In [1]:

```
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
```

In [2]:

```
times = [0.0,1.0,2.0,3.0,4.0,5.0,6.0]
distances = [0.00,44.05,107.16,148.44,196.40,254.44,300.00]
```

One common use of modeling is **interpolation** to determine a value "inside" or "in between" the measured data points. Here we will make a prediction for the value of the dependent variable `distances`

for a given independent variable `times`

that falls "in between" two measurements from a road trip, where the distances are those traveled for the given elapse times.

In [3]:

```
plt.scatter(times, distances)
plt.plot(times, np.poly1d(np.polyfit(times, distances, 1))(times), color='r')
plt.grid(True)
plt.xlabel('X Data Time Driven (hours)')
plt.ylabel('Y Data Distance Traveled (miles)');
```

In [4]:

```
# Compute the total change in distance and change in time
total_distance = distances[-1] - distances[0]
total_time = times[-1] - times[0]
# Estimate the slope of the data from the ratio of the changes
average_speed = total_distance / total_time
# Predict the distance traveled for a time not measured
elapse_time = 2.5
distance_traveled = average_speed * elapse_time
print(f"Total Distance {total_distance}")
print(f"\nTotal Time {total_time}")
print(f"\nAverage Speed {average_speed}")
print(f"\nThe distance traveled is {distance_traveled}")
```

Notice that the answer distance is 'inside' that range of data values, so, less than the max(distances) but greater than the min(distances)

Another common use of modeling is **extrapolation** to estimate data values "outside" or "beyond" the range (min and max values of `time`

) of the measured data. Here, we have measured distances for times $0$ through $5$ hours, but we are interested in estimating how far we'd go in $8$ hours. Using the same data set from the previous exercise, we have prepared a linear model `distance = model(time)`

. Use that `model()`

to make a prediction about the distance traveled for a time much larger than the other times in the measurements.

In [5]:

```
# Plot the observed data
plt.scatter(times, distances)
# Fit a model and predict future dates
predict_vals = [0.0,1.0,2.0,3.0,4.0,5.0,6.0,8.0,9.0,10.0]
model = np.polyfit(times, distances, 1)
predicted = np.polyval(model, predict_vals)
# Plot the model
plt.plot(predict_vals, predicted, color='r')
plt.grid(True)
plt.xlabel('X Data Time Driven (hours)')
plt.ylabel('Y Data Distance Traveled (miles)');
```

In [6]:

```
def model(time, a0=0, a1=average_speed):
"""
Purpose:
For a given value of time, compute the model value for distance
Args:
time (float, np.ndarray): elapse time in units of hours
a0 (float): default=0, coefficient for the Zeroth order term in the model, i.e. a0 + a1*x
a1 (float): default=50, coefficient for the 1st order term in the model, i.e. a0 + a1*x
Returns:
distance (float, np.ndarray): model values corresponding to input time array, with the same length/size.
"""
distance = a0 + (a1*time)
return distance
```

In [7]:

```
# Select a time not measured.
time = 8
# Use the model to compute a predicted distance for that time.
distance = model(time)
# Inspect the value of the predicted distance traveled.
print(distance)
# Determine if you will make it without refueling.
answer = (distance <= 400)
print(answer)
```

Notice that the car can travel just to the range limit of $400$ miles, so you'd run out of gas just as you completed the trip

Another common application of modeling is to compare two data sets by building models for each, and then comparing the models. Here, we are given data for a road trip two cars took together. The cars stopped for gas every $50$ miles, but each car did not need to fill up the same amount, because the cars do not have the same fuel efficiency (MPG). The function `efficiency_model(miles, gallons)`

is created to estimate efficiency as average miles traveled per gallons of fuel consumed.

In [8]:

```
car1 = {'miles': [1. , 50.9, 100.8, 150.7, 200.6, 250.5, 300.4, 350.3, 400.2, 450.1,500.],
'gallons': [0.03333333, 1.69666667, 3.36, 5.02333333, 6.68666667, 8.35, 10.01333333, 11.67666667,
13.34, 15.00333333, 16.66666667]}
car2 = {'miles': [1. , 50.9, 100.8, 150.7, 200.6, 250.5, 300.4, 350.3, 400.2, 450.1, 500. ],
'gallons': [0.02 , 1.018, 2.016, 3.014, 4.012, 5.01, 6.008, 7.006, 8.004, 9.002, 10. ]}
```

In [9]:

```
plt.scatter(car1['miles'], car1['gallons'], label='car1')
plt.scatter(car2['miles'], car2['gallons'], label='car2')
plt.grid(True)
plt.ylabel('Y Data Fuel Consumed (gallons)')
plt.xlabel('X Data Distance Traveled (miles)')
plt.legend();
```

In [10]:

```
# Complete the function to model the efficiency.
def efficiency_model(miles, gallons):
return np.mean(miles/gallons)
# Use the function to estimate the efficiency for each car.
car1['mpg'] = efficiency_model(np.array(car1['miles']), np.array(car1['gallons']))
car2['mpg'] = efficiency_model(np.array(car2['miles']), np.array(car2['gallons']))
# Finish the logic statement to compare the car efficiencies.
if car1['mpg'] > car2['mpg'] :
print('car1 is the best')
elif car1['mpg'] < car2['mpg'] :
print('car2 is the best')
else:
print('the cars have the same efficiency')
```

Notice the original plot that visualized the raw data, the slope is $1$/MPG and so `car1`

is steeper than `car2`

, but if you plot(gallons, miles) the slope is MPG, and so `car2`

has a steeper slope than `car1`

In [11]:

```
plt.scatter(car1['gallons'], car1['miles'], label='car1')
plt.scatter(car2['gallons'], car2['miles'], label='car2')
plt.grid(True)
plt.xlabel('Y Data Fuel Consumed (gallons)')
plt.ylabel('X Data Distance Traveled (miles)')
plt.legend();
```

The mean describes the center of the data. The standard deviation describes the spread of the data. But to compare two variables, it is convenient to normalize both. Here, we are provided with two arrays of data, which are highly correlated, and we will compute and visualize the normalized deviations of each array.

In [12]:

```
num_samples = 50
# The desired mean values of the sample.
mu = [6.01327144, 301.09351247]
# The desired covariance matrix.
cov = [[1.44408412e+00, 7.11018554e+01],
[7.11018554e+01, 3.62713035e+03]]
# Generate the random samples.
Y = np.random.multivariate_normal(mu, cov, size=num_samples)
# x and y data
x = Y[:,0]
y = Y[:,1]
```

In [13]:

```
# Compute the deviations by subtracting the mean offset
dx = x - np.mean(x)
dy = y - np.mean(y)
# Normalize the data by dividing the deviations by the standard deviation
zx = dx / np.std(x)
zy = dy / np.std(y)
```

In [14]:

```
plt.figure(figsize=(10,1))
plt.plot(dx, color='r')
plt.plot(dy, color='b')
plt.xlabel('Array Index')
plt.ylabel('Deviations of x & y')
plt.show()
plt.figure(figsize=(10,1))
plt.plot(zx, color='r')
plt.plot(zy, color='b')
plt.xlabel('Array Index')
plt.ylabel('Normalized Deviations of x & y')
plt.show()
```

Notice how hard it is to compare $dx$ and $dy$, versus comparing the normalized $zx$ and $zy$.

Covariance is a measure of whether two variables change ("vary") together. It is calculated by computing the products, point-by-point, of the deviations seen in the previous exercise, `dx[n]*dy[n]`

, and then finding the average of all those products.

Correlation is in essence the normalized covariance. Here, we have with two arrays of data, which are highly correlated, and we will visualize and compute both the `covariance`

and the `correlation`

.

In [15]:

```
# Compute the covariance from the deviations.
dx = x - np.mean(x)
dy = y - np.mean(y)
covariance = np.mean(dx * dy)
print("Covariance: ", covariance)
# Compute the correlation from the normalized deviations.
zx = dx / np.std(x)
zy = dy / np.std(y)
correlation = np.mean(zx * zy)
print("Correlation: ", correlation)
```

In [16]:

```
plt.plot(zx*zy)
plt.axhline(0, linestyle='dashed', color='r')
plt.xlabel('Array Index')
plt.ylabel('Product of Normalized Deviations')
plt.title(f'Correlation = np.mean(zx*zy) = {round(correlation,2)*100}');
```

Notice that we've plotted the product of the normalized deviations, and labeled the plot with the correlation, a single value that is the mean of that product. The product is always positive and the mean is typical of how the two vary together.

Intuitively, we can look at the plots provided and "see" whether the two variables seem to "vary together".

**Data Set A:**$x$ and $y$ change together and appear to have a strong relationship.**Data Set B:**there is a rough upward trend; $x$ and $y$ appear only loosely related.**Data Set C:**looks like random scatter; $x$ an $y$ do not appear to change together and are unrelated.

In [17]:

```
# Generate correlated random variables
# The desired number of samples
num_samples = 50
# The desired mean values of the sample.
mu = [4.71815762, 9.44076045, 5.09321393, 10.16904155, 5.0165893, 10.6883259]
# The desired covariance matrix.
cov = [[1.36208968, 2.70163528, 1.37835545, 2.78133499, 1.73359544, 0.95056748],
[2.70163528, 5.40596017, 2.73195338, 5.65673871, 3.43406231, 2.30481703],
[1.37835545, 2.73195338, 1.47904869, 3.23872554, 1.77708497, 1.88996037],
[2.78133499, 5.65673871, 3.23872554, 24.2142715, 3.65117281, 14.5615414],
[1.73359544, 3.43406231, 1.77708497, 3.65117281, 2.25638144, 2.41304719],
[0.95056748, 2.30481703, 1.88996037, 14.5615414, 2.41304719, 295.880271]]
# Generate the random samples.
y = np.random.multivariate_normal(mu, cov, size=num_samples)
# Create a dictionaty from the random samples split into 3 sets each with an x and y array of values
data_sets = {'A': {'x':y[:,0],
'y':y[:,1]},
'B': {'x':y[:,2],
'y':y[:,3]},
'C': {'x':y[:,4],
'y':y[:,5]}}
```

Recall that deviations differ from the mean, and we normalized by dividing the deviations by standard deviation. Here we will compare the $3$ data sets by computing correlation, and determining which data set has the most strongly correlated variables $x$ and $y$.

In [18]:

```
for name, data in data_sets.items():
plt.scatter(data['x'], data['y'], color='k')
plt.axhline(0, linestyle='dashed', color='b')
plt.grid(True)
plt.xlabel('X')
plt.ylabel('Y')
plt.title(f'Data set {name} has correlation = ?')
plt.show()
```

In [19]:

```
# Complete the function that will compute correlation.
def correlation(x,y):
x_dev = x - np.mean(x)
y_dev = y - np.mean(y)
x_norm = x_dev / np.std(x)
y_norm = y_dev / np.std(y)
return np.mean(x_norm * y_norm)
# Compute and store the correlation for each data set in the list.
for name, data in data_sets.items():
data['correlation'] = correlation(data['x'], data['y'])
print('\ndata set {} has correlation {:.2f}'.format(name, data['correlation']))
```

In [20]:

```
# Assign the data set with the best correlation.
best_data = data_sets['A']
```

Note that the strongest relationship is in Dataset $A$, with correlation closest to $1.0$ and the weakest is Datatset $C$ with correlation value closest to zero.

The Taylor series of a function is an infinite sum of terms, where each term has a larger exponent like x, x2, x3, etc. A Taylor Series is expressed in terms of the function's derivatives at a single point. For most common functions, the function and the sum of its Taylor series are equal near this point. Taylor's series are named after **Brook Taylor** who introduced them in $1715$.

- approximate any curve
- polynomial form
- often, first order is enough

- The
**derivative**of a function of a real variable measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). Derivatives are a fundamental tool of calculus. For example, the derivative of the position of a moving object with respect to time is the object's velocity: this measures how quickly the position of the object changes when time advances.

- A polynomial in a single indeterminate $x$ can always be written (or rewritten) in the form

- where $a_{0},\ldots ,a_{n}$ are constants and $x$ is the indeterminate. The word "indeterminate" means that $x$ represents no particular value, although any value may be substituted for it. The mapping that associates the result of this substitution to the substituted value is a function, called a
*polynomial function*.

Previously, we have been given a ** pre-defined model** to work with. In this Here, we will implement a model function that returns model values for $y$, computed from input $x$ data, and any input coefficients for the "zero-th" order term

`a0`

, the "first-order" term `a1`

, and a quadratic term `a2`

of a model.Recall that "first order" is linear, so we'll set the defaults for this general linear model with `a2=0`

, but later, we will change this for comparison.

In [21]:

```
# Define the general model as a function
def model(x, a0=3, a1=2, a2=0):
return a0 + (a1*x) + (a2*x*x)
# Generate array x, then predict y values for specific, non-default a0 and a1
x = np.linspace(-10, 10, 21)
y = model(x)
plt.scatter(x, y)
plt.plot(x, np.poly1d(np.polyfit(x, y, 1))(x), color='r')
plt.grid(True)
plt.title('Plot of modeled Y for given data X')
plt.xlabel('X')
plt.ylabel('Y');
```

Notice that we used `model()`

to compute predicted values of `y`

for given possibly measured values of `x`

. The model takes the independent data and uses it to generate a model for the dependent variables corresponding values.

Now that we've built a ** general** model, let's "optimize" or "fit" it to a new measured data set,

`xd`

, `yd`

, by finding the `a0`

, `a1`

for which the model data and the measured data line up on a plot.This is an iterative visualization strategy, where we start with a *guess* for model parameters, pass them into the `model()`

, over-plot the resulting modeled data on the measured data, and visually check that the line passes through the points. If it doesn't, we change the model parameters and try again.

In [22]:

```
df_xy = pd.read_csv('df_xy.csv').drop('Unnamed: 0', axis=1)
df_xy.info()
xd = df_xy['X'].values
yd = df_xy['Y'].values
```

In [23]:

```
plt.scatter(xd, yd)
plt.grid(True)
plt.xlim(-5,15)
plt.ylim(-250,750)
plt.axvline(0, color='k')
plt.axhline(0, color='k')
plt.title('Hiking Trip')
plt.xlabel('Step distance (km)')
plt.ylabel('Altitude (meters)');
```

In [24]:

```
# Complete the plotting function definition
def plot_data_with_model(xd, yd, ym):
plt.scatter(xd, yd)
plt.grid(True)
plt.xlim(-5,15)
plt.ylim(-250,750)
plt.axvline(0, color='k')
plt.axhline(0, color='k')
plt.title('Hiking Trip')
plt.xlabel('Step distance (km)')
plt.ylabel('Altitude (meters)');
plt.plot(xd, ym, color='red') # over-plot modeled y data
plt.show()
return
# Select new model parameters a0, a1, and generate modeled `ym` from them.
a0 = 150
a1 = 25
ym = model(xd, a0, a1)
# Plot the resulting model to see whether it fits the data
fig = plot_data_with_model(xd, yd, ym)
```

Notice again that the measured x-axis data `xd`

is used to generate the modeled y-axis data `ym`

so to plot the model, you are plotting `ym`

vs `xd`

, which may seem counter-intuitive at first. But we are modeling the $y$ response to a given $x$; we are not modeling $x$.

The definition of temperature scales is related to the linear expansion of certain liquids, such as mercury and alcohol. Originally, these scales were literally rulers for measuring length of fluid in the narrow marked or "graduated" tube as a proxy for temperature. The alcohol starts in a bulb, and then expands linearly into the tube, in response to increasing temperature of the bulb or whatever surrounds it.

Here, we will explore the conversion between the Fahrenheit and Celsius temperature scales as a demonstration of interpreting slope and intercept of a linear relationship within a physical context.

In [25]:

```
# Complete the function to convert C to F
def convert_scale(temps_C):
(freeze_C, boil_C) = (0, 100)
(freeze_F, boil_F) = (32, 212)
change_in_C = boil_C - freeze_C
change_in_F = boil_F - freeze_F
slope = change_in_F / change_in_C
intercept = freeze_F - freeze_C
temps_F = intercept + (slope * temps_C)
return temps_F
# Use the convert function to compute values of F and plot them
temps_C = np.linspace(0, 100, 101)
temps_F = convert_scale(temps_C)
plt.plot(temps_C, temps_F)
plt.grid(True)
plt.ylabel('Temperature (Fahrenheit)')
plt.xlabel('Temperature (Celsius)');
```

The rate of change - ROC - is the speed at which a variable changes over a specific period of time. ROC is often used when speaking about momentum, and it can generally be expressed as a ratio between a change in one variable relative to a corresponding change in another; graphically, the rate of change is represented by the slope of a line. The ROC is often illustrated by the Greek letter delta $\Delta$.

Here, we will model the motion of a car driving (roughly) constant velocity by computing the average velocity over the entire trip. The linear relationship modeled is between the time elapsed and the distance traveled.

In this case, the model parameter `a1`

, or slope, is approximated or "estimated", as the mean velocity, or put another way, the "rate-of-change" of the distance ("rise") divided by the time ("run").

In [26]:

```
df_td = pd.read_csv('time_distance.csv').drop('Unnamed: 0', axis=1)
df_td.info()
times = df_td['times']
distances = df_td['distances']
```

In [27]:

```
# Compute an array of velocities as the slope between each point
diff_distances = np.diff(distances)
diff_times = np.diff(times)
velocities = diff_distances / diff_times
# Chracterize the center and spread of the velocities
v_avg = np.mean(velocities)
v_max = np.max(velocities)
v_min = np.min(velocities)
v_range = v_max - v_min
```

In [28]:

```
plt.scatter(times[1:], velocities, label='Velocities')
plt.axhline(v_avg, color='r', lw=5, alpha=0.4, label='Mean velocity')
plt.ylim(0,100)
plt.grid(True)
plt.ylabel('Instantaneous velocity (Kilometers/Hours)')
plt.xlabel('Time (hours)')
plt.legend();
```

Generally we might use the average velocity as the slope in our model. But notice that there is some random variation in the instantaneous velocity values when plotted as a time series. The range of values `v_max - v_min`

is one measure of the scale of that variation, and the standard deviation of velocity values is another measure.

In statistics, ordinary least squares (OLS) is a type of linear least squares method for estimating the unknown parameters in a linear regression model. OLS chooses the parameters of a linear function of a set of explanatory variables by the principle of least squares: minimizing the sum of the squares of the differences between the observed dependent variable (values of the variable being observed) in the given dataset and those predicted by the linear function.

**Simple linear regression model**

If the data matrix $X$ contains only two variables, a constant and a scalar regressor $x_i$, then this is called the "simple regression model". It provides much simpler formulas even suitable for manual calculation. The parameters are commonly denoted as ($\alpha$, $\beta$):

$$y_{i}=\alpha +\beta x_{i}+\varepsilon _{i}$$The least squares estimates in this case are given by simple formulas

$${\begin{aligned}{\hat {\beta }}&={\frac {\sum {x_{i}y_{i}}-{\frac {1}{n}}\sum {x_{i}}\sum {y_{i}}}{\sum {x_{i}^{2}}-{\frac {1}{n}}(\sum {x_{i}})^{2}}}={\frac {\operatorname {Cov} [x,y]}{\operatorname {Var} [x]}}\\{\hat {\alpha }}&={\overline {y}}-{\hat {\beta }}\,{\overline {x}}\ ,\end{aligned}}$$where Var(.) and Cov(.) are sample parameters.

Here, we will see the intercept and slope parameters in the context of modeling measurements taken of the volume of a solution contained in a large glass jug. The solution is composed of water, grains, sugars, and yeast. The total mass of both the solution and the glass container was also recorded, but the empty container mass was not noted.

Our job is to use the pandas DataFrame `df`

, with data columns `volumes`

and `masses`

, to build a linear model that relates the `masses`

(y-data) to the `volumes`

(x-data). The slope will be an estimate of the density (change in mass / change in volume) of the solution, and the intercept will be an estimate of the empty container weight (mass when volume=$0$).

In [29]:

```
fpath_csv = 'https://assets.datacamp.com/production/repositories/1480/datasets/e4c4cdd076de27d3c1bccfe1d0019d279c08d2fc/solution_data.csv'
df = pd.read_csv(fpath_csv, comment='#')
print(df.info())
print(df.head())
```

In [30]:

```
# Import ols from statsmodels, and fit a model to the data
from statsmodels.formula.api import ols
model_fit = ols(formula="masses ~ volumes", data=df)
model_fit = model_fit.fit()
# Extract the model parameter values, and assign them to a0, a1
a0 = model_fit.params['Intercept']
a1 = model_fit.params['volumes']
# Print model parameter values with meaningful names, and compare to summary()
print( "container_mass = {:0.4f}".format(a0) )
print( "\nsolution_density = {:0.4f}".format(a1) )
print('\n',model_fit.summary())
```

A residual is the vertical distance between a data point and the regression line. Each data point has one residual. They are positive if they are above the regression line and negative if they are below the regression line. If the regression line actually passes through the point, the residual at that point is zero.

The **error** (or **disturbance**) of an observed value is the *deviation* of the observed value from the (unobservable) *true* value of a quantity of interest (for example, a population mean), and the **residual** of an observed value is the difference between the observed value and the *estimated* value of the quantity of interest (for example, a sample mean). The distinction is most important in regression analysis, where the concepts are sometimes called the **regression errors** and **regression residuals**

The residual sum of squares measures the amount of error remaining between the regression function and the data set. A smaller residual sum of squares figure represents a regression function. Residual sum of squares–also known as the sum of squared residuals–essentially determines how well a regression model explains or represents the data in the model.

$$ SS_{\text{res}}=\sum _{i}(y_{i}-f_{i})^{2}=\sum _{i}e_{i}^{2}\ $$Previously we saw that the altitude along a hiking trail was roughly fit by a linear model, and we introduced the concept of ** differences** between the model and the data as a

Here, we'll work with the same measured data, and quantifying how well a model fits it by computing the sum of the square of the "differences", also called "residuals".

In [31]:

```
# Function to generate x and y data
def load_data():
num_pts=21; a0=3.0*50; a1=0.5*50; mu=0; sigma=1; ae=0.5*50; seed=1234;
np.random.seed(seed)
xmin = 0
xmax = 10
x1 = np.linspace(xmin, xmax, num_pts)
e1 = np.array([np.random.normal(mu, sigma) for n in range(num_pts)])
y1 = a0 + (a1*x1) + ae*e1
return x1, y1
```

In [32]:

```
# Load the data
x_data, y_data = load_data()
# Model the data with specified values for parameters a0, a1
y_model = model(x_data, a0=150, a1=25)
# Compute the RSS value for this parameterization of the model
rss = np.sum(np.square(y_data - y_model))
print(f"RSS = {rss}")
```

The value we compute for RSS is not meaningful by itself, but later it becomes meaningful in context when we compare it to other values of RSS computed for other parameterizations of the model.

Here, we will complete a function to visually compare model and data, and compute and print the RSS. We will call it more than once to see how RSS changes when you change values for `a0`

and `a1`

. We'll see that the values for the parameters we found earlier are the ones needed to ** minimize** the RSS.

In [33]:

```
# Complete function to load data, build model, compute RSS, and plot
def compute_rss_and_plot_fit(a0, a1):
xd, yd = load_data()
ym = model(xd, a0, a1)
residuals = ym - yd
rss = np.sum(np.square(residuals))
summary = "Parameters a0={}, a1={} yield RSS={:0.2f}".format(a0, a1, rss)
plt.scatter(xd, yd)
plt.grid(True)
plt.xlim(-5,15)
plt.ylim(-250,750)
plt.axvline(0, color='k')
plt.axhline(0, color='k')
plt.title(summary)
plt.xlabel('Step distance (km)')
plt.ylabel('Altitude (meters)')
plt.plot(xd, ym, color='red') # over-plot modeled y data
plt.show()
return rss, summary
# Chose model parameter values and pass them into RSS function
rss, summary = compute_rss_and_plot_fit(a0=150, a1=25)
print(summary)
```

As stated earlier, the significance of RSS is in context of other values. More specifically, the minimum RSS is of value in identifying the specific set of parameter values for our model which yield the smallest residuals in an overall sense.

Here we will compute and visualize how RSS varies for different values of model parameters. Start by holding the intercept constant, but vary the slope: and for each slope value, we'll compute the model values, and the resulting RSS. Once you have an array of RSS values, we will determine minimal RSS value, in code, and from that minimum, determine the slope that resulted in that minimal RSS.

In [34]:

```
x_data = df_xy['X'].values
y_data = df_xy['Y'].values
```

In [35]:

```
# Function to compute RSS
def compute_rss(yd, ym):
rss = np.sum(np.square(yd-ym))
return rss
```

In [36]:

```
# Empty list for RSS array
rss_list = []
# Loop over all trial values in a1_array, computing rss for each
a1_array = np.linspace(15, 35, 101)
for a1_trial in a1_array:
y_model = model(x_data, a0=150, a1=a1_trial)
rss_value = compute_rss(y_data, y_model)
rss_list.append(rss_value)
# Find the minimum RSS and the a1 value from whence it came
rss_array = np.array(rss_list)
best_rss = np.min(rss_array)
best_a1 = a1_array[np.where(rss_array==best_rss)]
print('The minimum RSS = {}, came from a1 = {}'.format(best_rss, best_a1[0]))
# Plot your rss and a1 values to confirm answer
plt.scatter(a1_array, rss_array)
plt.scatter(best_a1, best_rss, color='r', s=150)
plt.grid(True)
plt.title(f'Minimum RSS = {best_rss}\n came from a1 = {best_a1[0]}')
plt.xlabel('Slope a1')
plt.ylabel('RSS');
```

The best slope is the one out of an array of slopes than yielded the minimum RSS value out of an array of RSS values. Python tip: notice that we started with `rss_list`

to make it easy to `.append()`

but then later converted to`numpy.array()`

to gain access to all the `numpy`

methods.

`numpy`

¶The formulae below are the result of working through the calculus. Here, we'll trust that the calculus correct, and implement these formulae in code using `numpy`

.

In [37]:

```
x = df_xy['X'].values
y = df_xy['Y'].values
```

In [38]:

```
# prepare the means and deviations of the two variables
x_mean = np.mean(x)
y_mean = np.mean(y)
x_dev = x - x_mean
y_dev = y - y_mean
# Complete least-squares formulae to find the optimal a0, a1
a1 = np.sum(x_dev * y_dev) / np.sum( np.square(x_dev) )
a0 = y_mean - (a1 * x_mean)
# Use the those optimal model parameters a0, a1 to build a model
y_model = model(np.array(x), a0, a1)
# plot to verify that the resulting y_model best fits the data y
fig, rss = compute_rss_and_plot_fit(a0, a1)
```

Notice that the optimal slope a1, according to least-squares, is a ratio of the covariance to the variance. Also, note that the values of the parameters obtained here are NOT exactly the ones used to generate the pre-loaded data (a1=$25$ and a0=$150$), but they are close to those. Least-squares does not guarantee zero error; there is no perfect solution, but in this case, least-squares is the best we can do.

`Scipy`

¶It is possible to write a `numpy`

implementation of the analytic solution to find the minimal RSS value. But for more complex models, finding analytic formulae is not possible, and so we turn to other methods.

Here we will use `scipy.optimize`

to employ a more general approach to solve the same optimization problem.

In so doing, we will see additional return values from the method that tell answer us "how good is best". Here we will use the same measured data and parameters as seen in the last exercise for ease of comparison of the new `scipy`

approach.

In [39]:

```
from scipy import optimize
# Define a model function needed as input to scipy
def model_func(x, a0, a1):
return a0 + (a1*x)
# Load the measured data you want to model
x_data, y_data = load_data()
# call curve_fit, passing in the model function and data; then unpack the results
param_opt, param_cov = optimize.curve_fit(model_func, x_data, y_data)
a0 = param_opt[0] # a0 is the intercept in y = a0 + a1*x
a1 = param_opt[1] # a1 is the slope in y = a0 + a1*x
# test that these parameters result in a model that fits the data
fig, rss = compute_rss_and_plot_fit(a0, a1)
```

Notice that we passed the function object itself, `model_func`

into `curve_fit`

, rather than passing in the model data. The model function object was the input, because the optimization wants to know what form in general it's solve for; had we passed in a `model_func`

with more terms like an `a2*x**2`

term, we would have seen different results for the parameters output.

`statsmodels`

¶Several python libraries provide convenient abstracted interfaces so that you need not always be so explicit in handling the machinery of optimization of the model.

As an example, in this exercise, we will use the `statsmodels`

library in a more high-level, generalized work-flow for building a model using least-squares optimization (minimization of RSS).

In [40]:

```
x_data, y_data = load_data()
df = pd.DataFrame(dict(x_column=x_data, y_column=y_data))
# Pass data and `formula` into ols(), use and `.fit()` the model to the data
model_fit = ols(formula="y_column ~ x_column", data=df).fit()
# Use .predict(df) to get y_model values, then over-plot y_data with y_model
y_model = model_fit.predict(df)
# Extract the a0, a1 values from model_fit.params
a0 = model_fit.params['Intercept']
a1 = model_fit.params['x_column']
# Visually verify that these parameters a0, a1 give the minimum RSS
fig, rss = compute_rss_and_plot_fit(a0, a1)
```

Note that the params container always uses 'Intercept' for the `a0`

key, but all higher order terms will have keys that match the column name from the pandas DataFrame that we passed into `ols()`

.

If you found part of a skeleton, from an adult human that lived thousands of years ago, how could you estimate the height of the person that it came from? This exercise is in part inspired by the work of forensic anthropologist Mildred Trotter, who built a regression model for the calculation of stature estimates from human "long bones" or femurs that is commonly used today.

In this exercise, we'll use data from many living people, and the python library `scikit-learn`

, to build a linear model relating the length of the femur (thigh bone) to the "stature" (overall height) of the person. Then, we'll apply our model to make a prediction about the height of an ancient ancestor.

In [41]:

```
fpath_csv = 'https://assets.datacamp.com/production/repositories/1480/datasets/a1871736d7829e85ec4ead2212d621df69bb3977/femur_data.csv'
df = pd.read_csv(fpath_csv)
df.info()
legs = df['length'].values
heights = df['height'].values
```

In [42]:

```
plt.scatter(df['length'], df['height'])
plt.grid(True)
plt.xlim(-10,100)
plt.ylim(-30,300)
plt.axvline(0, color='k')
plt.axhline(0, color='k')
plt.title('Relating long bone length and hieght')
plt.xlabel('Femur Length (cm)')
plt.ylabel('Height (cm)');
```

In [43]:

```
# import the sklearn class LinearRegression and initialize the model
from sklearn.linear_model import LinearRegression
model = LinearRegression(fit_intercept=False)
# Prepare the measured data arrays and fit the model to them
legs = legs.reshape(len(legs),1)
heights = heights.reshape(len(heights),1)
model.fit(legs, heights)
# Use the fitted model to make a prediction for the found femur
fossil_leg = [[50.7]]
fossil_height = model.predict(fossil_leg)
print("Predicted fossil height = {:0.2f} cm".format(fossil_height[0,0]))
```

Notice that we used the pre-loaded data to fit or "train" the model, and then applied that model to make a prediction about newly collected data that was not part of the data used to fit the model. Also notice that `model.predict()`

returns the answer as an array of `shape = (1,1)`

, so we had to index into it with the `[0,0]`

syntax when printing.

Time-series data provides a context in which the "slope" of the linear model represents a "rate-of-change".

In this exercise, we will use measurements of sea level change from $1970$ to $2010$, build a linear model of that changing sea level and use it to make a prediction about the future sea level rise.

In [44]:

```
fpath_csv = 'https://assets.datacamp.com/production/repositories/1480/datasets/0c1231fa1e777d3d7dd2d0ab877158a0dd219a5b/sea_level_data.csv'
df0 = pd.read_csv(fpath_csv, comment='#')
df0.info()
years = df0['year'].values
levels = df0['sea_level_inches'].values
```

In [45]:

```
plt.scatter(df0['year'], df0['sea_level_inches'])
plt.grid(True)
plt.title('Global Average Seal Level Change')
plt.xlabel('Seal Level Change (inches)')
plt.ylabel('Time (years)');
```

In [47]:

```
# Prepare the measured data arrays and fit the model to them
years = years.reshape(len(years),1)
levels = levels.reshape(len(years),1)
# Build a model, fit to the data
model = LinearRegression(fit_intercept=True)
model.fit(years, levels)
# Use model to make a prediction for one year, 2100
future_year = 2100
future_level = model.predict([[future_year]])
print("Prediction: year = {}, level = {:.02f}".format(future_year, future_level[0,0]))
# Use model to predict for many years, and over-plot with measured data
years_forecast = np.linspace(1970, 2100, 131).reshape(-1, 1)
levels_forecast = model.predict(years_forecast)
plt.scatter(df0['year'], df0['sea_level_inches'])
plt.plot(years_forecast, levels_forecast, color='r')
plt.grid(True)
plt.title('Global Average Seal Level Change')
plt.xlabel('Seal Level Change (inches)')
plt.ylabel('Time (years)');
```

Notice also that although our model is linear, the actual data appears to have an up-turn that might be better modeled by adding a quadratic or even exponential term to our model. The linear model forecast may be underestimating the rate of increase in sea level.

Less than $100$ years ago, the universe appeared to be composed of a single static galaxy, containing perhaps a million stars. Today we have observations of hundreds of billions of galaxies, each with hundreds of billions of stars, all moving.

The beginnings of the modern physical science of cosmology came with the publication in 1929 by Edwin Hubble that included use of a linear model.

In this exercise, we will build a model whose slope will give Hubble's Constant, which describes the velocity of galaxies as a linear function of distance from Earth.

In [48]:

```
fpath_csv = 'https://assets.datacamp.com/production/repositories/1480/datasets/fffe3864d24c1864a5df0f0985a11d07bc1cd526/hubble_data.csv'
df = pd.read_csv(fpath_csv, comment='#')
df.info()
```

In [49]:

```
plt.scatter(df['distances'], df['velocities'])
plt.grid(True)
plt.xlim(-1.00,4.00)
plt.ylim(-400,1200)
plt.xlabel('Distances')
plt.ylabel('Velocities');
```

In [50]:

```
# Fit the model, based on the form of the formula
model_fit = ols(formula="velocities ~ distances", data=df).fit()
# Extract the model parameters and associated "errors" or uncertainties
a0 = model_fit.params['Intercept']
a1 = model_fit.params['distances']
e0 = model_fit.bse['Intercept']
e1 = model_fit.bse['distances']
# Print the results
print('For slope a1={:.02f}, the uncertainty in a1 is {:.02f}'.format(a1, e1))
print('For intercept a0={:.02f}, the uncertainty in a0 is {:.02f}'.format(a0, e0))
```

In this exercise, we will build a linear model by fitting monthly time-series data for the Dow Jones Industrial Average (**DJIA**) and then use that model to make predictions for daily data (in effect, an interpolation). Then we will compare that daily prediction to the real daily **DJIA** data.

A few notes on the data. "OHLC" stands for "Open-High-Low-Close", which is usually daily data, for example the opening and closing prices, and the highest and lowest prices, for a stock in a given day. `DayCount`

is an integer number of days from start of the data collection.

In [51]:

```
from pandas_datareader import data as web
import warnings
warnings.filterwarnings('ignore')
start = pd.Timestamp('2013')
end = pd.Timestamp('2015-01-01')
df_daily = web.DataReader('DJIA', 'yahoo', start, end)[['Close']]
# Resample to monthly
df_monthly = df_daily.resample('M').last()
# Count number of days elapsed since a certain date
basedate_d = df_daily.index[0]
df_daily['DayCount'] = df_daily.apply(lambda x: (x.name - basedate_d).days, axis=1)
basedate_m = df_monthly.index[0]
df_monthly['DayCount'] = df_monthly.apply(lambda x: (x.name - basedate_m).days, axis=1)
print(df_daily.head())
print('\n',df_monthly.head())
```

In [52]:

```
plt.figure(figsize=(10,5))
plt.scatter(df_monthly.index, df_monthly['Close'], label='Mpnthly Data')
plt.grid(True)
plt.xlabel('Date')
plt.ylabel('DJIA Close ($)')
plt.legend();
```

In [53]:

```
# fit a model to the df_monthly data
model_fit = ols('Close ~ DayCount', data=df_monthly).fit()
# Use the model to make a predictions for both monthly and daily data
df_monthly['Model'] = model_fit.predict(df_monthly.DayCount)
df_daily['Model'] = model_fit.predict(df_daily.DayCount)
# Calculate the RSS
rss_d = compute_rss(df_daily['Close'], df_daily['Model'])
rss_m = compute_rss(df_monthly['Close'], df_monthly['Model'])
plt.figure(figsize=(10,5))
plt.scatter(df_monthly.index, df_monthly['Close'], label='Mpnthly Data')
plt.plot(df_monthly.index, df_monthly['Model'], label='Model', color='r')
plt.grid(True)
plt.title(f'RSS {rss_m}')
plt.xlabel('Date')
plt.ylabel('DJIA Close ($)')
plt.legend();
plt.figure(figsize=(10,5))
plt.scatter(df_daily.index, df_daily['Close'], label='Daily Data')
plt.plot(df_daily.index, df_daily['Model'], label='Model', color='r')
plt.grid(True)
plt.title(f'RSS {rss_d}')
plt.xlabel('Date')
plt.ylabel('DJIA Close ($)')
plt.legend();
```

Notice the monthly data looked linear, but the daily data clearly has additional, nonlinear trends. Under-sampled data often misses real-world features in the data on smaller time or spatial scales. Using the model from the under-sampled data to make interpolations to the daily data can result is large residuals. Notice that the RSS value for the daily plot is more than 30 times worse than the monthly plot.

In this exercise, we consider the perils of extrapolation. Shown here is the profile of a hiking trail on a mountain. One portion of the trail, marked in black, looks linear, and was used to build a model. But we see that the best fit line, shown in red, does not fit outside the original "domain", as it extends into this new outside data, marked in blue.

If we want use the model to make predictions for the altitude, but still be accurate to within some tolerance, what are the smallest and largest values of independent variable `x`

that we can allow ourselves to apply the model to?"

In [54]:

```
df = pd.read_csv('hiking.csv').drop('Unnamed: 0', axis=1)
df.info()
x_data = df['x_data'].values
y_data = df['y_data'].values
y_model = df['y_model'].values
```

In [55]:

```
col = np.where(np.array(x_data)<=0,'k', np.where(np.array(x_data)<10,'b','k'))
plt.scatter(x_data, y_data, c=col)
plt.plot(x_data, y_model, color='r')
plt.grid(True)
plt.xlim(-15,25)
plt.ylim(-250,750)
plt.axvline(0, color='k')
plt.axhline(0, color='k')
plt.title('Hiking Trip')
plt.xlabel('Step distance (kilometers)')
plt.ylabel('Altitude (meters)');
```

In [56]:

```
# Compute the residuals, "data - model", and determine where [residuals < tolerance]
residuals = np.abs(y_data - y_model)
tolerance = 100
x_good = x_data[residuals < tolerance]
# Find the min and max of the "good" values, and plot y_data, y_model, and the tolerance range
print(f'Minimum good x value = {np.min(x_good)}')
print(f'Maximum good x value = {np.max(x_good)}')
col = np.where(np.array(x_data)<=0,'k', np.where(np.array(x_data)<10,'b','k'))
plt.scatter(x_data, y_data, c=col)
plt.plot(x_data, y_model, color='r')
a = x_data <= 13
b = x_data >= -6
mask = (a & b)
plt.fill_between(x_data, y_model-tolerance, y_model+tolerance, where=mask, alpha=0.2, color='g')
plt.grid(True)
plt.xlim(-15,25)
plt.ylim(-250,750)
plt.axvline(0, color='k')
plt.axhline(0, color='k')
plt.title('Hiking Trip')
plt.xlabel('Step distance (kilometers)')
plt.ylabel('Altitude (meters)');
```

**RSS:**- Used to help find the optimal values for model parameters

- The total sum of squares (proportional to the variance of the data):

- The sum of squares of residuals, also called the residual sum of squares:

**RMSE:**- Root Mean Square Error (RMSE) is the standard deviation of the residuals (prediction errors). Residuals are a measure of how far from the regression line data points are; RMSE is a measure of how spread out these residuals are. In other words, it tells you how concentrated the data is around the line of best fit.

- The RMSE of an estimator $\hat{\theta}$ with respect to an estimated parameter $\theta$ is defined as the square root of the mean square error:

- The RMSE of predicted values $\hat y_t$ for times $t$ of a regression's dependent variable $y_{t}$ with variables observed over $T$ times, is computed for $T$ different predictions as the square root of the mean of the squares of the deviations:

**R-squared**- R-squared or
**the coefficient of determination**is a statistical measure that represents the proportion of the variance for a dependent variable that's explained by an independent variable or variables in a regression model. Whereas correlation explains the strength of the relationship between an independent and dependent variable, R-squared explains to what extent the variance of one variable explains the variance of the second variable.

- R-squared or

- The most general definition of the coefficient of determination is

In this exercise, we will quantify the over-all model "goodness-of-fit" of a pre-built model, by computing one of the most common quantitative measures of model quality, the RMSE, step-by-step.

In [57]:

```
x_data = df_xy['X'].values
y_data = df_xy['Y'].values
```

In [61]:

```
def model_fit_and_predict(x, y):
a0=150
a1=25
ym = a0 + (a1*x)
return ym
```

In [62]:

```
# Build the model and compute the residuals "model - data"
y_model = model_fit_and_predict(np.array(x_data), np.array(y_data))
residuals = y_model - y_data
# Compute the RSS, MSE, and RMSE and print the results
RSS = np.sum(np.square(residuals))
MSE = RSS/len(residuals)
RMSE = np.sqrt(MSE)
print('RMSE = {:0.2f}, MSE = {:0.2f}, RSS = {:0.2f}'.format(RMSE, MSE, RSS))
```

In this exercise we'll compute another measure of goodness, **R-squared**. R-squared is the ratio of the variance of the residuals divided by the variance of the data we are modeling, and in so doing, is a measure of how much of the variance in your data is "explained" by your model, as expressed in the spread of the residuals.

In [63]:

```
# Compute the residuals and the deviations
residuals = y_model - y_data
deviations = np.mean(y_data) - y_data
# Compute the variance of the residuals and deviations
var_residuals = np.mean(np.square(residuals))
var_deviations = np.mean(np.square(deviations))
# Compute r_squared as 1 - the ratio of RSS/Variance
r_squared = 1 - (var_residuals / var_deviations)
print('R-squared is {:0.2f}'.format(r_squared))
```

The **standard error (SE)** of a statistic (usually an estimate of a parameter) is the standard deviation of its sampling distribution or an estimate of that standard deviation. If the parameter or the statistic is the mean, it is called the **standard error of the mean (SEM)**.

- The standard error(SE) is very similar to standard deviation. Both are measures of spread. To put it simply, the two terms are essentially equal — but there is one important difference. While the standard error uses statistics (sample data) standard deviations use parameters (population data).

- A statistic and a parameter are very similar. They are both descriptions of groups, like $50\%$ of dog owners prefer $X$ Brand dog food. The difference between a statistic and a parameter is that statistics describe a
**sample**. A parameter describes an entire**population**.

The data need not be perfectly linear, and there may be some random variation or "spread" in the measurements, and that does translate into variation of the model parameters. This variation is in the parameter is quantified by "standard error", and interpreted as "uncertainty" in the estimate of the model parameter.

In this exercise, we will use `ols`

from `statsmodels`

to build a model and extract the standard error for each parameter of that model.

In [64]:

```
# Store data in a DataFrame, and use ols() to fit a model to it.
df = pd.DataFrame(dict(times=times, distances=distances))
model_fit = ols(formula="distances ~ times", data=df).fit()
# Extact the model parameters and their uncertainties
a0 = model_fit.params['Intercept']
e0 = model_fit.bse['Intercept']
a1 = model_fit.params['times']
e1 = model_fit.bse['times']
# Print the results with more meaningful names
print('Estimate of the intercept = {:0.2f}'.format(a0))
print('Uncertainty of the intercept = {:0.2f}'.format(e0))
print('Estimate of the slope = {:0.2f}'.format(a1))
print('Uncertainty of the slope = {:0.2f}'.format(e1))
```

The size of the parameters standard error only makes sense in comparison to the parameter value itself. In fact the units are the same! So `a1`

and `e1`

both have units of velocity (meters/second), and `a0`

and `e0`

both have units of distance (meters).

Given two data sets of distance-versus-time data, one with very small velocity and one with large velocity. Notice that both may have the same standard error of slope, but different R-squared for the model overall, depending on the size of the slope ("effect size") as compared to the standard error ("uncertainty").

If we plot both data sets as scatter plots on the same axes, the contrast is clear. Variation due to the slope is different than variation due to the random scatter about the trend line. In this exercise, our goal is to compute the standard error and R-squared for two data sets and compare.

In [65]:

```
df = pd.read_csv('trip2.csv').drop('Unnamed: 0', axis=1)
df.info()
```

In [66]:

```
plt.scatter(df['times'], df['distances1'], c='k')
plt.scatter(df['times'], df['distances2'], c='r')
plt.grid(True)
plt.title('Driving Trip')
plt.xlabel('Travel Distance (kilometers)')
plt.ylabel('Elapse Time (hours)');
```

In [67]:

```
# Build and fit two models, for columns distances1 and distances2 in df
model_1 = ols(formula="distances1 ~ times", data=df).fit()
model_2 = ols(formula="distances2 ~ times", data=df).fit()
# Extract R-squared for each model, and the standard error for each slope
se_1 = model_1.bse['times']
se_2 = model_2.bse['times']
rsquared_1 = model_1.rsquared
rsquared_2 = model_2.rsquared
# Print the results
print('Model 1: SE = {:0.3f}, R-squared = {:0.3f}'.format(se_1, rsquared_1))
print('Model 2: SE = {:0.3f}, R-squared = {:0.3f}'.format(se_2, rsquared_2))
```

Notice that the standard error is the same for both models, but the r-squared changes. The uncertainty in the estimates of the model parameters is independent from R-squared because that uncertainty is being driven not by the linear trend, but by the inherent randomness in the data. This serves as a transition into looking at statistical inference in linear models.

In this exercise we will work with a generated population. We will construct a sample by drawing points at random from the population. We will compute the mean standard deviation of the `sample`

taken from that `population`

to test whether the sample is representative of the population. Our goal is to see where the sample statistics are the same or very close to the population statistics.

In [68]:

```
sns.set(style="whitegrid", color_codes=True)
# Seed random number generator
np.random.seed(123)
population = np.random.normal(99.98, 9.73, 310)
sns.distplot(population, bins=25, color='r', kde=False);
```

In [69]:

```
# Compute the population statistics
print("Population mean {:.1f}, stdev {:.2f}".format(population.mean(), population.std()))
# Set random seed for reproducibility
np.random.seed(42)
# Construct a sample by randomly sampling 31 points from the population
sample = np.random.choice(population, size=31)
# Compare sample statistics to the population statistics
print("Sample mean {:.1f}, stdev {:.2f}".format(sample.mean(), sample.std()))
```

Notice that the sample statistics are similar to the population statistics, but not the identical. If you were to compute the `len()`

of each array, it is very different, but the means are not that much different as you might expect.

If we create one sample of `size=1000`

by drawing that many points from a population. Then compute a sample statistic, such as the mean, a single value that summarizes the sample itself.

If you repeat that sampling process `num_samples=100`

times, you get $100$ samples. Computing the sample statistic, like the mean, for each of the different samples, will result in a distribution of values of the mean. The goal then is to compute the mean of the means and standard deviation of the means.

Here we will use `population`

, `num_samples`

, and `num_pts`

, and note that the `means`

and `deviations`

arrays have been initialized to zero to give us containers to use for the for loop.

In [70]:

```
# Seed random number generator
np.random.seed(12)
population = np.random.normal(99.9840024355, 10.001880368, 1000000)
sns.distplot(population, bins=50, color='r', kde=False);
```