared, however, could rise or fall. If the additional variable x3 is does not have significant explanatory power for y, the adjusted R-squared will fall since the increase in k by one will work to reduce the adjusted R-squared and will more than offset the increase in R-squared. Adjusted R-squared thus incorporates a penalty term that penalises large models with insignificant variables. If the estimated value of alpha3 were exactly zero, if would be as if the variable x3 were not included the second model. Therefore, in that case, the two values of R-squared would be equal. Applying the same logic as above, if the estimated value of alpha3 were exactly zero, even though the R-squared values were equal for the two models, the adjusted R-squared would be lower for model 2 since the number of explanatory variables has still been increased by one.

Incorrect! If the two models have identical values of R-squared, this means that the variable x3 has exactly zero explanatory power for y, and that the estimate of alpha3 must be exactly zero. The two models would not have identical adjusted R-squared since the raw R-squared values are the same, but k is higher for model 2 so that it will definitely have a lower adjusted R-squared.