data8-sp23-final

3 (j) (2.0 pt) The median of a set of 15 integers will always be an integer. *$ True *$ False (k) (2.0 pt) Suppose a hypothesis test is proposed and we already know that the null hypothesis is true. If 100 researchers each independently collect a large random sample of the same size to carry out an experiment and they all use 5% as their p-value cutoff, we should expect around 95% of them to fail to reject the null. *$ True *$ False (l) (2.0 pt) Suppose you fit a regression line to two data sets: (A) the original data set; and (B) the dataset with a few outliers (with respect to the y-axis) removed. The line for (A) will have a larger slope (in absolute value) than the line for (B). Assume both data sets are standardized just prior to fitting the lines. *$ True *$ False (m) (2.0 pt) If we use linear regression to predict y -values based on our x -values, the median of our residuals will always be zero. *$ True *$ False (n) (2.0 pt) Given a function error(a, b) which computes some error based on its input arguments, a valid output from the call minimize(error) could be an array containing two elements: 4 and 20. *$ True *$ False

6 (b) (4.0 points) Note: This section does not depend on the previous parts. Armond thinks it could be a good idea to predict guest spend from how far away they live from the hotel. He collects a random sample of guest bookings and creates a table with two columns: • Spend : (float) the amount spent (in dollars) by a guest during their stay • Distance : (float) the number of miles the guest traveled to get to the hotel Suppose that: • the ' Spend ' column has a mean of 5,000 and a standard deviation of 500 • the ' Distance ' column has a mean of 4,000 and a standard deviation of 1,000 • the correlation between the ' Spend ' and ' Distance ' columns is 0.8 i. (2.0 pt) Armonds fits a regression line using the table above to predict ' Spend ' from ' Distance ' . What is the intercept of this regression line? *$ 21,000 *$ 12,000 *$ 6,600 *$ 3,400 *$ -4,000 *$ -11,000 *$ None of the above ii. (2.0 pt) For a guest from Berkeley, who would have to travel 2,000 miles to get to Hawaii, what would this regression line predict as their spend? *$ 5,800 *$ 4,200 *$ 4,000 *$ 2,500 *$ 2,000 *$ None of the above

9 (d) (8.0 points) Note: This section does not depend on the previous parts. Belinda notices that guests’ nationality seems to differ quite a bit depending on whether they request a late check out. She creates a table with the following columns: • ' Nationality ' : (string) the country where the guests reside (e.g. ' Italy ' , ' France ' ) • ' Late Checkout ' : (string) whether the guests requested a late check out ( ' Yes ' or ' No ' ). She decides to conduct a hypothesis test to determine whether the observed difference between guests who request a late checkout and those who do not is due to chance. i. (2.0 pt) Which of the following is a null hypothesis that Belinda could use for her test? Select all that apply. 939 The ' Late Checkout ' distribution is the same among guests of all nationalities. 939 The ' Late Checkout ' distribution is more varied among domestic guests compared to international guests. 939 The ' Nationality ' distribution is the same between guests who request a late checkout and guest who do not. 939 The ' Nationality ' is more diverse among guests who request a late checkout compared to guest who do not. 939 The ' Nationality ' distribution is different between guests who request a late checkout and guest who do not. 939 None of the above ii. (2.0 pt) Which of the following could be used as a test statistic for this hypothesis test? Select all that apply. 939 The difference in standard deviation of ' Nationality ' between guests who request a late checkout and guests who do not. 939 The total variation distance between the ' Late Checkout ' distributions of domestic guests and international guests. 939 The difference of means in ' Nationality ' between guests who request a late checkout and guest who do not. 939 The total variation distance bewtween the ' Nationality ' distributions of guests who request a late checkout and guest who do not. 939 The absolute difference of means in ' Nationality ' between guests who request a late checkout and guests who do not. 939 None of the above

12 (e) (3.0 pt) Suppose Frank shows up late. What is the probability that the website correctly predicted that he was not on time? *$ 0 . 6 × 0 . 89 *$ 0 . 15 × 0 . 89 × 0 . 92 *$ 0 . 6 × 0 . 89 0 . 3 × 0 . 15 + 0 . 6 × 0 . 89 + 0 . 1 × 0 . 92 *$ 0 . 89 *$ 0 . 15 × 0 . 89 × 0 . 92 0 . 3 × 0 . 15 + 0 . 6 × 0 . 89 + 0 . 1 × 0 . 92 (f) (3.0 pt) What is the probability that the website returns ' yes ' and Frank Ocean is late or does not show up? *$ 0 . 6 × 0 . 11 + 0 . 10 × 0 . 08 *$ 1 − 0 . 3 × 0 . 85 0 . 3 × 0 . 85 + 0 . 6 × 0 . 11 + 0 . 1 × 0 . 08 *$ (0 . 6 + 0 . 10) × 0 . 11 *$ 0 . 6 × 0 . 11 + 0 . 1 × 0 . 08 0 . 3 × 0 . 85 + 0 . 6 × 0 . 11 + 0 . 1 × 0 . 08 *$ 0 . 11 + 0 . 08 2 (g) (3.0 pt) Suppose that the morning of a performance, the weather forecast isn’t looking too great, so Sarah & Rebecca think that there’s a 20% chance Frank is on time, 35% chance he is late, and 45% chance he does not show up. Suppose Sarah & Rebecca then consult the website and it returns ' no ' . Given this new information, what is Sarah & Rebecca’s subjective probability that Frank does not show up? *$ 0 . 45 *$ 1 − (0 . 2 × 0 . 35) *$ 0 . 45 × 0 . 08 *$ 0 . 45 × 0 . 92 0 . 2 × 0 . 15 + 0 . 35 × 0 . 89 + 0 . 45 × 0 . 92 *$ 0 . 15 × 0 . 92 0 . 25 × 0 . 15 + 0 . 6 × 0 . 89 + 0 . 15 × 0 . 92

15 (b) (15.0 points) Note: This section does not depend on the previous parts. Suppose that instead of classifying whether a business will get a ' Buy ' recommendation, Roman would like to use nearest neighbors to predict a business’s probability of getting a ' Buy ' recommendation. We will use the following method: i. Find the k nearest neighbors of the business with respect to ' Revenue ' and ' Profit ' . ii. For each of the k nearest neighbors, define their weight as follows. Let d r be the difference between the neighbor’s revenue and the unclassified businesses’s revenue. Let d p be the difference between the neighbor’s profit and the unclassified businesses’s profit. Calculate the neighbor’s weight as weight = 1 1 + q d 2 r + d 2 p i. Among the k nearest neighbors, find the weighted proportion that received a ' Buy ' rating. For example, if k = 3 and the 3 nearest neighbors have ' Buy ' values of 0 , 0 , & 1 with weights 0.9, 0.3, & 0.2, respectively, then the weighted proportion that received a ‘Buy’ recommendation is weight z}|{ 0 . 9 ∗ Buy z}|{ 0 + weight z}|{ 0 . 3 ∗ Buy z}|{ 0 + weight z}|{ 0 . 2 ∗ Buy z}|{ 1 0 . 9 + 0 . 3 + 0 . 2 | {z } sum of weights In order to estimate the ' Buy ' probability of a new business using the scheme above, Roman defines a function knn_prob that takes the following arguments: • train : A three-column table in which the first column is labeled ' Revenue ' , the second column is labeled ' Profit ' , and the third column is labeled ' Buy ' . Each row of the table represents a business in the training set. • business : An array of length two containing the business revenue and profit (in that order) of the business to predict on. • k : The value of k to use for k -nearest-neighbors. The function returns the weighted proportion that received a ' Buy ' recommendation among the k -nearest- neighbors, according to the scheme proposed above. Here is the function, partially completed: def knn_prob(train, business, k): revenue_diffs = train.column( ' Revenue ' ) - __________ (a) profit_diffs = train.column( ' Profit ' ) - __________ (b) train_dist = train.with_column( ' Distance ' , __________) (c) nn = train_dist.sort(__________).take(__________) (d) (e) weights = 1/(1 + __________) (f) return np.sum(__________) / np.sum(weights)

18 iii. (3.0 pt) Fill in blank (c). iv. (3.0 pt) Fill in blank (d). v. (3.0 pt) Fill in blank (e).