Answer: b

The function np.std directly calculated the standard deviation of array oren. Even though oren is sample of the population, its standard deviation is still a pretty good estimate for the standard deviation of the population because it is a random sample. The other options don’t really make sense in this context.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

Answer: a

Note that a is equal to the mean of oren, which is a pretty good estimator of the mean of the overall population as well as the mean of the distribution of sample means. The other options don’t really make sense in this context.

Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer: b / np.sqrt(c)

Note that we can use the Central Limit Theorem for this problem which states that the standard deviation (SD) of the distribution of sample means is equal to (population SD) / np.sqrt(sample size). Since the SD of the sample is also the SD of the population in this case, we can plug our variables in to see that b / np.sqrt(c) is the answer.

Difficulty: ⭐️

The average score on this problem was 91%.

Answer: (560 - a) / b

To convert a value to standard units, we take the value, subtract the mean from it, and divide by SD. In this case that is (560 - a) / b, because a is the mean of our dog prices sample array and b is the SD of the dog prices sample array.

Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer: True

True. The central limit theorem states that if you have a population and you take a sufficiently large number of random samples from the population, then the distribution of the sample means will be approximately normally distributed.

Difficulty: ⭐️

The average score on this problem was 91%.

Answer: Decrease by a factor of \sqrt{2}

Recall that the central limit theorem states that the STD of the sample distribution is equal to (population STD) / np.sqrt(sample size). So if we increase the sample size by a factor of 2, the STD of the sample distribution will decrease by a factor of \sqrt{2}.

Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer: None of the above

Again, from our formula given by the central limit theorem, the sample STD doesn’t depend on the number of bootstrap resamples so long as it’s “sufficiently large”. Thus increasing our bootstrap sample from 1000 to 4000 will have no effect on the std of boots

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

Answer: a + 1.75 * b / np.sqrt(c)

Recall that a 92% confidence interval means an interval that consists of the middle 92% of the distribution. In other words, we want to “chop” off 4% from either end of the ditribution. Thus to get the right endpoint, we want the value corresponding to the 96th percentile in the mean dog price distribution, or mean + 1.75 * (SD of population / np.sqrt(sample size) or a + 1.75 * b / np.sqrt(c) (we divide by np.sqrt(c) due to the central limit theorem). Note that the second line of information that was given stats.norm.cdf(1.4) is irrelavant to this particular problem.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 48%.

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%.

Answer: 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%.

Answer: 0.48

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%.

Answer: Left endpoint = 33, Right endpoint = 37

According to the Central Limit Theorem, the standard deviation of the distribution of the sample mean is \frac{\text{sample SD}}{\sqrt{\text{sample size}}} = \frac{10}{\sqrt{100}} = 1. Then using the fact that the distribution of the sample mean is roughly normal, since 95% of the area of a normal curve falls within two standard deviations of the mean, we can find the endpoints of the 95% CLT-based confidence interval as 35 - 2 = 33 and 35 + 2 = 37.

We can think of this as using the formula below: \left[\text{sample mean} - 2\cdot \frac{\text{sample SD}}{\sqrt{\text{sample size}}}, \: \text{sample mean} + 2\cdot \frac{\text{sample SD}}{\sqrt{\text{sample size}}} \right]. Plugging in the appropriate quantities yields [35 - 2\cdot\frac{10}{\sqrt{100}}, 35 - 2\cdot\frac{10}{\sqrt{100}}] = [33, 37].

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answer: Option 1

As we found in the previous part, the distribution of the sample mean should have a standard deviation of 1. We also know it should be centered at the mean of our sample, at 35, but since all the options are centered here, that’s not too helpful. Only Option 1, however, has a standard deviation of 1. Remember, we can approximate the standard deviation of a normal curve as the distance between the mean and either of the inflection points. Only Option 1 looks like it has inflection points at 34 and 36, a distance of 1 from the mean of 35.

If you chose Option 2, you probably confused the standard deviation of our original sample, 10, with the standard deviation of the distribution of the sample mean, which comes from dividing that value by the square root of the sample size.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

Answer: A CLT-based 90% confidence interval for the mean age of credit card applicants, based on the data in hundred_apps, would be narrower than the interval you gave in part (a).

Let’s analyze each of the options:

• Option 1: We are not using bootstrapping to compute sample means since we are sampling from the apps DataFrame, which is our population here. If we were bootstrapping, we’d need to sample from our first sample, which is hundred_apps.

• Option 2: We can’t be sure what the distribution of the ages of credit card applicants are. The Central Limit Theorem says that the distribution of sample_means is roughly normally distributed, but we know nothing about the population distribution.

• Option 3: The CLT-based 95% confidence interval that we calculated in part (a) was computed as follows: \left[\text{sample mean} - 2\cdot \frac{\text{sample SD}}{\sqrt{\text{sample size}}}, \text{sample mean} + 2\cdot \frac{\text{sample SD}}{\sqrt{\text{sample size}}} \right] A CLT-based 90% confidence interval would be computed as \left[\text{sample mean} - z\cdot \frac{\text{sample SD}}{\sqrt{\text{sample size}}}, \text{sample mean} + z\cdot \frac{\text{sample SD}}{\sqrt{\text{sample size}}} \right] for some value of z less than 2. We know that 95% of the area of a normal curve is within two standard deviations of the mean, so to only pick up 90% of the area, we’d have to go slightly less than 2 standard deviations away. This means the 90% confidence interval will be narrower than the 95% confidence interval.

• Option 4: The left endpoint of the interval from part (a) was calculated using the Central Limit Theorem, whereas using np.percentile(sample_means, 2.5) is calculated empirically, using the data in sample_means. Empirically calculating a confidence interval doesn’t necessarily always give the exact same endpoints as using the Central Limit Theorem, but it should give you values close to those endpoints. These values are likely very similar but they are not guaranteed to be the same. One way to see this is that if we ran the code to generate sample_means again, we’d probably get a different value for np.percentile(sample_means, 2.5).

• Option 5: The key observation is that if we used the data in hundred_apps to create 1,000 CLT-based 95% confidence intervals for the mean age of applicants in apps, all of these intervals would be exactly the same. Given a sample, there is only one CLT-based 95% confidence interval associated with it. In our case, given the sample hundred_apps, the one and only CLT-based 95% confidence interval based on this sample is the one we found in part (a). Therefore if we generated 1,000 of these intervals, either they would all contain the parameter or none of them would. In order for a statement like the one here to be true, we would need to collect 1,000 different samples, and calculate a confidence interval from each one.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 49%.

Answer: 625

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%.

Answer: \frac{17}{27}

A Central Limit Theorem-based 95% confidence interval for a population proportion is given by the following:

\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 this interval uses the fact that (about) 95% of values in a normal distribution are within 2 standard deviations of the mean. It’s key to divide by \sqrt{\text{Sample Size}} when computing the standard deviation because the distribution that is roughly normal is the distribution of the sample mean (and hence, sample proportion), not the distribution of the sample itself.

The width of the above interval – that is, the right endpoint minus the left endpoint – is

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

From the provided hint, we have that

\text{Sample SD} = \sqrt{(\text{Prop. of 0s}) \cdot (\text{Prop of 1s})} = \sqrt{\frac{3}{9} \cdot \frac{6}{9}} = \frac{\sqrt{18}}{9}

Then, since we know that the sample size is 9 and that \sqrt{18} is about \frac{17}{4}, we have

\text{width} = 4 \cdot \frac{\text{Sample SD}}{\sqrt{\text{Sample Size}}} = 4 \cdot \frac{\frac{\sqrt{18}}{9}}{\sqrt{9}} = 4 \cdot \frac{\sqrt{18}}{9 \cdot 3} = 4 \cdot \frac{\frac{17}{4}}{27} = \frac{17}{27}

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 51%.

Answer: Options 1 and 2

Option 1: We use Central Limit Theorem (CLT) for large random samples, and a sample of 9 is considered to be very small. This makes it difficult to use CLT for this problem.

Option 2: Recall CLT happens when our sample is drawn with replacement. When we are handed nine cards we are never replacing cards back into our deck, which means that we are sampling without replacement.

Option 3: This is wrong because CLT states that a large sample is approximately a normal distribution even if the data itself is not normally distributed. This means it doesn’t matter if our data had not been normally distributed if we had a large enough sample we could use CLT.

Option 4: This is wrong because CLT does apply to the sample proportion distribution. Recall that proportions can be treated like means.

Difficulty: ⭐️⭐️

The average score on this problem was 77%.

Answer: 0.025

Answer: 400

Answer: The proportion of students who did use Hulu minus the proportion of students who did not use Hulu.

Answer: 0.04

Answer: We fail to reject the null hypothesis.

Answer: data/data.sum()

To calculate proportion for each group, we divide each value in the array (tickets sold to each group) by the sum of all values (total tickets sold). Remember values in an array can be processed as a whole.

Difficulty: ⭐️

The average score on this problem was 95%.

Answer: Child, adult, and senior tickets are purchased in proportions 0.25, 0.6, and 0.15. (Option 2)

Recall, null hypothesis is the hypothesis that there is no significant difference between specified populations, any observed difference being due to sampling or experimental error. So, we assume the distribution is the same as the model.

Difficulty: ⭐️⭐️

The average score on this problem was 88%.

Answer: sum of squared differences in proportions, mean of squared differences in proportions (Option 2, 4)

We need to use squared difference to avoid the case that large positive and negative difference cancel out in the process of calculating sum or mean, resulting in small sum of difference or mean of difference that does not reflect the actual deviation. So, we eliminate Option 1 and 3.

Difficulty: ⭐️⭐️

The average score on this problem was 77%.

Answer: 0.02

We first calculate the proportion for each value in admissions_data \frac{550}{550+1550+400} = 0.22 \frac{1550}{550+1550+400} = 0.62 \frac{400}{550+1550+400} = 0.16 So, we have the distribution of the admissions_data

Then, we calculate the observed value of the test statistic (the mean of the absolute differences in proportions) \frac{|0.22-0.25|+|0.62-0.6|+|0.16-0.15|}{number\ of\ goups} =\frac{0.03+0.02+0.01}{3} = 0.02

Difficulty: ⭐️⭐️

The average score on this problem was 82%.

Answer: (a) admissions_data.sum() (b) model (c) np.abs(simulated_proportions - model).mean() (d) np.count_nonzero(simulated_stats >= observed_stat) / 10000

Recall, in np.random.multinomial(n, [p_1, ..., p_k]), n is the number of experiments, and [p_1, ..., p_k] is a sequence of probability. The method returns an array of length k in which each element contains the number of occurrences of an event, where the probability of the ith event is p_i.

We want our simulated_proportion to have the same data size as admissions_data, so we use admissions_data.sum() in (a).

Since our null hypothesis is based on model, we simulate based on distribution in model, so we have model in (b).

In (c), we compute the mean of the absolute differences in proportions. np.abs(simulated_proportions - model) gives us a series of absolute differences, and .mean() computes the mean of the absolute differences.

In (d), we calculate the p_value. Recall, the p_value is the chance, under the null hypothesis, that the test statistic is equal to the value that was observed in the data or is even further in the direction of the alternative. np.count_nonzero(simulated_stats >= observed_stat) gives us the number of simulated_stats greater than or equal to the observed_stat in the 10000 times simulations, so we need to divide it by 10000 to compute the proportion of simulated_stats greater than or equal to the observed_stat, and this gives us the p_value.

Difficulty: ⭐️⭐️

The average score on this problem was 79%.

Answer: False

Recall, the p-value is the chance, under the null hypothesis, that the test statistic is equal to the value that was observed in the data or is even further in the direction of the alternative. It only gives us the strength of evidence in favor of the null hypothesis, which is different from “the probability that the null hypothesis is true”.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 64%.

Answer: 1.5

Recall, the total variation distance (TVD) is the sum of the absolute differences in proportions, divided by 2. When we compute the mean of the absolute differences in proportions, we are computing the sum of the absolute differences in proportions, divided by the number of groups (which is 3). Thus, to get TVD, we first multiply our current statistics (the mean of the absolute differences in proportions) by 3, we get the sum of the absolute differences in proportions. Then according to the definition of TVD, we divide this value by 2. Thus, we have \text{current statistics}\cdot 3 / 2 = \text{current statistics}\cdot 1.5.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 65%.

Answer: Among 2500 beds and outdoor furniture items, the proportion of beds.
Among 2500 beds and outdoor furniture items, the number of beds.

Our test statistic needs to be able to distinguish between the two hypotheses. The first option does not do this, because it includes an absolute value. If the absolute difference between the proportion of beds and the proportion of outdoor furniture were large, it could be because IKEA sells more beds than outdoor furniture, but it could also be because IKEA sells more outdoor furniture than beds.

The second option is a valid test statistic, because if the proportion of beds is large, that suggests that the alternative hypothesis may be true.

Similarly, the third option works because if the number of beds (out of 2500) is large, that suggests that the alternative hypothesis may be true.

The fourth option is invalid because out of 2500 beds and outdoor furniture items, the number of beds plus the number of outdoor furniture items is always 2500. So the value of this statistic is constant regardless of whether the alternative hypothesis is true, which means it does not help you distinguish between the two hypotheses.

Difficulty: ⭐️⭐️

The average score on this problem was 78%.

Answer: Reading the table from top to bottom, the five expressions should be used in the following blanks: None, (b), (a), (c), None.

The correct way to define obs_diff is

    outdoor = (app_data.get('category')=='outdoor') 
    bed = (app_data.get('category')=='bed')
    obs_diff = (app_data[outdoor].shape[0] - app_data[bed].shape[0]) / app_data[outdoor | bed].shape[0]

The first provided line of code defines a boolean Series called outdoor with a value of True corresponding to each outdoor furniture item in app_data. Using this as the condition in a query results in a DataFrame of outdoor furniture items, and using .shape[0] on this DataFrame gives the number of outdoor furniture items. So app_data[outdoor].shape[0] represents the number of outdoor furniture items in app_data. Similarly, app_data[bed].shape[0] represents the number of beds in app_data. Likewise, app_data[outdoor | bed].shape[0] represents the total number of outdoor furniture items and beds in app_data. Notice that we need to use an or condition (|) to get a DataFrame that contains both outdoor furniture and beds.

We are told that the test statistic should be the proportion of outdoor furniture minus the proportion of beds. Translating this directly into code, this means the test statistic should be calculated as

    obs_diff = app_data[outdoor].shape[0]/app_data[outdoor | bed].shape[0] - app_data[bed].shape[0]) / app_data[outdoor | bed].shape[0]

Since this is a difference of two fractions with the same denominator, we can equivalently subtract the numerators first, then divide by the common denominator, using the mathematical fact \frac{a}{c} - \frac{b}{c} = \frac{a-b}{c}.

This yields the answer

    obs_diff = (app_data[outdoor].shape[0] - app_data[bed].shape[0]) / app_data[outdoor | bed].shape[0]

Notice that this is the observed value of the test statistic because it’s based on the real-life data in the app_data DataFrame, not simulated data.

Difficulty: ⭐️

The average score on this problem was 90%.

Answer: Way 1, Way 3, Way 4, Way 6

Let’s consider each way in order.

Way 1 is a correct solution. This code begins by defining a variable multi which will evaluate to an array with two elements representing the number of items in each of the two categories, after 2500 items are drawn randomly from the two categories, with each category being equally likely. In this case, our categories are beds and outdoor furniture, and the null hypothesis says that each category is equally likely, so this describes our scenario accurately. We can interpret multi[0] as the number of outdoor furniture items and multi[1] as the number of beds when we draw 2500 of these items with equal probability. Using the same mathematical fact from the solution to Problem 8.2, we can calculate the difference in proportions as the difference in number divided by the total, so it is correct to calculate the test statistic as (multi[0] - multi[1])/2500.

Way 2 is an incorrect solution. Way 2 is based on a similar idea as Way 1, except it calls np.random.multinomial twice, which corresponds to two separate random processes of selecting 2500 items, each of which is equally likely to be a bed or an outdoor furniture item. However, is not guaranteed that the number of outdoor furniture items in the first random selection plus the number of beds in the second random selection totals 2500. Way 2 calculates the proportion of outdoor furniture items in one random selection minus the proportion of beds in another. What we want to do instead is calculate the difference between the proportion of outdoor furniture and beds in a single random draw.

Way 3 is a correct solution. Way 3 does the random selection of items in a different way, using np.random.choice. Way 3 creates a variable called choice which is an array of 2500 values. Each value is chosen from the list [0,1] with each of the two list elements being equally likely to be chosen. Of course, since we are choosing 2500 items from a list of size 2, we must allow replacements. We can interpret the elements of choice by thinking of each 1 as an outdoor furniture item and each 0 as a bed. By doing so, this random selection process matches up with the assumptions of the null hypothesis. Then the sum of the elements of choice represents the total number of outdoor furniture items, which the code saves as the variable choice_sum. Since there are 2500 beds and outdoor furniture items in total, 2500 - choice_sum represents the total number of beds. Therefore, the test statistic here is correctly calculated as the number of outdoor furniture items minus the number of beds, all divided by the total number of items, which is 2500.

Way 4 is a correct solution. Way 4 is similar to Way 3, except instead of using 0s and 1s, it uses the strings 'bed' and 'outdoor' in the choice array, so the interpretation is even more direct. Another difference is the way the number of beds and number of outdoor furniture items is calculated. It uses np.count_nonzero instead of sum, which wouldn’t make sense with strings. This solution calculates the proportion of outdoor furniture minus the proportion of beds directly.

Way 5 is an incorrect solution. As described in the solution to Problem 8.2, app_data[outdoor|bed] is a DataFrame containing just the outdoor furniture items and the beds from app_data. Based on the given information, we know app_data[outdoor|bed] has 2500 rows, 1000 of which correspond to beds and 1500 of which correspond to furniture items. This code defines a variable samp that comes from sampling this DataFrame 2500 times with replacement. This means that each row of samp is equally likely to be any of the 2500 rows of app_data[outdoor|bed]. The fraction of these rows that are beds is 1000/2500 = 2/5 and the fraction of these rows that are outdoor furniture items is 1500/2500 = 3/5. This means the random process of selecting rows randomly such that each row is equally likely does not make each item equally likely to be a bed or outdoor furniture item. Therefore, this approach does not align with the assumptions of the null hypothesis.

Way 6 is a correct solution. Way 6 essentially modifies Way 5 to make beds and outdoor furniture items equally likely to be selected in the random sample. As in Way 5, the code involves the DataFrame app_data[outdoor|bed] which contains 1000 beds and 1500 outdoor furniture items. Then this DataFrame is grouped by 'category' which results in a DataFrame indexed by 'category', which will have only two rows, since there are only two values of 'category', either 'outdoor' or 'bed'. The aggregation function .count() is irrelevant here. When the index is reset, 'category' becomes a column. Now, randomly sampling from this two-row grouped DataFrame such that each row is equally likely to be selected does correspond to choosing items such that each item is equally likely to be a bed or outdoor furniture item. The last line simply calculates the proportion of outdoor furniture items minus the proportion of beds in our random sample drawn according to the null model.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 59%.

Answer: <=

To answer this question, we need to know whether small values or large values of the test statistic indicate the alternative hypothesis. The alternative hypothesis is that IKEA sells more beds than outdoor furniture. Since we’re calculating the proportion of outdoor furniture minus the proportion of beds, this difference will be small (negative) if the alternative hypothesis is true. Larger (positive) values of the test statistic mean that IKEA sells more outdoor furniture than beds. A value near 0 means they sell beds and outdoor furniture equally.

The p-value is defined as the proportion of simulated test statistics that are equal to the observed value or more extreme, where extreme means in the direction of the alternative. In this case, since small values of the test statistic indicate the alternative hypothesis, the correct answer is <=.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 43%.