# Winter 2021 Final Exam

Instructor(s): Janine Tiefenbruck

This exam was administered remotely via Gradescope. The exam was open-internet, and students were able to use Jupyter Notebooks. They had 3 hours to work on it.

## Problem 1

One way to use np.arange to produce the sequence [2, 6, 10, 14] is np.arange(2, 15, 4). This gives three inputs to np.arange.

Fill in the blanks below to show a different way to produce the same sequence, this time using only one input to np.arange. Each blank below must be filled in with a single number only, and the final result, x*np.arange(y)+z, must produce the sequence [2, 6, 10, 14].

Using x*np.arange(y)+z fill in the missing values:

x = _

y = _

z = _

x = 4, y = 4, z = 2

The question states that we are trying to create the sequence [2, 6, 10, 14] by filling in the blanks for the statement x*np.arange(y)+z. If we look at the sequence we are attempting to derive, we can see that each step in the sequence increments by 4. Similarly, we can see that the sequence begins at 2. We know that passing an argument by itself in np.arange will increment up to that number (for example np.arange(4) will produce the sequence [0,1,2,3]). Knowing this, we can create this sequence by setting y to 4. Attempting to reach the desired sequence of [2, 6, 10, 14] from [0, 1, 2, 3] we can multiply each number by 4 by setting x to 4 and instantiate the sequence at 2 by setting z as 2.

##### Difficulty: ⭐️

The average score on this problem was 96%.

## Problem 2

The command .set_index can take as input one column, to be used as the index, or a sequence of columns to be used as a nested index (sometimes called a MultiIndex). A MultiIndex is the default behavior of the dataframe returned by .groupby with multiple columns.

You are given a dataframe called restaurants that contains information on a variety of local restaurants’ daily number of customers and daily income. There is a row for each restaurant for each date in a given five-year time period.

The columns of restaurants are 'name' (str), 'year' (int), 'month' (int), 'day' (int), 'num_diners' (int), and 'income' (float).

Assume that in our data set, there are not two different restaurants that go by the same name (chain restaurants, for example).

Which of the following would be the best way to set the index for this dataset?

• restaurants.set_index('name')

• restaurants.set_index(['year', 'month', 'day'])

• restaurants.set_index(['name', 'year', 'month', 'day'])

Answer: restaurants.set_index(['name', 'year', 'month', 'day'])

The correct answer is to create an index with the 'name', 'year', ‘month’, and ‘day’ columns. The question provides that there is a row for each restaurant for each data in the five year span. Therefore, we are interested in the granularity of a specific day (the day, the month, and the year). In order to have this information available in this index, we must set the index to be a multi index with columns ['name', 'year', 'month', 'day']. Looking at the other options, simply looking at the 'name' column would not account for the fact the dataframe contains daily data on customers and income for each restaurant. Similarly, the second option of ['name', 'month', 'day'] would not account for the fact that the data comes in a five year span so there will naturally be five overlaps (one for each year) for each unique date that must be accounted for.

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 53%.

## Problem 3

If we merge a table with n rows with a table with m rows, how many rows does the resulting table have?

• n

• m

• max(m,n)

• m * n

• not enough information to tell

Answer: not enough information to tell

The question does not provide enough information to know the resulting table size with certainty. When merging two tables together, the tables can be merged with a inner, left, right, and outer join. Each of these joins will produce a different amount of rows. Since the question does not provide the type of join, it is impossible to tell the resulting table size.

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

## Problem 4

You sample from a population by assigning each element of the population a number starting with 1. You include element 1 in your sample. Then you generate a random number, n, between 2 and 5, inclusive, and you take every nth element after element 1 to be in your sample. For example, if you select n=2, then your sample will be elements 1, 3, 5, 7, and so on.

### Problem 4.1

True or False: Before the sample is drawn, you can calculate the probability of selecting each subset of the population.

The answer is true since someone can easily sketch each sample to view the probability of selecting a certain subset. For example, when n = 2 we know the elements are 1, 3, 5, 7, and so on. Similarly we know this information for n = 3, 4 and 5. Using this information we could calculate the probability of selecting a subset.

##### Difficulty: ⭐️

The average score on this problem was 97%.

### Problem 4.2

True or False: Each subset of the population is equally likely to be selected.

No, each subset of the population is not equally likely to be selected since the element assigned as element 1 will always be selected due to the way sampling is conducted as defined in the question. That is, the question says we always include element one in the sample which will over represent it in samples as compared to other parts of the population.

##### Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 46%.

## Problem 5

You are given a table called books that contains columns 'author' (str), 'title' (str), 'num_chapters' (int), and 'publication_year' (int).

### Problem 5.1

What will be the output of the following code? books.groupby(“publication_year”).mean().shape

• 1

• 2

• 3

• 4

Answer: 1

The output will return 1. Notice that the final function call is to .shape. We know that .shape is a call to see how many columns are in the resulting data frame. When we group by publication year, there is only one column that will be aggregated by the groupby call (which is the 'num_chapters' column). The other columns are string, and therefore, will not be aggregated in the groupby call (since you can’t take the mean of a string). Consequently .shape will only result one column for the mean of the 'num_chapters' column.

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

### Problem 5.2

Which of the following strategies would work to compute the absolute difference in the average number of chapters per book for authors “Dean Koontz” and “Charles Dickens”?

• group by 'author', aggregate with .mean(), use get on 'num_chapters' column compute the absolute value of the difference between iloc["Charles Dickens"] and iloc["Dean Koontz"]

• do two queries to get two separate tables (one for each of “Dean Koontz” and “Charles Dickens”), use get on the 'num_chapters' column of each table, use the Series method .mean() on each, compute the absolute value of the difference in these two means

• group by both 'author' and 'title', aggregate with .mean(), use get on 'num_chapters' column, use loc twice to find values in that column corresponding to “Dean Koontz” and “Charles Dickens”, compute the absolute value of the difference in these two values

• query using a compound condition to get all books corresponding to “Dean Koontz” or “Charles Dickens”, group by 'author', aggregate with .mean(), compute absolute value of the difference in index and index

Answer: do two queries to get two separate tables (one for each of “Dean Koontz” and “Charles Dickens”), use get on the 'num_chapters' column of each table, use the Series method .mean() on each, compute the absolute value of the difference in these two means

Logically, we want to somehow separate data for author “Dean Koontz” and “Charles Dickens”. (If we don’t we’ll be taking a mean that includes the chapters of books from both authors.) To achieve this separation, we can create two separate tables with a query that specifies a value on the 'author' column. Now having two separate tables, we can aggregate on the 'num_chapters' (the column of interest). To get the 'num_chapters' column we can use the get method. To actually acquire the mean of the 'num_chapters' column we can evoke the .mean() call.

##### Difficulty: ⭐️⭐️

The average score on this problem was 80%.

### Problem 5.3

Which of the following will produce the same value as the total number of books in the table?

• books.groupby('Title').count().shape

• books.groupby('Author').count().shape

• books.groupby(['Author, 'Title']).count().shape

Answer: books.groupby(['Author, 'Title']).count().shape

The key in this question is to understand that different authors can create books with the same name. The first two options check for each unique book title (the first response) and check for each unique other (the second response). To ensure we have all unique author and title pairs we must group based on both 'Author' and 'Title'. To actually get the number of rows we can take .shape.

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 56%.

## Problem 6

Suppose you have a dataset of 29 art galleries that includes the number of pieces of art in each gallery.

A histogram of the number of art pieces in each gallery, as well as the code that generated it, is shown below. ### Problem 6.1

How many galleries have at least 80 but less than 100 art pieces? Input your answer below. Make sure your answer is an integer and does not include any text or symbols.

Through looking at the graph we can find the total number of art galleries by taking 0.012 (height of bin) * 20 (the size of the bin) * 29 (total number of art galleries). This will yield an anwser of 6.96 which should be rounded to the nearest integer (7).

##### Difficulty: ⭐️

The average score on this problem was 94%.

### Problem 6.2

If we added to our dataset two more art galleries, each containing 24 pieces of art, and plotted the histogram again for the larger dataset, what would be the height of the bin [20,45)? Input your answer as a number rounded to six decimal places.

Taking the area of the bin [20,45] we can find the number of art galleries already within this bin 0.0055 * 25 = 0.1375 (estimation based on the visualization). To find the number take this proportion x the total number of art galleries. 0.1375 * 29 = about 4 art galleries. If we add two art galleries to this total we get 4 art galleries in the [20,45] bin to get 6 art galleries. To find the frequency of 6 art galleries to the entire data set we can take 6/31. Note that the question asks for the height of the bin. Therefore, we can take (6/31) / 25 due to the size of the bin which will give an answer of 0.007742 upon rounding to six decimal places.

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 66%.

## Problem 7

Assume df is a DataFrame with distinct rows. Which of the following best describes df.sample(10)?

• an array of length 10, where some of the entries might be the same

• an array of length 10, where no two entries can be the same

• a DataFrame with 10 rows, where some of the rows might be the same

• a DataFrame with 10 rows, where no two rows can be the same

Answer: a DataFrame with 10 rows, where no two rows can be the same

Looking at the documentation for .sample() we can see that it accepts a few arguments. The first argument specifies the number of rows (which is why we specify 10). The next argument is a boolean that specifies if the sampling happens with or without replacement. By default, the sampling will occur without replacement (which happens in this question since no argument is specified so the default is evoked). Looking at the return, we can see that since we are sampling a dataframe, a dataframe will also be returned which is why a DataFrame with 10 rows, where no two rows can be the same is correct.

##### Difficulty: ⭐️

The average score on this problem was 94%.

## Problem 8

True or False: If you roll two dice, the probability of rolling two fives is the same as the probability of rolling a six and a three.

The probability of rolling two fives can be found with 1/6 * 1/6 = 1/36. The probability of rolling a six and a three can be found with 2/6 (can roll either a 3 or 6) * 1/6 (roll a different side form 3 or 6, depending on what you rolled first) = 1/18. Therefore, the probabilities are not the same.

##### Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 33%.

## Problem 9

Suppose you do an experiment in which you do some random process 500 times and calculate the value of some statistic, which is a count of how many times a certain phenomenon occurred out of the 500 trials. You repeat the experiment 10,000 times and draw a histogram of the 10,000 statistics.

### Problem 9.1

Is this histogram a probability histogram or an empirical histogram?

• probability histogram

• empirical histogram

Empirical histograms refer to distributions of observed data. Since the question at hand is conducting an experiment and creating a histogram of observed data from these trials the correct anwser is an empirical histogram.

##### Difficulty: ⭐️

The average score on this problem was 90%.

### Problem 9.2

If you instead repeat the experiment 100,000 times, how will the histogram change?

• it will become wider

• it will become narrower

• it will barely change at all

Answer: it will barely change at all

Doing more of an experiment will barely change the histogram. The parameter we are trying to estimate through our experiment is some statistic. The number of experiments has no effect on the histograms distribution since the value of some statistic is not becoming more random.

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

### Problem 9.3

For each experiment, if you instead do the random process 5,000 times, how will the histogram change?

• it will become wider

• it will become narrower

• it will barely change at all

By increasing the number of random process we increase the possible range of values from 500 to 5000. The statistic being calculated is the count of how many times a phenomenon occurs. If the number of random process increases 10x the statistic can now take values ranging from [0, 5000] instead of [0, 500] which will clearly make the histogram width wider (due to the wider range of values it can take).

##### Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 39%.

## Problem 10

Give an example of a dataset and a question you would want to answer about that dataset which you could answer by performing a permutation test (also known as an A/B test).

Creative responses that are different than ones we’ve already seen in this class will earn the most credit.

Answer: Responses vary. For this question we looked for creative responses. One good example includes

A dataset about prisoners in the US with the sentence times, race, and crime. Do White people who commit homicide get shorter sentence times than Black people who commit homicide? We can clearly perform an A/B test to compare black and white populations as they correlated to shorter sentence times.

##### Difficulty: ⭐️

The average score on this problem was 93%.

## Problem 11

Suppose you draw a sample of size 100 from a population with mean 50 and standard deviation 15. What is the probability that your sample has a mean between 50 and 53? Input the probability below, as a number between 0 and 1, rounded to two decimal places.

This problem is testing our understanding of the Central Limit Theorem and normal distributions. Recall, the Central Limit Theorem tells us that the distribution of the sample mean is roughly normal, with the following characteristics:

\begin{align*} \text{Mean of Distribution of Possible Sample Means} &= \text{Population Mean} = 50 \\ \text{SD of Distribution of Possible Sample Means} &= \frac{\text{Population SD}}{\sqrt{\text{Sample Size}}} = \frac{15}{\sqrt{100}} = 1.5 \end{align*}

Given this information, it may be easier to express the problem as “We draw a value from a normal distribution with mean 50 and SD 1.5. What is the probability that the value is between 50 and 53?” Note that this probability is equal to the proportion of values between 50 and 53 in a normal distribution whose mean is 50 and 1.5 (since probabilities can be thought of as proportions).

In class, we typically worked with the standard normal distribution, in which the mean was 0, the SD was 1, and the x-axis represented values in standard units. Let’s convert the quantities of interest in this problem to standard units, keeping in mind that the mean and SD we’re using now are the mean and SD of the distribution of possible sample means, not of the population.

• 50 converted to standard units is \frac{50 - \text{mean}}{\text{SD}} = \frac{50 - 50}{1.5} = 0 (no calculation was necessary – 0 in standard units is equal to the mean in original units).
• 53 converted to standard units is \frac{53 - \text{mean}}{\text{SD}} = \frac{53 - 50}{1.5} = 2.

Now, our problem boils down to finding the proportion of values in a standard normal distribution that are between 0 and 2, or the proportion of values in a normal distribution that are in the interval [\text{mean}, \text{mean} + 2 \text{ SDs}].

From class, we know that in a normal distribution, roughly 95% of values are within 2 standard deviations of the mean, i.e. the proportion of values in the interval [\text{mean} - 2 \text{ SDs}, \text{mean} + 2 \text{ SDs}] is 0.95. Since the normal distribution is symmetric about the mean, half of the values in this interval are to the right of the mean, and half are to the left. This means that the proportion of values in the interval [\text{mean}, \text{mean} + 2 \text{ SDs}] is \frac{0.95}{2} = 0.475, which rounds to 0.48, and thus the desired result is 0.48.

##### Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 48%.

## Problem 12

You need to estimate the proportion of American adults who want to be vaccinated against Covid-19. You plan to survey a random sample of American adults, and use the proportion of adults in your sample who want to be vaccinated as your estimate for the true proportion in the population. Your estimate must be within 0.04 of the true proportion, 95% of the time. Using the fact that the standard deviation of any dataset of 0’s and 1’s is no more than 0.5, calculate the minimum number of people you would need to survey. Input your answer below, as an integer.

Note: Before reviewing these solutions, it’s highly recommended to revisit the lecture on “Choosing Sample Sizes,” since this problem follows the main example from that lecture almost exactly.

While this solution is long, keep in mind from the start that our goal is to solve for the smallest sample size necessary to create a confidence interval that achieves certain criteria.

The Central Limit Theorem tells us that the distribution of the sample mean is roughly normal, regardless of the distribution of the population from which the samples are drawn. At first, it may not be clear how the Central Limit Theorem is relevant, but remember that proportions are means too – for instance, the proportion of adults who want to be vaccinated is equal to the mean of a collection of 1s and 0s, where we have a 1 for each adult that wants to be vaccinated and a 0 for each adult who doesn’t want to be vaccinated. What this means (😉) is that the Central Limit Theorem applies to the distribution of the sample proportion, so we can use it here too.

Not only do we know that the distribution of sample proportions is roughly normal, but we know its mean and standard deviation, too:

\begin{align*} \text{Mean of Distribution of Possible Sample Means} &= \text{Population Mean} = \text{Population Proportion} \\ \text{SD of Distribution of Possible Sample Means} &= \frac{\text{Population SD}}{\sqrt{\text{Sample Size}}} \end{align*}

Using this information, we can create a 95% confidence interval for the population proportion, using the fact that in a normal distribution, roughly 95% of values are within 2 standard deviations of the mean:

\left[ \text{Population Proportion} - 2 \cdot \frac{\text{Population SD}}{\sqrt{\text{Sample Size}}}, \: \text{Population Proportion} + 2 \cdot \frac{\text{Population SD}}{\sqrt{\text{Sample Size}}} \right]

However, this interval depends on the population proportion (mean) and SD, which we don’t know. (If we did know these parameters, there would be no need to collect a sample!) Instead, we’ll use the sample proportion and SD as rough estimates:

\left[ \text{Sample Proportion} - 2 \cdot \frac{\text{Sample SD}}{\sqrt{\text{Sample Size}}}, \: \text{Sample Proportion} + 2 \cdot \frac{\text{Sample SD}}{\sqrt{\text{Sample Size}}} \right]

Note that the width of this interval – that is, its right endpoint minus its left endpoint – is: \text{width} = 4 \cdot \frac{\text{Sample SD}}{\sqrt{\text{Sample Size}}}

In the problem, we’re told that we want our interval to be accurate to within 0.04, which is equivalent to wanting the width of our interval to be less than or equal to 0.08 (since the interval extends the same amount above and below the sample proportion). As such, we need to pick the smallest sample size necessary such that:

\text{width} = 4 \cdot \frac{\text{Sample SD}}{\sqrt{\text{Sample Size}}} \leq 0.08

We can re-arrange the inequality above to solve for our sample’s size:

\begin{align*} 4 \cdot \frac{\text{Sample SD}}{\sqrt{\text{Sample Size}}} &\leq 0.08 \\ \frac{\text{Sample SD}}{\sqrt{\text{Sample Size}}} &\leq 0.02 \\ \frac{1}{\sqrt{\text{Sample Size}}} &\leq \frac{0.02}{\text{Sample SD}} \\ \frac{\text{Sample SD}}{0.02} &\leq \sqrt{\text{Sample Size}} \\ \left( \frac{\text{Sample SD}}{0.02} \right)^2 &\leq \text{Sample Size} \end{align*}

All we now need to do is pick the smallest sample size that satisfies the above inequality. But there’s an issue – we don’t know what our sample SD is, because we haven’t collected our sample! Notice that in the inequality above, as the sample SD increases, so does the minimum necessary sample size. In order to ensure we don’t collect too small of a sample (which would result in the width of our confidence interval being larger than desired), we can use an upper bound for the SD of our sample. In the problem, we’re told that the largest possible SD of a sample of 0s and 1s is 0.5 – this means that if we replace our sample SD with 0.5, we will find a sample size such that the width of our confidence interval is guaranteed to be less than or equal to 0.08. This sample size may be larger than necessary, but that’s better than it being smaller than necessary.

By substituting 0.5 for the sample SD in the last inequality above, we get

\begin{align*} \left( \frac{\text{Sample SD}}{0.02} \right)^2 &\leq \text{Sample Size} \\\ \left( \frac{0.5}{0.02} \right)^2 &\leq \text{Sample Size} \\ 25^2 &\leq \text{Sample Size} \implies \text{Sample Size} \geq 625 \end{align*}

We need to pick the smallest possible sample size that is greater than or equal to 625; that’s just 625.

##### Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 40%.

## Problem 13

• Hector earned a 77 on an exam where the mean was 70 and the standard deviation was 5.
• Clara earned an 80 on an exam where the mean was 75 and the standard deviation was 3.
• Vivek earned an 83 on an exam where the mean was a 75 and the standard deviation was 6.

Rank these three students in ascending order of their exam performance relative to their classmates.

• Hector, Clara, Vivek

• Vivek, Hector, Clara

• Clara, Hector, Vivek

• Vivek, Clara, Hector

To compare Vivek, Hector, and Clara’s relative performance we want to compare their Z scores to handle standardization. For Vivek, his Z score is (83-75) / 6 = 4/3. For Hector, his score is (77-70) / 5 = 7/5. For Clara, her score is (80-75) / 3 = 5/3. Ranking these, 5/3 > 7/5 > 4/3 which yields the result of Vivek, Hector, Clara.

##### Difficulty: ⭐️⭐️

The average score on this problem was 76%.

## Problem 14

The data visualization below shows all Olympic gold medals for women’s gymnastics, broken down by the age of the gymnast. ### Problem 14.1

Based on this data, rank the following three quantities in ascending order: the median age at which gold medals are earned, the mean age at which gold medals are earned, the standard deviation of the age at which gold medals are earned.

• mean, median, SD

• median, mean, SD

• SD, mean, median

• SD, median, mean

The standard deviation will clearly be the smallest of the three values as most of the data is encompassed between the range of [14-26]. Intuitively, the standard deviation will have to be about a third of this range which is around 4 (though this is not the exact standard deviation, but is clearly much less than the mean and median with values closer to 19-25). Comparing the median and mean, it is important to visualize that this distribution is skewed right. When the data is skewed right it pulls the mean towards a higher value (as the higher values naturally make the average higher). Therefore, we know that the mean will be greater than the median and the ranking is SD, median, mean.

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

### Problem 14.2

Which of the following is larger for this dataset?

• the difference between the 50th percentile of ages and the 25th percentile of ages

• the difference between the 75th percentile of ages and the 50th percentile of ages

• both are the same

Answer: the difference between the 75th percentile of ages and the 50th percentile of ages

Since the distribution is right skewed, the 75th percentile will have a larger difference from the 50th percentile than the 25th percentile. With right skewness, values above the 50th percentile will be more different than those smaller than the 50th percentile (and thus more spread out according to the graph).

##### Difficulty: ⭐️⭐️

The average score on this problem was 78%.

## Problem 15

In a board game, whenever it is your turn, you roll a six-sided die and move that number of spaces. You get 10 turns, and you win the game if you’ve moved 50 spaces in those 10 turns. Suppose you create a simulation, based on 10,000 trials, to show the distribution of the number of spaces moved in 10 turns. Let’s call this distribution Dist10. You also wonder how the game would be different if you were allowed 15 turns instead of 10, so you create another simulation, based on 10,000 trials, to show the distribution of the number of spaces moved in 15 turns, which we’ll call Dist15

### Problem 15.1

What can we say about the shapes of Dist10 and Dist15?

• both will be roughly normally distributed

• only one will be roughly normally distributed

• neither will be roughly normally distributed

Answer: both will be roughly normally distributed

By the central limit theorem, both simulations will appear to be roughly normally distributed.

##### Difficulty: ⭐️

The average score on this problem was 90%.

### Problem 15.2

What can we say about the centers of Dist10 and Dist15?

• both will have approximately the same mean

• the mean of Dist10 will be smaller than the mean of Dist15

• the mean of Dist15 will be smaller than the mean of Dist10

Answer: the mean of Dist10 will be smaller than the mean of Dist15

The distribution which moves in 10 turns will have a smaller mean as there are less turns to move spaces. Therefore, the mean movement from turns will naturally be higher for the distribution with more turns.

##### Difficulty: ⭐️⭐️

The average score on this problem was 83%.

### Problem 15.3

• both will have approximately the same standard deviation

• the standard deviation of Dist10 will be smaller than the standard deviation of Dist15

• the standard deviation of Dist15 will be smaller than the standard deviation of Dist10

Answer: the standard deviation of Dist10 will be smaller than the standard deviation of Dist15

Since taking more turns allows for more possible values, the spread of Dist10 will be smaller than the standard deviation of Dist15. (ie. consider the possible range of values that are attainable for each case)

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 65%.

## Problem 16

True or False: The slope of the regression line, when both variables are measured in standard units, is never more than 1.

Standard units standardize the data into z scores. When converting to Z scores the scale of both the dependent and independent variables are the same, and consequently, the slope can at most increase by 1. Alternatively, according to the reference sheet, the slope of the regression line, when both variables are measured in standard units, is also equal to the correlation coefficient. And by definition, the correlation coefficient can never be greater than 1 (since you can’t have more than a ‘perfect’ correlation).

##### Difficulty: ⭐️

The average score on this problem was 93%.

## Problem 17

True or False: The slope of the regression line, when both variables are measured in original units, is never more than 1.

Original units refers to units as they are. Clearly, regression slopes can be greater than 1 (for example if for every change in 1 unit of x corresponds to a change in 20 units of y the slope will be 20).

##### Difficulty: ⭐️

The average score on this problem was 96%.

## Problem 18

True or False: Suppose that from a sample, you compute a 95% bootstrapped confidence interval for a population parameter to be the interval [L, R]. Then the average of L and R is the mean of the original sample.

False, a 95% confidence interval indicates we are 95% confident that the true population parameter falls within the interval [L, R]. Note that the problem specifies that the confidence interval is bootstrapped. Since the interval is found using bootstrapping, L anr R averaged will not be the mean of the original sample since the mean of the original sample is not what is used in calculating the bootstrapped confidence interval. The bootstrapped confidence interval (as noted on the reference sheet is created by re-sampling the data with replacement over and over again). Thus, the interval is not centered around the mean because bootstrapping and random sampling will lead to variation in the confidence interval.

##### Difficulty: ⭐️⭐️

The average score on this problem was 87%.

## Problem 19

True or False: Suppose that from a sample, you compute a 95% normal confidence interval for a population parameter to be the interval [L, R]. Then the average of L and R is the mean of the original sample.

True, a 95% confidence interval indicates we are 95% confident that the true population parameter falls within the interval [L, R]. Looking at how a confidence interval is calculated is by adding/ subtracting a confidence level value (z) by the standard error. Since the top and bottom of the interval will be different from the mean by the same amount, the average will be the mean. (For more information, refer to the reference sheet)

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 68%.

## Problem 20

You order 25 large pizzas from your local pizzeria. The pizzeria claims that these pizzas are 16 inches in diameter, but you’re not so sure. You measure each pizza’s diameter and collect a dataset of 25 actual pizza diameters. You want to run a hypothesis test to determine whether the pizzeria’s claim is accurate.

### Problem 20.1

What would your Null Hypothesis be?

What would your Alternative Hypothesis be?

Null Hypothesis: The mean pizza diameter at the local pizzeria is 16 inches.

Alternative Hypothesis: The mean pizza diameter at the local pizzeria is not 16 inches.

The null hypothesis is the hypothesis where there is no significant difference from some statement. In this case, the statement of interest is the pizzeria’s claims of pizzas are 16 inches in diameter.

The alternative hypothesis is a statement that contradicts the null hypothesis. In this case this statement is that the mean pizza diameter at the local pizzeria is not 16 inches. (ie. the other option to the null hypothesis)

##### Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 35%.

### Problem 20.2

What test statistic would you use?

Answer: Mean Pizza Diameter, or other valid statistics such as the (absolute) difference between each pizza’s diameter and 16 (expected value)

Looking at the null and alternative hypothesis we can see we are directly interested in the mean pizza diameter, so it is most likely the best measurement for the test statistic. The main idea is that we somehow want to show the difference in distribution of the pizza diameters.

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 70%.

### Problem 20.3

Explain how you would do your hypothesis test and how you would draw a conclusion from your results.

• Generate confidence interval for population mean (or equivalent) by bootstrapping (or by calculating directly with the sample).

• Correctly describe how to reject or fail to reject the null hypothesis (depending on whether interval contains 16, for example).

When conducting the hypothesis test, we first want to create a confidence interval either by using bootstrapping or constructing a 95% confidence interval to understand the true mean diameter of pizzas. The next step is to define the rejection criteria, failing to reject the null if 16 is within the 95% confidence interval (since we believe the true population mean) is within this range with 95% confidence. We will reject the null hypothesis if 16 is not within the 95% confidence interval. Note that this assumes you used true mean diameter of pizzas as your test statistic.

##### Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 38%.

## Problem 21

A restaurant keeps track of each table’s number of people (average 3; standard deviation 1) and the amount of the bill (average $60, standard deviation$12). If the number of people and amount of the bill are linearly associated with correlation 0.8, what is the predicted bill for a table of 5 people? Input your answer below, to the nearest cent. Make sure your answer is just a number and does not include the \$ symbol or any text.

To answer this question, first find the z score for a table of 5 people. Z = (5-3)/1 = 2. Now having this Z score, find the price that correlated in the bill distribution by finding the value for 2 standard deviations larger than the mean while also accounting for the correlation between the two variables. This is calculated with mean + ((ZSD) r) which is 60 + ((12 * 2) * 0.8) = 79.20.

Alternatively, we could solve for the regression line and plug our values in according to the reference sheet:

m = (0.8) * (12/1) and b = 60 - (48/5) * 3 (where m is the slope and b is the y-intercept)

Thus plugging the appropriate values in our regression line yields

y = (48/5) * 5 + 60 - (48/5)*3 = 79.2

##### Difficulty: ⭐️⭐️

The average score on this problem was 88%.

## Problem 22

From a population with mean 500 and standard deviation 50, you collect a sample of size 100. The sample has mean 400 and standard deviation 40. You bootstrap this sample 10,000 times, collecting 10,000 resample means.

### Problem 22.1

Which of the following is the most accurate description of the mean of the distribution of the 10,000 bootstrapped means?

• The mean will be exactly equal to 400.

• The mean will be exactly equal to 500.

• The mean will be approximately equal to 400.

• The mean will be approximately equal to 500.

Answer: The mean will be approximately equal to 400.

The distribution of bootstrapped means’ mean will be approximately 400 since that is the mean of the sample and bootstrapping is taking many samples of the original sample. The mean will not be exactly 400 do to some randomness though it will be very close.

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 54%.

### Problem 22.2

Which of the following is closest to the standard deviation of the distribution of the 10,000 bootstrapped means?

• 400

• 40

• 4

• 0.4

To find the standard deviation of the distribution, we can take the sample standard deviation S divided by the square root of the sample size. From plugging in, we get 40 / 10 = 4.

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 51%.

## Problem 23

Recall the mathematical definition of percentile and how we calculate it.

Let p be a number between 0 and 100. The pth percentile of a collection is the smallest
value in the collection that is at least as large as p% of all the values.

By this definition, any percentile between 0 and 100 can be computed for any collection of values and is always an element of the collection. Suppose there are n elements in the collection. To find the pth percentile:

1. Sort the collection in increasing order.
2. Find p% of n: (p/100) n. Call that h. If h is an integer, define k=h. Otherwise, let k be the smallest integer greater than h.
3. Take the kth element of the sorted collection.

You have a dataset of 7 values, which are [3, 6, 7, 9, 10, 15, 18]. Using the mathematical definition of percentile above, find the smallest and largest integer values of p so that the pth percentile of this dataset corresponds to the value 10. Input your answers below, as integers between 0 and 100.

### Problem 23.1

Smallest = _

From the definition provided in the question, we want all values of (p/100) * n which will yield an integer larger than 4, but less than or equal to 5 because we want the 5th element (10) in the dataset. To approach this problem we can find how many percentiles each piece of data falls within by taking 100 / 7 which yields around 14.3. Wanting to find the percentiles for the range of 4 to 5 we can multiple (100/7) by 4 to get our lower bound. (100/7) * 4 = 57.14 which is rounded up to 58 since the 57th percentile belongs to the 4th element while 58 belongs to the fifth element.

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

### Problem 23.2

Largest = _

To find the largest we will take (100/7) * 5 which yields 71.43. We will round down since the 72th percentile belongs to the sixth element in the data set. For more information look at the solution above.

##### Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

## Problem 24

Are nonfiction books longer than fiction books?

• hypothesis testing

• permutation (A/B) testing

• Central Limit Theorem

• regression

The question Are nonfiction books longer than fiction books? is investigating the difference between two underlying populations (nonfiction books and fiction books). A permutation test is the best data science tool when investigating differences between two underlying distributions.

##### Difficulty: ⭐️

The average score on this problem was 90%.

## Problem 25

Do people have more friends as they get older?

• hypothesis testing

• permutation (A/B) testing

• Central Limit Theorem

• regression

The question at hand is investigating two continuous variables (time and number of friends). Regression is the best data science tool as it is dealing with two continuous variables and we can understand correlations between time and the number of friends.

##### Difficulty: ⭐️

The average score on this problem was 90%.

## Problem 26

Does an ice cream shop sell more chocolate or vanilla ice cream cones?

• hypothesis testing

• permutation (A/B) testing

• Central Limit Theorem

• regression