Answer: n

The size argument in np.random.choice provides the number of samples we draw. In the manual_multinomial function, we randomly draw n items, and so the size should be n.

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Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: True

Here, we are using np.random.choice to simulate picking n elements from values. We draw with replacement since we are allowed to have repeated elements. For example, if we were flipping a coin five times, we would need to have repeated elements, since there are only two possible outcomes of a coin flip but we are flipping the coin more than two times.

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Difficulty: ⭐️⭐️

The average score on this problem was 79%.

Answer: np.count_nonzero(choices == value)

The choices variable contains an array of the n randomly drawn values selected from values. In each iteration of the for-loop, we want to count the number of elements in choices that are equal to the given value. To do this, we can use np.count_nonzero(choices == value). In the end, value_counts is an array that says how many times we selected 0, how many times we selected 1, and so on.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 37%.

Answer: 0.5

From the given percentiles, we can notice that since the distribution is symmetric around the mean, the mean should be around the 50th percentile. Given the symmetry and the percentiles around 0.5, we can infer that the mean should be very close to 0.5.

Another way we can look at it is by noticing that prop is pulled from a [0.5, 0.5] distribution (because we are simulating under the null hypotheses) in np.random.multinomial(). This means that its expected for most of the distribution to be from around 0.5.

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Difficulty: ⭐️⭐️

The average score on this problem was 84%.

Answer: 0.05

If we look again at the percentiles, we notice that it seems to resemble a normal distribution. So by taking the mean and the 97.5th percentile, we can solve for the standard deviation. Since [2.5, 97.5] is the 95% confidence interval, we can say that the 97.5th percentile is two standard deviations away from the mean (2.5 too!). Thus,

0.5 + 2 \cdot \text{SD} = 0.6

\therefore Solving for SD, we get \text{SD} = 0.05

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 45%.

Answer: 0.1

Each time through our for-loop, we execute the following lines of code:

prop_1br = np.append(prop_1br, prop)

abs_diff = np.append(abs_diff, np.abs(prop-0.5))

Additionally, we’re told the following statements evaluate to True:

np.percentiles(prop_1br, 2.5) == 0.4

np.percentiles(prop_1br, 5) == 0.42

np.percentiles(prop_1br, 95) == 0.58

np.percentiles(prop_1br, 97.5) == 0.6

We can combine these pieces of information to find the answer to this question.

First, consider the shape of the distribution of prop_1br. We know it’s symmetrical around 0.5, and beyond that, we can infer that it’s a normal distribution.

Now, think about how this relates to the distribution of abs_diff. abs_diff is generated by finding the absolute difference between prop_1br and 0.5. Because of this, abs_diff is an array of distances (which are nonnegative by definition) from 0.5.

We know that prop_1br is normal, and symmetrical about 0.5. So, the distribution of how far away prop_1br is from 0.5 will look like we took the distribution of prop_1br, moved it to be centered at 0, and folded it in half so that all negative values become positive. This is because the previous center at 0.5 represents a distance of 0 from 0.5. Similarly, a value of 0.6 would represent a distance of 0.1 from 0.5, and a value of 0.4 would also represent a distance of 0.1 from 0.5.

Now, the problem becomes much simpler to solve. Before, we were told that 95% of our the in prop_1br lies between 0.4 and 0.6 (Thanks to the lines of code that evaluate to True). This is the same as telling us that 95% of the data in prop_1br lies within a distance of 0.1 to 0.5 (Because 0.4 and 0.6 are both 0.1 away from 0.5).

Because of this, the 95% percentile of abs_diff is 0.1, since 95% of the data in prop_1br lies within a distance of 0.1 to 0.5 (meaning that 95% of the data in abs_diff is between 0 and 0.1).

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 10%.

Answer: prop_1br

Our first pair of hypotheses’ alternative hypothesis asks if one number is greater than the other. Because of this, we can’t use an absolute value test statistic to answer the question, since all absolute value cares about is the distance the simulation is from the null assumption, not whether one value is greater than the other.

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Difficulty: ⭐️⭐️

The average score on this problem was 82%.

Answer: abs_diff

Our first pair of hypotheses’ alternative hypothesis asks if one number is not equal to the other. Because of this, we have to use a test statistic that sees the distance both ways, not just in one direction. Therefore, we use the absolute value.

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Difficulty: ⭐️⭐️

The average score on this problem was 83%.

Answer: 0.59

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 20%.

Answer: Option 3: 0.1

Since, we’re looking at a proportion of UCSD students that love dogs, we’ll set a “yes” vote to a value of 1 and a “no” vote to a value of 0. (Try to see why this makes the mean of “yes”/“no” votes also the proportion of “yes” votes). Also by central limit theorem, the distribution of the sample mean is approximately normal. Now recall that a 95% confidence interval of a sample mean is given by [sample mean - 2 * (sample std / np.sqrt(sample size)), sample mean + 2 * (sample std / np.sqrt(sample size))]. As a result, we realize that the width of a 95% confidence interval is 4 * (sample std / np.sqrt(sample size)). Now, the sample size is already constant, which was given to be 400. However, we can attempt to maximize the sample std. It’s not hard to see that the maximum std we could achieve is by recieving an equal number of yes/no votes (aka 200 of each vote). Calculating the standard deviation in this case is just 0.5*, and so the widest possible width for a 95% confidence interval is just 4 * 0.5/np.sqrt(400) which evaluates to 0.1.

*To make the calculation of the standard deviation faster, try to see why calculating the std of a dataset with 200 1’s and 200 0’s is the same as calculating the std of a data set with only a single 1 and a single 0.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 34%.

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.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 40%.

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.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 48%.

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}

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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.

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Difficulty: ⭐️⭐️

The average score on this problem was 77%.

Answer: (a): [p, 1-p], (b): falls + result[1] OR (a): [1-p, p], (b): falls + result[0]

1. (a) First, we should think about what np.random.multinomial is trying to do here. It’s trying to make an array of how many times each scenario happened. There are 2 possible scenarios here: Ylesia succeeds or Ylesia fails. In this code, p is the probability that Ylesia succeeds a skill, and therefore the probabilty that Ylesia does not succeed (she fails) will be 1-p. So to properly simulate how many times she falls, we should put [p, 1-p] in blank (a).
2. (b) Our answer from (a) will make an array stored in result, with index 0 being how many times she succeeded (corresponds to p), and index 1 being how many times she fell (corresponds to 1-p). Since index 1 corresponds to the scenario in which she falls, in order to correctly increase the number of falls, we add falls by result[1]. Therefore, blank (b) is falls + result[1].

Likewise, you can change the order with (a): [1-p, p] and (b): falls + result[0] and it would still correctly simulate how many times she falls.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 59%.

Answer: (a): 1, (b): prob_no_falls * p

1. (a) We start with the initial value of prob_no_falls. This should be set to 1 because we’re computing a probability product, and starting with 1 ensures the initial value doesn’t affect the multiplication of subsequent probabilities.
2. (b) Inside the for-loop, we want to update prob_no_falls by multiplying it by each probability of success (p) in skill_success. This is because the probability of Ylesia not falling throughout multiple independent skills is the product of her not falling during each skill.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer:(a): count_falls(skill_success), (b): np.count_nonzero(results == 0) / 10000, though there are many other correct solutions

1. (a) For this question, we are doing a simulation where we calculate the probability of Ylesia not falling during her routine based on 10,000 trials. To do so, we want to find out the number of times that Yelsia did not fall any skill during her routine out of the 10,000 trials. Based on the given codes, we have an array where we are appending something into that array for each trial. We can utilize the function defined in part a to calculate the number of times Ylesia falls during a single trial so blank a will be count_falls(skill_success).
2. (b) After 10,000 iterations, we have an array of the number of falls for each trial. Then, we want to count the number of times that we get 0 in that array, which means Ylesia did not fall. Lastly, to get the probability, we will need to divide by the total number of trials which is 10,000. This gives us the answer for blank b: np.count_nonzero(results == 0) / 10000.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 66%.

Answer: -1.5

The 86% confidence interval for the population mean is given by:

\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]

Since the width is equal to the difference between the right and left endpoints,

\text{width} = 2 \cdot |z| \cdot \frac{\text{sample SD}}{\sqrt{\text{sample size}}}

We solve for |z|. The maximum width of our CI is given to be 0.15, so we must also use the maximum possible standard deviation, 0.5. we substitute the known values to obtain:

0.15 = 2 \cdot |z| \cdot \frac{0.5}{\sqrt{100}}

which leaves |z| = 1.5 after computation. To find the z such that scipy.stats.norm.cdf(z) evaluates to 0.07, we realize that z is the point under the normal curve, in standard units, left of which represents 7\% of the area under the entire curve. Note that scipy.stats.norm.cdf(0) evaluates to 0.5 (Recall: half of the area is left of the mean, which is zero in standard units). We must therefore take a negative value for z. Thus z = -1.5.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 55%.

Answer: [0.01, \ 0.07]

Recall the formula for the width of an 86\% confidence interval:

\text{width} = 2 \cdot |z| \cdot \frac{\text{sample SD}}{\sqrt{\text{sample size}}}

where we found |z| = 1.5 in part (a). Instead of using the maximum sample SD, we will now use 0.2 and compute the new width of the confidence interval. This results in

\text{width} = 2 \cdot 1.5 \cdot \frac{0.2}{\sqrt{10}} = 0.06

Since this is a CLT-based confidence interval for the population mean, the interval must be centered at the mean. We compute the interval using the structure from part (a), which leaves

\left[ 0.04 - \frac{1}{2} \cdot 0.06, \ 0.04 + \frac{1}{2} \cdot 0.06 \right] = [0.01, \ 0.07]

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 37%.

Answer: 25 boxes

Recall the formula for the width of an 86\% confidence interval:

\text{width} = 2 \cdot |z| \cdot \frac{\text{sample SD}}{\sqrt{\text{sample size}}}

where we must again use the fact that |z| = 1.5 from part (a). Here, we want a width that is no larger than 0.012, given that our sample SD remains 0.2. Plugging everything in:

0.012 \geq 2 \cdot 1.5 \cdot \frac{0.2}{\sqrt{n}}

Rearranging the expression to solve for n, we get

\begin{align*} n &\geq \left( \frac{3 \cdot 0.2}{0.012} \right)^2 \\ n &\geq \left( \frac{600}{12} \right)^2 \\ n &\geq (50)^2 \\ n &\geq 2500 \end{align*}

However, 2500 isn’t our final answer. The question asks for the number of boxes Hermione must buy, given that each box contains 100 beans. The bound we computed above for n corresponds to the minimum number of beans Hermione must observe. To get the minimum number of boxes, we simply divide the bound by 100. The final answer is 25 boxes.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 37%.