Answer: \frac{365}{5}\cdot(v_1 + v_2 + v_3 + v_4 + v_5)

Our sample is the number of visitors on the five days, and our population is the number of visitors in all 365 days.

First, we calculate the sample mean, the average number of visitors in the 5 days, which is m = \frac{1}{5}\cdot(v_1 + v_2 + v_3 + v_4 + v_5). We use this statistic to estimate the population mean, the average number of visitors in this year.

Then, we use the estimated population mean to calculate the estimated pupulation sum, so we multiply the number of days in a year (365) with the estimated population mean. We get 365 m = \frac{365}{5}\cdot(v_1 + v_2 + v_3 + v_4 + v_5)

Difficulty: ⭐️

The average score on this problem was 92%.

Answer: 10.500

Recall, we can make predictions in standard units with the following formula

\text{predicted}\ y_{su} = r \cdot x_{su}

We’re given that the correlation coefficient, r, between visitors and admission cost is 0.25. Here, we’re using the number of visitors (x) to predict admission cost (y). Given that the number of visitors at the Museum of Natural History is 6 standard deviations below average,

\text{predicted}\ y_{su} = r \cdot x_{su} = 0.25 \cdot -6 = -1.5

We then compute y, which is the admission cost (in dollars) at the Museum of Natural History.

\begin{align*} y_{su} &= \frac{y- \text{Mean of } y}{\text{SD of } y}\\ -1.5 &= \frac{y-15}{3} \\ -1.5 \cdot 3 &= y-15\\ -4.5 + 15 &= y\\ y &= \boxed{10.5} \end{align*}

So, the regression line predicts that the admission cost at the Museum of Natural History is $10.50.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 62%.

Answer: r = \frac{3}{10}

To find the correlation coefficient r we use the equation of the regression line in standard units and solve for r as follows. \begin{align*} \text{predicted } y_{\text{(su)}} &= r \cdot x_{\text{(su)}} \\ \frac{4}{5} &= r \cdot \frac{8}{3} \\ r &= \frac{4}{5} \cdot \frac{3}{8} \\ r &= \frac{3}{10} \end{align*}

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 52%.

Answer: s = \frac{6}{25}

We again use the equation of the regression line in standard units, with the value of r we found in the previous part. \begin{align*} \text{predicted } y_{\text{(su)}} &= r \cdot x_{\text{(su)}} \\ s &= \frac{3}{10} \cdot \frac{4}{5} \\ s &= \frac{6}{25} \end{align*}

Notice that when we predict income based on age, our predictions are different than when we predict age based on income. That is, the answer to this question is not \frac{8}{3}. We can think of this phenomenon as a consequence of regression to the mean which means that the predicted variable is always closer to average than the original variable. In part (a), we start with an income of \frac{8}{3} standard units and predict an age of \frac{4}{5} standard units, which is closer to average than \frac{8}{3} standard units. Then in part (b), we start with an age of \frac{4}{5} and predict an income of \frac{6}{25} standard units, which is closer to average than \frac{4}{5} standard units. This happens because whenever we make a prediction, we multiply by r which is less than one in magnitude.

Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 21%.

Answer: I < \text{mean income}, \dfrac{| I - \text{mean income}|}{\text{standard deviation of incomes}} \leq \dfrac{| \text{mean age} + 2 |}{\text{standard deviation of ages}}

To understand this answer, we will investigate each option.

• I < 0:

This option asks whether income is guaranteed to be negative. This is not necessarily true. For example, it’s possible that the slope of the regression line is 2 and the intercept is 10, in which case the income associated with a -2 year old would be 6, which is positive.

• I < \text{mean income}:

This option asks whether the predicted income is guaranteed to be lower than the mean income. It helps to think in standard units. In standard units, the regression line goes through the point (0, 0) and has slope r, which we are told is positive. This means that for a below-average x, the predicted y is also below average. So this statement must be true.

• | I - \text{mean income}| \leq | \text{mean age} + 2 |:

First, notice that | \text{mean age} + 2 | = | -2 - \text{mean age}|, which represents the horizontal distance betweeen these two points on the regression line: (\text{mean age}, \text{mean income}), (-2, I). Likewise, | I - \text{mean income}| represents the vertical distance between those same two points. So the inequality can be interpreted as a question of whether the rise of the regression line is less than or equal to the run, or whether the slope is at most 1. That’s not guaranteed when we’re working in original units, as we are here, so this option is not necessarily true.

• \dfrac{| I - \text{mean income}|}{\text{standard deviation of incomes}} \leq \dfrac{| \text{mean age} + 2 |}{\text{standard deviation of ages}}:

Since standard deviation cannot be negative, we have \dfrac{| I - \text{mean income}|}{\text{standard deviation of incomes}} = \left| \dfrac{I - \text{mean income}}{\text{standard deviation of incomes}} \right| = I_{\text{(su)}}. Similarly, \dfrac{|\text{mean age} + 2|}{\text{standard deviation of ages}} = \left| \dfrac{-2 - \text{mean age}}{\text{standard deviation of ages}} \right| = -2_{\text{(su)}}. So this option is asking about whether the predicted income, in standard units, is guaranteed to be less (in absolute value) than the age. Since we make predictions in standard units using the equation of the regression line \text{predicted } y_{\text{(su)}} = r \cdot x_{\text{(su)}} and we know |r|\leq 1, this means |\text{predicted } y_{\text{(su)}}| \leq | x_{\text{(su)}}|. Applying this to ages (x) and incomes (y), this says exactly what the given inequality says. This is the phenomenon we call regression to the mean.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 69%.

Answer: k = 4

To find this answer, we’ll use the definition of the regression line in original units, which is \text{predicted } y = mx+b, where m = r \cdot \frac{\text{SD of } y}{\text{SD of }x}, \: \: b = \text{mean of } y - m \cdot \text{mean of } x

Next we substitute these value for m and b into \text{predicted } y = mx + b, interpret x as age and y as income, and use the given information to find k. \begin{align*} \text{predicted } y &= mx+b \\ \text{predicted } y &= r \cdot \frac{\text{SD of } y}{\text{SD of }x} \cdot x+ \text{mean of } y - r \cdot \frac{\text{SD of } y}{\text{SD of }x} \cdot \text{mean of } x\\ \text{predicted income}&= r \cdot \frac{\text{SD of income}}{\text{SD of age}} \cdot \text{age}+ \text{mean income} - r \cdot \frac{\text{SD of income}}{\text{SD of age}} \cdot \text{mean age} \\ \frac{31}{2}&= -\frac{1}{3} \cdot k \cdot 24+ \frac{7}{2} + \frac{1}{3} \cdot k \cdot 33 \\ \frac{31}{2}&= -8k+ \frac{7}{2} + 11k \\ \frac{31}{2}&= 3k+ \frac{7}{2} \\ 3k &= \frac{31}{2} - \frac{7}{2} \\ 3k &= 12 \\ k &= 4 \end{align*}

Another way to solve this problem uses the equation of the regression line in standard units and the definition of standard units.

\begin{align*} \text{predicted } y_{\text{(su)}} &= r \cdot x_{\text{(su)}} \\ \frac{\text{predicted income} - \text{mean income}}{\text{SD of income}} &= r \cdot \frac{\text{age} - \text{mean age}}{\text{SD of age}} \\ \frac{\frac{31}{2} - \frac{7}{2}}{k\cdot \text{SD of age}} &= -\frac{1}{3} \cdot \frac{24 - 33}{\text{SD of age}} \\ \frac{12}{k\cdot \text{SD of age}} &= -\frac{1}{3} \cdot \frac{-9}{\text{SD of age}} \\ \frac{12}{k\cdot \text{SD of age}} &= \frac{3}{\text{SD of age}} \\ \frac{k\cdot \text{SD of age}}{\text{SD of age}} &= \frac{12}{3}\\ k &= 4 \end{align*}

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 45%.

Answer: a, d

When performing linear regression analysis, the slope (m) and intercept (b) of the regression line can be calculated as follows:

m = r \cdot \frac{\text{SD of } y}{\text{SD of } x} = 0.6 \times \left(\frac{4}{200}\right) = 0.012

b = (\text{mean of } y) - m \cdot (\text{mean of } x) = 15 - 0.012 \times 500 = 9

To predict the dependent variable (y_i) based on the independent variable (x_i): \text{predicted } y_i = 0.012 \cdot x_i + 9

1. True. The regression line will always pass through the point (\bar{x}, \bar{y}) due to the way we calculate it, which is (500, 15) in this case.

2. False. y = 0.012 \cdot 200 + 9 = 2.4 + 9 = 11.4 \neq 4

3. False. As calculated, the slope of the regression line is 0.012, not 0.6. The correlation coefficient is not the slope.

4. True. The intercept, based on the mean values and the calculated slope, is 9.

5. False. The least squares regression line minimizes the root mean square error (RMSE) compared to any other possible regression lines.

6. False. Regression provides a prediction or estimation, not an exact accounting. Not all books will fall exactly on the regression line due to the inherent variability and other influencing factors not accounted for in the simple linear model.

Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer: 0.36

Given the regression equation(calculated in previous question): \text{predicted } y_i = 0.012 \cdot x_i + 9

Assume:

• x_i for “The Simple Wild” is x.

• x_j for “The Martian” is x + 30.

• Predicted price for “The Simple Wild” (y_i): y_i = 0.012 \cdot x + 9

• Predicted price for “The Martian” (y_j): y_j = 0.012 \cdot (x + 30) + 9 y_j = 0.012 \cdot x + 0.36 + 9 y_j = 0.012 \cdot x + 9.36

The price of “The Martian” minus the price of “The Simple Wild”: y_j - y_i = (0.012 \cdot x + 9.36) - (0.012 \cdot x + 9) = 9.36 - 9 = 0.36

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 39%.

Answer: 17.4

Given the regression equation(calculated in (a)): \text{predicted } y_i = 0.012 \cdot x_i + 9

“The Goldfinch” has 700 pages. Substitute into the regression equation: y_i = 0.012 \cdot 700 + 9 = 8.4 + 9 = 17.4

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 51%.

Answer: "Roadside Picnic" has fewer than 500 pages.

The linear regression line has to pass through the point (500, 15).

• Fewer Than 500 Pages: If “Roadside Picnic” has fewer than 500 pages, then this point is toward the left side of the pivot point of the regression line, and price -12 will drag the line down(towards the lower left corner) more than 12 does, so the slope will decrease if price changes from -12 to 12.
  Plot for your reference:

  

• Exactly 500 Pages: If the book has exactly 500 pages, which is the mean or center of the data distribution, the incorrect price would mainly influence the intercept rather than the slope. Correcting the price would adjust the intercept but have a lesser effect on the slope.

• More Than 500 Pages: If “Roadside Picnic” has more than 500 pages, then it’s toward the right side of the regression line, and price -12 will drag the line down(towards the lower right corner) more than 12 does, so correcting this error would increase the slope.
  Plot for your reference:

  

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 49%.

Answer: No

The regression line would not be the same due to the change in data and the resulting impact on mean and standard deviation of the price and pages.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 65%.

Answer: No

95% of bootstrapped slopes and intercepts fall within specific intervals does not imply that 95% of predicted prices for a book with 500 pages will also fall within the interval [500a + c, 500b + d]. Because the intervals for slopes and intercepts are calculated independently, their combination does not account for the covariance between these regression parameters or the error variance around predictions. Additionally, combining slope and intercept independently can result in regression lines that do not pass through the pivotal point (mean(X), mean(Y)), making them unreliable for accurate prediction.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 55%.

Answer: hypothesis testing, bootstrapping, Central Limit Theorem, confidence interval

The average time to build the LAPPLAND TV storage unit is an unknown population parameter. We’re trying to figure out if this parameter could be equal to the specific value of 2 hours and 30 minutes. We can use the framework we learned in class to set this up as a hypothesis test via confidence interval. When we have a null hypothesis of the form “the parameter equals the specific value” and an alternative hypothesis of “it does not,” this framework applies, and conducting the hypothesis test is equivalent to constructing a confidence interval for the parameter and seeing if the specific value falls in the interval.

There are two ways in which we could construct the confidence interval. One is through bootstrapping, and the other is through the Central Limit Theorem, which applies in this case because our statistic is the mean.

The only listed tools that could not be used here are permutation testing and regression. Permutation testing is used to determine whether two samples could have come from the same population, but here we only have one sample. Permutation testing would be helpful to answer a question like, “Does it take the same amount of time to assemble the LAPPLAND TV storage as it does to assemble the HAUGA TV storage unit?”

Regression is used to predict one numerical quantity based on another, not to estimate a parameter as we are doing here. Regression would be appropriate to answer a question like, “How does the assembly time for the LAPPLAND TV storage unit change with the assembler’s age?”

Difficulty: ⭐️⭐️

The average score on this problem was 78%.

Answer: Option 1

Since r = \frac{2}{5}, the correct option must be a scatter plot with a mild positive (up and to the right) linear association. Option 3 can be ruled out immediately, since the linear association in it is negative (down and to the right). Option 2’s linear association is too strong for r = \frac{2}{5}, and Option 4’s linear association is too weak for r = \frac{2}{5}, which leaves Option 1.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

Answer: 24

At a high level, we’ll start with the formula for the regression line in standard units, and re-write it in a form that will allow us to use the information provided to us in the question.

Recall, the regression line in standard units is

\text{predicted }y_{\text{(su)}} = r \cdot x_{\text{(su)}}

Using the definitions of \text{predicted }y_{\text{(su)}} and x_{\text{(su)}} gives us

\frac{\text{predicted } y - \text{mean of }y}{\text{SD of }y} = r \cdot \frac{x - \text{mean of }x}{\text{SD of }x}

Here, the x variable is sunshine hours in January and the y variable is sunshine hours in July. Given that the standard deviation of January and July sunshine hours are equal, we can simplifies our formula to

\text{predicted } y - \text{mean of }y = r \cdot (x - \text{mean of }x)

Since we’re asked how much more sunshine Santa Clarita will have in July compared to the average, we’re interested in the difference y - \text{mean of} y. We were given that Santa Clarita had 60 more sunshine hours in January than the average, and that the correlation between the two variables(correlation coefficient) is \frac{2}{5}. In terms of the variables above, then, we know:

• x - \text{mean of }x = 60.

• r = \frac{2}{5}.

Then,

\text{predicted } y - \text{mean of }y = r \cdot (x - \text{mean of }x) = \frac{2}{5} \cdot 60 = 24

Therefore, the regression line predicts that Santa Clarita will have 24 more sunshine hours than the average California city in July.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 68%.

Answer: \frac{2}{5}

To determine the slope of Anthony’s new regression line, we need to understand how the modifications he made to the dataset (subtracting 5 hours from each x and y value) affect the slope. In simple linear regression, the slope of the regression line (m in y = mx + b) is calculated using the formula:

m = r \cdot \frac{\text{SD of y}}{\text{SD of x}}

r, the correlation coefficient between the two variables, remains unchanged in Anthony’s modifications. Remember, the correlation coefficient is the mean of the product of the x values and y values when both are measured in standard units; by subtracting the same constant amount from each x value, we aren’t changing what the x values convert to in standard units. If you’re not convinced, convert the following two arrays in Python to standard units; you’ll see that the results are the same.

x1 = np.array([5, 8, 4, 2, 9])
x2 = x1 - 5

Furthermore, Anthony’s modifications also don’t change the standard deviations of the x values or y values, since the xs and ys aren’t any more or less spread out after being shifted “down” by 5. So, since r, \text{SD of }y, and \text{SD of }x are all unchanged, the slope of the new regression line is the same as the slope of the old regression line, pre-modification!

Given the fact that the correlation coefficient is \frac{2}{5} and the standard deviation of sunshine hours in January (\text{SD of }x) is equal to the standard deviation of sunshine hours in July (\text{SD of }y), we have

m = r \cdot \frac{\text{SD of }y}{\text{SD of }x} = \frac{2}{5} \cdot 1 = \frac{2}{5}

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer: 7

Let’s denote the original intercept as b and the new intercept in the new dataset as b'. The equation for the original regression line is y = mx + b, where:

• y is a predicted number of sunshine hours in July, before 5 was subtracted from each number of hours.
• m is the slope of the line, which we know is \frac{2}{5} from the previous part.
• x is a number of sunshine hours in January, before 5 was subtracted from each number of hours.
• b is the original intercept. This is 10.

When Anthony subtracts 5 from each x and y value, the new regression line becomes y - 5 = m \cdot (x - 5) + b'

Expanding and rearrange this equation, we have

y = mx - 5m + 5 + b'

Remember, x and y here represent the number of sunshine hours in January and July, respectively, before Anthony subtracted 5 from each number of hours. This means that the equation for y above is equivalent to y = mx + b. Comparing, we see that

-5m + 5 + b' = b

Since m = \frac{2}{5} (from the previous part) and b = 10, we have

-5 \cdot \frac{2}{5} + 5 + b' = 10 \implies b' = 10 - 5 + 2 = 7

Therefore, the intercept of Anthony’s new regression line is 7.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 34%.

Answer: Option 1

Since r = \frac{2}{5}, the correct option must be a scatter plot with a mild positive (up and to the right) linear association. Option 3 can be ruled out immediately, since the linear association in it is negative (down and to the right). Option 2’s linear association is too strong for r = \frac{2}{5}, and Option 4’s linear association is too weak for r = \frac{2}{5}, which leaves Option 1.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

Answer: 24

At a high level, we’ll start with the formula for the regression line in standard units, and re-write it in a form that will allow us to use the information provided to us in the question.

Recall, the regression line in standard units is

\text{predicted }y_{\text{(su)}} = r \cdot x_{\text{(su)}}

Using the definitions of \text{predicted }y_{\text{(su)}} and x_{\text{(su)}} gives us

\frac{\text{predicted } y - \text{mean of }y}{\text{SD of }y} = r \cdot \frac{x - \text{mean of }x}{\text{SD of }x}

Here, the x variable is sunshine hours in January and the y variable is sunshine hours in July. Given that the standard deviation of January and July sunshine hours are equal, we can simplifies our formula to

\text{predicted } y - \text{mean of }y = r \cdot (x - \text{mean of }x)

Since we’re asked how much more sunshine Santa Clarita will have in July compared to the average, we’re interested in the difference y - \text{mean of} y. We were given that Santa Clarita had 60 more sunshine hours in January than the average, and that the correlation between the two variables(correlation coefficient) is \frac{2}{5}. In terms of the variables above, then, we know:

• x - \text{mean of }x = 60.

• r = \frac{2}{5}.

Then,

\text{predicted } y - \text{mean of }y = r \cdot (x - \text{mean of }x) = \frac{2}{5} \cdot 60 = 24

Therefore, the regression line predicts that Santa Clarita will have 24 more sunshine hours than the average California city in July.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 68%.

Answer: \frac{2}{5}

To determine the slope of Anthony’s new regression line, we need to understand how the modifications he made to the dataset (subtracting 5 hours from each x and y value) affect the slope. In simple linear regression, the slope of the regression line (m in y = mx + b) is calculated using the formula:

m = r \cdot \frac{\text{SD of y}}{\text{SD of x}}

r, the correlation coefficient between the two variables, remains unchanged in Anthony’s modifications. Remember, the correlation coefficient is the mean of the product of the x values and y values when both are measured in standard units; by subtracting the same constant amount from each x value, we aren’t changing what the x values convert to in standard units. If you’re not convinced, convert the following two arrays in Python to standard units; you’ll see that the results are the same.

x1 = np.array([5, 8, 4, 2, 9])
x2 = x1 - 5

Furthermore, Anthony’s modifications also don’t change the standard deviations of the x values or y values, since the xs and ys aren’t any more or less spread out after being shifted “down” by 5. So, since r, \text{SD of }y, and \text{SD of }x are all unchanged, the slope of the new regression line is the same as the slope of the old regression line, pre-modification!

Given the fact that the correlation coefficient is \frac{2}{5} and the standard deviation of sunshine hours in January (\text{SD of }x) is equal to the standard deviation of sunshine hours in July (\text{SD of }y), we have

m = r \cdot \frac{\text{SD of }y}{\text{SD of }x} = \frac{2}{5} \cdot 1 = \frac{2}{5}

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer: 7

Let’s denote the original intercept as b and the new intercept in the new dataset as b'. The equation for the original regression line is y = mx + b, where:

• y is a predicted number of sunshine hours in July, before 5 was subtracted from each number of hours.
• m is the slope of the line, which we know is \frac{2}{5} from the previous part.
• x is a number of sunshine hours in January, before 5 was subtracted from each number of hours.
• b is the original intercept. This is 10.

When Anthony subtracts 5 from each x and y value, the new regression line becomes y - 5 = m \cdot (x - 5) + b'

Expanding and rearrange this equation, we have

y = mx - 5m + 5 + b'

Remember, x and y here represent the number of sunshine hours in January and July, respectively, before Anthony subtracted 5 from each number of hours. This means that the equation for y above is equivalent to y = mx + b. Comparing, we see that

-5m + 5 + b' = b

Since m = \frac{2}{5} (from the previous part) and b = 10, we have

-5 \cdot \frac{2}{5} + 5 + b' = 10 \implies b' = 10 - 5 + 2 = 7

Therefore, the intercept of Anthony’s new regression line is 7.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 34%.

Answer: A distinctive pattern in the residual plot

We’re told in the problem that the number of sunshine hours per month in Chicago increases from the winter (January) to the summer (July/August) and then decreases again to the winter (December). Here’s a real plot of this data; we don’t need real data to answer this question, but this is the kind of plot you could sketch out in the exam given the description in the question. (The gold shaded area is irrelevant for our purposes.)

The points in this plot aren’t tightly clustered around a straight line, and that’s because there’s a non-linear relationship between month and number of sunshine hours. As such, when we draw a straight line through this scatter plot, it won’t be able to fully capture the relationship being shown. It’ll likely start off in the bottom left and increase to the top right, which will lead to the sunshine hours for summer months being underestimated and the sunshine hours for later winter months (November, December) being overestimated. This will lead to a distinctive pattern in our residual plot, which means that linear regression as-is isn’t the right tool for modeling this data (because ideally, the residual plot would be a patternless cloud of points).

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 47%.

Answer: Option 3.

The largest slope out of these four options will be the slope that represents the greatest increase in y-units per x-unit: m = \dfrac{\Delta y}{\Delta x}.

Option 1, which predicts "Rent" from "Sqft", has large values for its y-variable ("Rent"), but also has large values for its x-variable ("Sqft"). The resulting slope is not that big, as it is a fraction of large values over large values.

Option 2, which predicts "Sqft" from "Rent", is also not that big of a slope for the same reasons as Option 1 (slope is a fraction of large values over large values).

Option 3, which predicts "Rent" from "Bed", has large values for its y-variable ("Rent"), but has small values for its x-variable ("Bed"). The resulting slope is incredibly big, as it is a fraction of large values over small values.

Option 4, which predicts "Bed" from "Rent", has small values for its y-variable ("Bed"), but has large values for its x-variable ("Rent"). The resulting slope is incredibly small, as it is a fraction of small values over large values.

Of all four options, Option 3 is the largest slope.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 71%.

Answer: Option 4.

As explained above, Option 4 is the smallest slope.

Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: 3 hours

We know that \bar{x} = 200, \sigma_x = 80, r = \frac{2}{3}, and \bar{y} = 6. In this problem we are given x = 160 and y = 5. In order to find the standard deviation of time here, we can start by standardizing our values:

x_{su} = \frac{160-200}{80} = -\frac{1}{2}

Then according to the formula: \text{Predicted} \: y_{\text{su}} = r \cdot x_{\text{su}}

\text{Predicted} \: y_{\text{su}} = \frac{2}{3} \cdot -\frac{1}{2} = \frac{1}{3}

Now that we have y in standard units, we can plug it into this formula to solve for the standard deviation of y:

y_{su} = \frac{y - \bar{y}}{\sigma_y}

- \frac{1}{3}= \frac{5-6}{\sigma_y}

\text{SD}_y = 3

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 61%.

Answer: \frac{3}{4} hours longer.

Since we are interested in how much longer stage 14 is than stage 20 based on this linear association, we first want to calculate the slope. Note that r=\frac{2}{3}, \text{SD}_y=1.5, and \text{SD}_x=80:

m = r \cdot \frac{\text{SD}_y}{\text{SD}_x}

m = \frac{2}{3} \cdot \frac{1.5}{80} = \frac{1}{80}

This means that for every additional 1km, time increases by \frac{1}{80}.

Since Stage 14 is 60km longer than Stage 20, we simply multiply our slope by 60, giving \frac{60}{80} = \frac{3}{4}. Thus we expect Stage 14 to take \frac{3}{4} hours longer.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 48%.

Answer: It would stay the same.

Adding a 30 minute break to all the stages simply increases each stage’s time by an additional 30 minutes. This would not change the slope since adding time simply shifts the time data points right, but doesn’t change the relationship between distance and time.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 55%.

Answer: r=1

We can follow a similar process to part 1, but instead solve for r now. First, we calculate x in standard units:

x_{\text{su}} = \frac{x - \bar{x}}{\text{SD}_x}

x_{\text{su}} = \frac{360-200}{80} = \frac{160}{80} = 2

Now find y in standard units:

y_{\text{su}} = \frac{y - \bar{y}}{\text{SD}_y}

y_{\text{su}} = \frac{9-6}{1.5} = \frac{3}{1.5} = 2

Now we can solve for r simply by y_{\text{su}} = r * x_{\text{su}}

2 = r * 2

r=1

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 35%.

Answer: hypothesis testing

The question at hand is dealing with differences between sales of different flavors of ice cream, which is the same thing as the total of ice cream cones sold. We can use hypothesis testing to test our null hypothesis that the count of Vanilla cones sold is higher than Chocolate, and our alternative hypothesis that the count of Chocolate cones sold is more than Vanilla. A permutation test is not suitable here because we are not comparing any numerical quantity associated with each group. A permutation test could be used to answer questions like “Are chocolate ice cream cones more expensive than vanilla ice cream cones?” or “Do chocolate ice cream cones have more calories than vanilla ice cream cones?”, or any other question where you are tracking a number (cost or calories) along with each ice cream cone. In our case, however, we are not tracking a number along with each individual ice cream cone, but instead tracking a total of ice cream cones sold.

An analogy to this hypothesis test can be found in the “fair or unfair coin” problem in Lectures 20 and 21, where our null hypothesis is that the coin is fair and our alternative hypothesis is that the coin is unfair. The “fairness” of the coin is not a numerical quantity that we can track with each individual coin flip, just like how the count of ice cream cones sold is not a numerical quantity that we can track with each individual ice cream cone.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

Answer: permutation (A/B) testing

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%.

Answer: 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%.

Answer: 79.20

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%.

Answer: False

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%.

Answer: r > 0

To answer this problem, it’s useful to recall the regression line in standard units:

\text{predicted } y_{\text{(su)}} = r \cdot x_{\text{(su)}}

If a value is positive in standard units, it means that it is above the average of the distribution that it came from, and if a value is negative in standard units, it means that it is below the average of the distribution that it came from. Since we’re told that Wingspan has a "Complexity" that is 2 points higher than the average, we know that x_{\text{(su)}} is positive. Since we’re told that the predicted "Rating" is 3 points higher than the average, we know that \text{predicted } y_{\text{(su)}} must also be positive. As a result, r must also be positive, since you can’t multiply a positive number (x_{\text{(su)}}) by a negative number and end up with another positive number.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

Answer: SD of "Complexity" < SD of "Rating"

Since the distance of the predicted "Rating" from its average is larger than the distance of the "Complexity" from its average, it might be reasonable to guess that the values in the "Rating" column are more spread out. This is true, but let’s see concretely why that’s the case.

Let’s start with the equation of the regression line in standard units from the previous subpart. Remember that here, x refers to "Complexity" and y refers to "Rating".

\text{predicted } y_{\text{(su)}} = r \cdot x_{\text{(su)}}

We know that to convert a value to standard units, we subtract the value by the mean of the column it came from, and divide by the standard deviation of the column it came from. As such, x_{\text{(su)}} = \frac{x - \text{mean of } x}{\text{SD of } x}. We can substitute this relationship in the regression line above, which gives us

\frac{\text{predicted } y - \text{mean of } y}{\text{SD of } y} = r \cdot \frac{x - \text{mean of } x}{\text{SD of } x}

To simplify things, let’s use what we were told. We were told that the predicted "Rating" was 3 points higher than average. This means that the numerator of the left side, \text{predicted } y - \text{mean of } y, is equal to 3. Similarly, we were told that the "Complexity" was 2 points higher than average, so x - \text{mean of } x is 2. Then, we have:

\frac{3}{\text{SD of } y} = \frac{2r}{\text{SD of }x}

Note that for convenience, we included r in the numerator on the right-hand side.

Remember that our goal is to compare the SD of "Rating" (y) to the SD of "Complexity" (x). We now have an equation that relates these two quantities! Since they’re both currently on the denominator, which can be tricky to work with, let’s take the reciprocal (i.e. “flip”) both fractions.

\frac{\text{SD of } y}{3} = \frac{\text{SD of }x}{2r}

Now, re-arranging gives us

\text{SD of } y \cdot \frac{2r}{3} = \text{SD of }x

Since we know that r is somewhere between 0 and 1, we know that \frac{2r}{3} is somewhere between 0 and \frac{2}{3}. This means that \text{SD of } x is somewhere between 0 and two-thirds of the value of \text{SD of } y, which means that no matter what, \text{SD of } x < \text{SD of } y. Remembering again that here "Complexity" is our x and "Rating" is our y, we have that the SD of "Complexity" is less than the SD of "Rating".

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 42%.

Answer: r > 0

To answer this problem, it’s useful to recall the regression line in standard units:

\text{predicted } y_{\text{(su)}} = r \cdot x_{\text{(su)}}

If a value is positive in standard units, it means that it is above the average of the distribution that it came from, and if a value is negative in standard units, it means that it is below the average of the distribution that it came from. Since we’re told that Wingspan has a "Complexity" that is 2 points higher than the average, we know that x_{\text{(su)}} is positive. Since we’re told that the predicted "Rating" is 3 points higher than the average, we know that \text{predicted } y_{\text{(su)}} must also be positive. As a result, r must also be positive, since you can’t multiply a positive number (x_{\text{(su)}}) by a negative number and end up with another positive number.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

Answer: SD of "Complexity" < SD of "Rating"

Since the distance of the predicted "Rating" from its average is larger than the distance of the "Complexity" from its average, it might be reasonable to guess that the values in the "Rating" column are more spread out. This is true, but let’s see concretely why that’s the case.

Let’s start with the equation of the regression line in standard units from the previous subpart. Remember that here, x refers to "Complexity" and y refers to "Rating".

\text{predicted } y_{\text{(su)}} = r \cdot x_{\text{(su)}}

We know that to convert a value to standard units, we subtract the value by the mean of the column it came from, and divide by the standard deviation of the column it came from. As such, x_{\text{(su)}} = \frac{x - \text{mean of } x}{\text{SD of } x}. We can substitute this relationship in the regression line above, which gives us

\frac{\text{predicted } y - \text{mean of } y}{\text{SD of } y} = r \cdot \frac{x - \text{mean of } x}{\text{SD of } x}

To simplify things, let’s use what we were told. We were told that the predicted "Rating" was 3 points higher than average. This means that the numerator of the left side, \text{predicted } y - \text{mean of } y, is equal to 3. Similarly, we were told that the "Complexity" was 2 points higher than average, so x - \text{mean of } x is 2. Then, we have:

\frac{3}{\text{SD of } y} = \frac{2r}{\text{SD of }x}

Note that for convenience, we included r in the numerator on the right-hand side.

Remember that our goal is to compare the SD of "Rating" (y) to the SD of "Complexity" (x). We now have an equation that relates these two quantities! Since they’re both currently on the denominator, which can be tricky to work with, let’s take the reciprocal (i.e. “flip”) both fractions.

\frac{\text{SD of } y}{3} = \frac{\text{SD of }x}{2r}

Now, re-arranging gives us

\text{SD of } y \cdot \frac{2r}{3} = \text{SD of }x

Since we know that r is somewhere between 0 and 1, we know that \frac{2r}{3} is somewhere between 0 and \frac{2}{3}. This means that \text{SD of } x is somewhere between 0 and two-thirds of the value of \text{SD of } y, which means that no matter what, \text{SD of } x < \text{SD of } y. Remembering again that here "Complexity" is our x and "Rating" is our y, we have that the SD of "Complexity" is less than the SD of "Rating".

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 42%.

Answer: 3.1

Let’s recall the formulas for the regression line in original units, since we’re given information in original units in this question (such as the fact that for a "Play Time" of 8 minutes, the predicted "Rating" is 4 stars). Remember that throughout this question, "Play Time" is our x and "Rating" is our y.

The regression line is of the form y = mx + b, where

m = r \cdot \frac{\text{SD of } y}{\text{SD of }x}, b = \text{mean of }y - m \cdot \text{mean of } x

There’s a lot of information provided to us in the question – let’s think about what it means in the context of our xs and ys.

• The first piece is that r is negative, so -1 \leq r < 0.
• The second piece is that \text{SD of } x = 2 \cdot (\text{SD of } y). Equivalently, we can say that \frac{\text{SD of } y}{\text{SD of } x} = \frac{1}{2}. This form is convenient, because it’s close to the definition of the slope of the regression line, m. Using this fact, the slope of the regression line is m = r \cdot \frac{\text{SD of } y}{\text{SD of }x} = r \cdot \frac{1}{2} = \frac{r}{2}.
• The \text{mean of } x is 10. This means that the intercept of the regression line, b, is b = \text{mean of }y - m \cdot \text{mean of } x = \text{mean of }y - \frac{r}{2} \cdot 10 = \text{mean of }y - 5r.
• If x is 8, the predicted y is 4.

Given all of this information, we need to find possible values for the \text{mean of } y. Substituting our known values for m and b into y = mx + b gives us

y = \frac{r}{2} x + \text{mean of }y - 5r

Now, using the fact that if if x = 8, the predicted y is 4, we have

\begin{align*}4 &= \frac{r}{2} \cdot 8 + \text{mean of }y - 5r\\4 &= 4r - 5r + \text{mean of }y\\ 4 + r &= \text{mean of} y\end{align*}

Cool! We now know that the \text{mean of } y is 4 + r. We know that r must satisfy the relationship -1 \leq r < 0. By adding 4 to all pieces of this inequality, we have that 3 \leq r + 4 < 4, which means that 3 \leq \text{mean of } y < 4. Of the four options provided, only one is greater than or equal to 3 and less than 4, which is 3.1.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 55%.