Quantitative Methods

• Topic 01

  Time Value of Money

  54 questions
• Topic 02

  Discounted Cash Flow Applications

  24 questions
• Topic 03

  Statistical Concepts and Data

  54 questions
• Topic 04

  Probability Concepts

  24 questions
• Topic 05

  Probability Distributions

  24 questions
• Topic 06

  Sampling and Estimation

  23 questions
• Topic 07

  Hypothesis Testing

  24 questions
• Topic 08

  Simple Linear Regression

  69 questions
• Topic 09

  Multiple Regression

  24 questions
• Topic 10

  Big Data and Machine Learning in Investment Analysis

  39 questions

A few questions from this unit, with the answer and a full explanation. The complete bank is available when you start practising.

1. A regression output table for a 48-observation sample reports: $b_0 = 2.1$, $SE(b_0) = 1.5$; $b_1 = 0.75$, $SE(b_1) = 0.12$; R² = 0.52; SEE = 3.4. At the 5% significance level (df = 46, critical t ≈ 2.01), which of the following conclusions is correct?

   • A
     The intercept is significant ($t = 2.1/1.5 = 1.40$) and the slope is not ($t = 0.12/0.75 = 0.16$)
   • B
     The slope is significant ($t = 0.75/0.12 = 6.25 > 2.01$), but the intercept is not ($t = 2.1/1.5 = 1.40 < 2.01$); R² = 0.52 means X explains 52% of variation in Y
     Correct answer
   • C
     Both the slope and intercept are statistically significant at the 5% level
   • D
     Neither the slope nor the intercept is statistically significant because R² = 0.52 is below 0.60
   Explanation

   Slope t-statistic: $t = 0.75/0.12 = 6.25$. Since $6.25 > 2.01$, the slope is statistically significant at the 5% level.

   Intercept t-statistic: $t = 2.1/1.5 = 1.40$. Since $1.40 < 2.01$, the intercept is **not statistically significant** at the 5$H_0: \beta_0 = 0$.

   R² = 0.52: 52% of variation in Y is explained by X. This is a useful but not outstanding fit; the threshold for 'useful' is context-dependent, not fixed at 0.60 or any other number.

   SEE = 3.4 is the typical prediction error magnitude, not a criterion for model validity.

2. Two banks offer savings accounts. Bank A offers a stated annual rate of 5% compounded semi-annually. Bank B offers a stated annual rate of 4.95% compounded monthly. Which bank has the higher effective annual rate, and by how many basis points?

   • A
     Bank B has the higher EAR by approximately 3 basis points.
   • B
     Bank A has the higher EAR by approximately 3 basis points; EAR_A ≈ 5.063% vs EAR_B ≈ 5.061%.
     Correct answer
   • C
     Both banks have the same EAR because the stated rates are almost identical.
   • D
     Bank A has the higher EAR by approximately 50 basis points because the stated rates differ by 5 basis points.
   Explanation

   $EAR_A = (1 + 0.05/2)^2 - 1 = (1.025)^2 - 1 = 1.050625 - 1 = 5.0625\%$. $EAR_B = (1 + 0.0495/12)^{12} - 1 = (1.004125)^{12} - 1 \approx 5.055\%$... more precisely: $(1.004125)^{12}$: using the approximation $EAR \approx r + \binom{m}{2}(r/m)^2 \approx 4.95\% + 11 \times (0.004125)^2/2$. More exactly: $(1.004125)^{12} \approx 1.05063$. So $EAR_B \approx 5.063\%$. In this case, both EARs are very close—the 5 basis point lower stated rate of Bank B is almost exactly offset by the higher frequency of monthly vs. semi-annual compounding. The difference is only about 1–3 basis points, with Bank A being slightly higher.

3. In k-fold cross-validation, a dataset is divided into k equal subsets. Which statement correctly describes how the model is trained and evaluated?

   • A
     Each of the k subsets is independently used to train a separate model, and the k models are averaged to produce a single prediction
   • B
     The first k−1 subsets are used for training and the kth subset is always held out as the test set, regardless of the number of runs
   • C
     The model is trained k times; each time, one different subset is held out for validation while the remaining k−1 subsets form the training set, and performance is averaged across all k runs
     Correct answer
   • D
     The dataset is split into k equally sized training sets; the model is trained k times and the best-performing run is selected as the final model
   Explanation

   In k-fold cross-validation, the dataset is split into k subsets (folds). The model is trained k separate times:

   1. In run 1, fold 1 is the validation set; folds 2 through k are the training set.
   2. In run 2, fold 2 is the validation set; all other folds are training data.
   3. This continues until each fold has served once as the validation set.

   Performance metrics (e.g., accuracy, R²) are averaged over all k runs to produce a robust estimate of out-of-sample performance. The holdout test set remains separate from this entire process.