dataset_name: machine_learning
description: The following are multiple choice questions (with answers) about machine
  learning.
fewshot_config:
  sampler: first_n
  samples:
  - question: 'Which image data augmentation is most common for natural images?

      (A) random crop and horizontal flip (B) random crop and vertical flip (C) posterization
      (D) dithering'
    target: Let's think step by step. Data augmentation is used to increase the diversity
      of images in the training dataset. It is important that natural images are kept
      natural after being augmented. Vertical flips of images are not natural, so
      (B) is false. Posterization makes the image look like a poster and and dithering
      increases color depth. None of these two preserve the natural property. The
      only natural data augmentation technique is (A). The answer is (A).
  - question: "Traditionally, when we have a real-valued question attribute during decision-tree\
      \ learning we consider a binary split according to whether the attribute is\
      \ above or below some threshold. Pat suggests that instead we should just have\
      \ a multiway split with one branch for each of the distinct values of the attribute.\
      \ From the list below choose the single biggest problem with Pat\u2019s suggestion:\n\
      (A) It is too computationally expensive. (B) It would probably result in a decision\
      \ tree that scores badly on the training set and a testset. (C) It would probably\
      \ result in a decision tree that scores well on the training set but badly on\
      \ a testset. (D) It would probably result in a decision tree that scores well\
      \ on a testset but badly on a training set."
    target: "Let's think step by step. Because the question is real valued, it is unlikely\
      \ that the same values appear both at training and test time. This means that\
      \ while such a decision tree could yield good performance on the training data,\
      \ when evaluated on the test data it will perform badly because the decision\
      \ tree won\u2019t know what to do with numbers that did not appear in the training\
      \ data. The answer is (C)."
  - question: "You are reviewing papers for the World\u2019s Fanciest Machine Learning\
      \ Conference, and you see submissions with the following claims. Which ones\
      \ would you consider accepting?\n(A) My method achieves a training error lower\
      \ than all previous methods! (B) My method achieves a test error lower than\
      \ all previous methods! (Footnote: When regularisation parameter \u03BB is chosen\
      \ so as to minimise test error.) (C) My method achieves a test error lower than\
      \ all previous methods! (Footnote: When regularisation parameter \u03BB is chosen\
      \ so as to minimise cross-validaton error.) (D) My method achieves a cross-validation\
      \ error lower than all previous methods! (Footnote: When regularisation parameter\
      \ \u03BB is chosen so as to minimise cross-validaton error.)"
    target: "Let's think step by step. In machine learning, we train with some data\
      \ and fixed hyperparameters and the training error can be arbitrarily low, so\
      \ (A) can\u2019t be right. Then, one compares different hyperparameters by selecting\
      \ the model with the lowest cross-validation error, this means that (B) and\
      \ (D) are not the right procedure. The only relevant number after these is the\
      \ test error and thus (C) is the right answer. The answer is (C)."
  - question: 'A 6-sided die is rolled 15 times and the results are: side 1 comes up
      0 times; side 2: 1 time; side 3: 2 times; side 4: 3 times; side 5: 4 times;
      side 6: 5 times. Based on these results, what is the probability of side 3 coming
      up when using Add-1 Smoothing?

      (A) 2.0/15 (B) 1.0/7 (C) 3.0/16 (D) 1.0/5'
    target: 'Let''s think step by step. Add-1 smoothing adds the value of one to the
      different counts and then normalizes the probabilities accordingly. The counts
      after adding one will be: side 1 comes up 1 time; side 2: 2 times; side 3: 3
      times; side 4: 4 times; side 5: 5 times; side 6: 6 times. The number of sum
      one die rolls will be 21, so the probability of drawing a three is 3/21 = 1/7.
      The answer is (B).'
  - question: 'To achieve an 0/1 loss estimate that is less than 1 percent of the true
      0/1 loss (with probability 95%), according to Hoeffding''s inequality the IID
      test set must have how many examples?

      (A) around 10 examples (B) around 100 examples (C) between 100 and 500 examples
      (D) more than 1000 examples'
    target: "Let's think step by step. By the Hoeffding\u2019s inequality, we expect\
      \ that with 95% probability the in-sample and out-of-sample errors differ by\
      \ epsilon when we have N samples if 2 exp(-2 epsilon^2 N)<0.05, this implies\
      \ that N > -1/(2*epsilon**2) log ( 0.05/2 )= log (40)*5000. Since log(40)>1,\
      \ we have that one needs more than 1000 examples. The answer is (D).\n\n"
tag: mmlu_flan_cot_fewshot_stem
include: _mmlu_flan_cot_fewshot_template_yaml
task: mmlu_flan_cot_fewshot_machine_learning
