Answer: str

When we do a groupby on the Chain column in hotels, this means that the values in the Chain column will be the indices of the DataFrame or Series we get as output, in this case the Series hotels.groupby("Chain")["Number of Rooms"].sum().

Since the values of Chain are strings, and since .idxmax() will return a value from the index of the aforementioned Series, summed is a string.

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Difficulty: ⭐️⭐️

The average score on this problem was 88%.

Answer: summed is the name of the hotel chain with the most total rooms

The result of the .groupby() and .sum() is a Series indexed by the unique Chains, whose values are the total number of rooms in hotels owned by each chain. The idxmax() function gets the index corresponding to the largest value in the Series, which will be the hotel chain name with the most total rooms.

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Difficulty: ⭐️

The average score on this problem was 97%.

Answer: hotel room

curious gets the most common value of Chain in the DataFrame frame. We already know that summed is the hotel chain with the most rooms in San Diego, so curious only equals summed if the most common Chain in frame is the hotel chain with the most total rooms; this occurs when each row of frame is a single hotel room.

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Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answers:

1. df.shape[0] >= 5 and df["Number of Rooms"].sum() >= 1000
2. "Location"
3. filter(f)
4. ["Location"].unique() or equivalent

We’d like to only consider certain neighborhoods according to group characteristics (having >= 5 hotels and >= 1000 hotel rooms), and .filter() allows us to do that by excluding groups not meeting those criteria. So, we can write a function that evaluates those criteria on one group at a time (the df of input to f is the subset of hotels containing just one Location value), and calling filter(f) means that the only remaining rows are hotels in neighborhoods that match those criteria. Finally, all we have to do is get the unique neighborhoods from this DataFrame, which are the neighborhoods for which f returned True.

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Difficulty: ⭐️⭐️

The average score on this problem was 83%.

Answers:

1. (cond1 & cond2).sum()
2. cond1.sum() + cond2.sum() - (cond1 & cond2).sum()

Note that cond1 and cond2 are boolean Series, and hotels[cond1] and hotels[cond2] are the subsets of hotels where Chain == "Marriott and "Location" == "Coronado", respectively.

1. When we perform an inner merge, we’re selecting every row where a Hotel Name appears in both hotels[cond1] and hotels[cond2]. This is the same set of indices (and therefore hotel names, since those are unique) as where (cond1 & cond2) == True. So, the length of combined will be the same as the number of Trues in (cond1 & cond2).

2. When we perform an outer merge, we’re selecting every row that appears in either DataFrame, although there will not be repeats for hotels that are both Marriott properties and are in Coronado. So, to find the total number of rows in either DataFrame, we take the sum of the sizes of each, and subtract rows that appear in both, which corresponds to answer cond1.sum() + cond2.sum() - (cond1 & cond2).sum().

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Difficulty: ⭐️⭐️

The average score on this problem was 79%.

Answer: missing at random

The description of loyalty indicates that most business travelers have a loyalty number – in other words, that depending on the value of is_business, the probability of a present value for loyalty can increase or decrease. This is therefore an example of MAR.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answer: missing by design

The value of company is missing if the trip is a business trip, and since this relationship should be deterministic, then this column is missing by design.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 60%.

Answers:

1. permutation test
2. either the (absolute) difference in means or the K-S statistic, depending on the shapes of the observed distributions

In part 1, we use a permutation test because we’re comparing two sampled distributions (the values of age for rows with missing vs. present values of is_business), rather than comparing our sample distribution to a known population distribution/metric.

For part 2, we should note that the test statistic, for each iteration of our permutation test, is meant to compare two distributions of the variable age, which is continuous. TVD is not well-suited for this, since it compares discrete, categorical distributions, and neither is “sample mean,” since we have two samples to compare. Differences in means or the K-S statistic work because they both compare two continuous distributions. However, there are cases in which the difference in means might not capture differences in distributions (for example, two distributions with the same mean but very different standard deviation); this is the reason for the caveat in the correct solution.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 68%.

Answers:

1. permutation test
2. total variation distance

For part 1, the reasoning is the same as the previous part – the only difference is that now, we’re comparing two categorical distributions rather than two continuous ones.

In part 2, we can note that we’re now comparing the categorical distributions of is_business, which can only have values of "yes", "no", and "partially", for rows with missing vs. present values of age. (Even though age is continuous, since we’re only using whether or not age is missing and not the actual values in the column, this doesn’t really matter.) The only test statistic listed that is meant to compare two categorical distributions is TVD, so this is the correct solution.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answers:

1. standard hypothesis test
2. less than
3. all primary occupants

For part 1, this is a standard hypothesis test, because instead of shuffling the values of a particular column, we are sampling entire rows of the DataFrame; the size of these samples is the same as the number of guests who are not (even partially) on business trips. By sampling randomly from the entire DataFrame, this assumes the null hypothesis that the distribution of age of primary occupants not on a business trip is equal to the distribution of age for occupants as a whole. Therefore, this procedure is meant to simulate values of the median age of a sample of size guests["is_business"] == "no").sum(), and sampling random rows means that we’re assuming that is_business == "no" and age are unrelated.

For part 2, the samples capture how likely the deviation of the observed statistic is to happen under the null hypothesis, but the p-value is being calculated by adding up whenever the observed statistic is greater or equal to the sample. This indicates that we’re doing a one-tailed test, and since to reject the null hypothesis and provide some evidence towards the alternate hypothesis, the p-value should be lower, the alternate hypothesis should be the complement of obs >= sm.

For part 3, the answer is “all primary occupants” because our sampling procedure doesn’t sample just from (full or partial) business travelers, but from the entire DataFrame. (In the line where we sample, the only arguments are the number of rows we’re sampling, and that we’re doing so without replacement – nothing about restricting which rows in guest to sample.) In other words, we’re not comparing primary occupants not on business trips to those on business trips, but rather we’re comparing primary occupants to the entire dataset of occupants.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 58%.

Answer: False

In impute, np.random.choice will return a single non-null value from s, and .fillna() will fill every null value with this single value. Meanwhile, probabilistic imputation draws a different value from a specified distribution to fill each missing value, making it such that there won’t be a single “spike” in the imputed distribution at a single chosen value.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 41%.

Answer: 4.6

The maximum possible value of impute(vals).mean() would occur when every single missing value in vals is filled in with the highest possible non-null value in vals. (As discussed in the previous solution, impute selects only one value from s to fill into every missing space.)

If this occurs, then the mean of the imputed Series will be weighted mean of the available data and the filled data, and given the numbers in the question, this is 0.8 \cdot 4 + 0.2 \cdot 7, or 4.6.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 50%.

Answer: None of the above

Since the value which impute() will choose to impute with is random, the effect that it has on the standard deviation of vals is unknown. If the missing values are filled with a value close to the mean, this could reduce standard deviation; if they are filled with a value far from the mean, this could increase standard deviation. (Of course, the imputation will also shift the mean, so without knowing details of the Series, it’s impossible to come up with thresholds.) In any case, since the value for imputation is chosen at random, none of these statements will always be true, and so the correct answer is “none of the above.”

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 38%.

Answer: 4

The answer simply requires counting the number of div elements, making sure not to double-count opening and closing tags.

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Difficulty: ⭐️⭐️

The average score on this problem was 82%.

Answer: find("div", attrs={"name": chain}).find("h3").get("style")

The solution finds all div elements whose name attribute is equal to the function input chain (which is how the chains are defined in the HTML document). Then, the colors are stored as part of the style attribute in an h3 tag within that div, so we find the first – and in this document, only – h3 tag within the div and get its style attribute. This returns a string like color: red, from which the given code .split()[-1] returns the last word, which is the correct color.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answers:

1. name = "Hyatt"
2. <h3 style="color: purple"></h3>

In order to make all of the three given statements true, we need to make it such that there are 2 div elements with class chain, the chain_color function returns purple for Hyatt, and the last div tag in the document has no text.

We do this by defining a div similar to the previous two, but without the chain class (because there are already 2 in the document). In order for chain_color("Hyatt") to return purple, we add "Hyatt" as a name attribute in the div, and add an h3 tag within that div with style="color: purple". However, since the last statement requires the last div in the document has no text, and the div we’re currently making will be the final one, we don’t include any text in the h3 tag. (Note that because the chain_color function doesn’t use the hotel name in the text to find a given chain’s name, but rather the name attribute in the enclosing div tag.)

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Difficulty: ⭐️

The average score on this problem was 90%.

Answer: ([A-Za-z]+): Available

In the above availability strings, we’re looking for a one-or-more-length sequence of letters (not numbers, as mentioned in the problem), followed by a colon and the word “Available.” However, since we only want to return the room type, and not the colon or the word “Available,” we use a capturing group (the parentheses) to just return the word before the colon.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

Answers:

1. ["$350.25"]
2. ["200.75", "402.99", "350.25"]

In the first part, the string that the pattern matches is a dollar sign ($), one or more digits (\d+), a period (\.), two digits (\d{2}), followed by the end of the string ($). Since the string we’re passing into re.findall is prices, which is the concatenation of one + ", " + two, the only option that matches is the last price displayed, "$350.25". (If we had not specified the match ending with the end of the string, "$200.75" would also have been a match.)

In the second part, the beginning of the pattern is "\$?", which means to match zero or one instance of the dollar sign character. In most cases, this means that if there is a dollar sign before the remainder of the pattern, it will be included in the match, but if not, the rest of the pattern will still match. However, the capturing group around the remainder of the pattern means that in either case, only the remainder of the pattern after the dollar sign is captured. (So, in this particular example, the \$? has no effect on the output.)

The rest of the pattern is structured similarly to the previous part, except now, the pattern does not require the end of the string after the price, so we’re just selecting sequences of digits followed by a period and two more digits, which is how we get our solution.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 65%.

Answer: \frac{3}{10}

The answer is simply the proportion of words in the sentence that are the word "beach". There are 10 words in the sentence, 3 of which are "beach".

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Difficulty: ⭐️

The average score on this problem was 97%.

Answer: 12

The TF-IDF is the product of the TF and IDF terms. So if the TF-IDF of this document is \frac{9}{10}, and the TF is \frac{3}{10}, as established in the last part, the IDF of the term "beach" is 3. The IDF for a word is the log of the inverse of the proportion of documents in which the word appears. So, since we know there are 96 total documents.

3 = log_{2}(\frac{96}{\text{\# documents containing "beach"}})

2^{3} = \frac{96}{\text{\# documents containing "beach"}}

\boxed{12} = {\text{\# documents containing "beach"}}

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: \vec{r}_1 and \vec{r}_2, and \vec{r}_2 and \vec{r}_3

The cosine similarity of two vectors \vec{a} and \vec{b} is \frac{\vec{a} \cdot \vec{b}}{||\vec{a}|| \cdot ||\vec{b}||}.

The cosine similarity of \vec{r}_1 and \vec{r}_2 is:

\frac{2 \cdot 4 + 3 \cdot 4}{\sqrt{13} \cdot \sqrt{32}} = \frac{20}{4 \cdot \sqrt{26}} = \frac{5}{\sqrt{26}}

The cosine similarity of \vec{r}_1 and \vec{r}_3 is:

\frac{2 \cdot 6 + 3 \cdot 4}{\sqrt{13} \cdot \sqrt{52}} = \frac{24}{26} = \frac{12}{13}

The cosine similarity of \vec{r}_2 and \vec{r}_3 is:

\frac{4 \cdot 6 + 4 \cdot 4}{\sqrt{32} \cdot \sqrt{52}} = \frac{40}{8 \cdot \sqrt{26}} = \frac{5}{\sqrt{26}}

\frac{12}{13} \approx 0.9231, and \frac{5}{\sqrt{26}} \approx 0.9806, so our answer is vector pairs \vec{r}_1 and \vec{r}_2, and \vec{r}_2 and \vec{r}_3

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

Answer: \frac{12}{65}

Using a bigram model, we find the probability of the first token, then for each remaining token, find its probability given just the previous token. The probability P(\text{X} | \text{Y}) is simply the proportion of times that token X is the next token after the previous token Y. So, for this corpus and input sequence, we have:

P(\text{A}) = \frac{6}{13}

P(\text{B} | \text{A}) = \frac{4}{5}

P(\text{R} | \text{B}) = \frac{2}{4} = \frac{1}{2}

P(\text{A} | \text{R}) = \frac{2}{2} = 1

So, we just have to multiply these probabilities together, and we get \frac{12}{65}.

Note that the denominator for P(\text{B} | \text{A}) is 5, not 6, even though there are 6 occurrences of the letter A. This is because the final one occurs at the end of the sentence, and we’re not using a stop token or anything else designating the end of the sequence. So there are only 5 instances of “A” where we have a subsequent token, which is how we make our probability model.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 47%.

Answer:

1. c = 2
2. d = 5
3. k = n-1

From our work on the previous question, we can see that 2p is equal to P(\text{A}) \cdot P(\text{B} | \text{A}), or in other words, the probability of the sequence AB.

The probability of sequence ABAB will be the probability of sequence AB, times P(\text{A} | \text{B}) \cdot P(\text{B} | \text{A}). More generally, we can see that after the first pair of tokens AB (which has a different probability since the first token is the first in the sequence, and therefore isn’t conditioned on B), each repetition of AB multiplies a probability of P(\text{A} | \text{B}) \cdot P(\text{B} | \text{A}).

Therefore, our resulting fraction \frac{c}{d} is simply this probability P(\text{A} | \text{B}) \cdot P(\text{B} | \text{A}). P(\text{A} | \text{B}) is \frac{1}{2}, and we already found that P(\text{B} | \text{A}) is \frac{4}{5}, so the resulting fraction is \frac{2}{5}. So c = 2, and d = 5.

This fraction is multiplied once for each repetition of AB after the first one (which has a different probability, represented by 2p), and so the exponent should be n-1.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 66%.

Answer: \frac{8}{13}

Precision is the proportion of predicted positives that actually were positives. So, given this confusion matrix, that value is \frac{8}{8 + 5}, or \frac{8}{13}.

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Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer:

1. B
2. 5

We already know that the precision is \frac{8}{13}. Recall is the proportion of true positives that were indeed classified positives, which in this matrix is \frac{8}{B + 8}. So, in order for precision to equal recall, B must be 5.

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Difficulty: ⭐️

The average score on this problem was 96%.

Answer:

1. A \cdot B
2. 40

We can solve this problem by simply stating recall and accuracy in terms of the values in the confusion matrix. As we already found, recall is \frac{8}{B+8}. Accuracy is the sum of correct predictions over total number of predictions, or \frac{A + 8}{A + B + 13}. Then, we simply set these equal to each other, and solve.

\frac{8}{B+8} = \frac{A + 8}{A + B + 13} 8(A + B + 13) = (A + 8)(B + 8) 8A + 8B + 104 = AB + 8A + 8B + 64 104 = AB + 64 AB = 40

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Difficulty: ⭐️⭐️

The average score on this problem was 79%.

Answer: Split 1’s “Yes” node, Split 1’s “No” node, and Split 2’s “No” node

The definition of entropy for a node is:

\text{entropy} = - \sum_C p_C \log_2 p_C

where C is each label class, and p_C is the proportion of points in the node in class C.

We can note that in binary classification, two nodes will have the same entropy if they have the same relative distribution of classes, or if the distributions of classes are complements of each other. (For example, a node with 75% positives and 25% negatives has the same entropy as a node with 25% positives and 75% negatives.) So, in this problem, we need only find nodes with the same proportions as the original node, which has 4 positives (\frac{1}{3}) and 8 negatives (\frac{2}{3}).

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Difficulty: ⭐️⭐️

The average score on this problem was 88%.

Answer: Split 3’s “Yes” node

The lowest possible entropy for a distribution occurs when it takes on only one value. In the definition of entropy, \log_2 p_C will be 0 when p_C is 1, so entropy will be zero when all points are of the same class. This only occurs in one of the given nodes, Split 3’s “Yes” node.

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Difficulty: ⭐️

The average score on this problem was 93%.

Answers:

1. 75
2. 900
3. False

If we plug in the weights from \vec{w}^* into the equation for H_1(x), noting that the values of the one-hot variables are either 0 or 1, we get -0.5 \cdot 250 + 200 \cdot 1 + 300 \cdot 0 + 50 \cdot 0, or -125 + 200, or 75.

For part 2, the difference between predicted and actual number of rooms is 75 - 45 = 30, and the squared loss is simply 30^2 = 900.

For part 3, the answer is False, since the fact that our best-fit line has an error at a given input does not mean that this input was not seen in the training data. In other words, the line of best fit does not always have to pass through all of the training points (and due to overfitting, in most cases you don’t want it to).

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Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answers:

1. True
2. True

In model H_1, there is redundant information in the one-hot encoded features, creating multicollinearity. The second model H_2 removes the is_Other feature, but the intercept term it is replaced with serves a similar function. When is_Other is true, the prediction of H_1 is w_1 \cdot \texttt{price} + w_4, and the prediction of H_2 is \beta_1 \cdot \texttt{price} + \beta_0. The multicollinearity in H_1 doesn’t affect model performance relative to model 2, but it will yield different parameter estimates since there will be multiple solutions, unlike with H_2. However, the models will have equivalent best solutions, even if the parameters differ, so both parts are True.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 53%.

Answers:

1. -0.5
2. 150
3. 250

As we saw in the previous question, these models should yield an equivalent best-fit line. The relationship between price and predicted outcome will be the same, so \beta_{1}^{*} will be -0.5, the same weight as in H_1.

If \beta_{0}^{*} = 50, this means we are adding 50 to all predictions, and that the “adjustments” we make to the prediction for is_Marriott and is_Hilton should therefore be reduced to compensate for this; therefore, the new weights \beta_{2}^{*} and \beta_{3}^{*} will be 150 and 250.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 69%.

Answers:

1. 5k
2. 6000(k-1)

When we do k-fold cross-validation for one single hyperparameter value, we split the dataset into k folds, and in each iteration i, train the model on the remaining k-1 folds and evaluate on fold i. Since every fold is left out and evaluated on once, the model is fit in total k times. We do this once for every hyperparameter value we want to test, so the total number of model fits required is 5k.

In part 2, we can note that each model fit is performed on the same size of data – the size of the remaining k-1 folds when we hold out a single fold. This size is 1 - \frac{1}{k} = \frac{k-1}{k} times the size of the entire dataset, in this case, 1200 \cdot \frac{k-1}{k}, and we fit a model on a dataset of this size 5k times. So, the sum of the training sizes for each model fit is:

5k \cdot \frac{k-1}{k} \cdot 1200 = 6000(k-1)

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 66%.

Answers:

1. decreases
2. 100
3. underfit the training data and have high bias

In part 1, we can see that as context_length increases, the training error increases, and the model performs worse. In general, higher model complexity leads to better model performance, so here, increasing context_length is reducing model complexity.

In part 2, we will choose a context_length of 100, since this parameterization leads to the best validation performance. If we increase context_length further, the validation error increases.

In part 3, since increased context_length indicates less complexity and worse training performance, increasing the context_length further would lead to underfitting, as the model would lack the expressiveness or number of parameters required to capture the data. Since training error represents model bias, and since high variance is associated with overfitting, a further increase in context_length would mean a more biased model.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 65%.