Statistical Thinking with Python (2)
How often do we get no-hitters?
The number of games played between each no-hitter in the modern era (1901-2015) of Major League Baseball is stored in the array nohitter_times. If we assume that no-hitters are described as a Poisson process, then the time between no-hitters is Exponentially distributed. As we have seen, the Exponential distribution has a single parameter, which we will call τ, the typical interval time. The value of the parameter τ that makes the exponential distribution best match the data is the mean interval time (where time is in units of number of games) between no-hitters.
Compute the value of this parameter from the data. Then, use np.random.exponential() to “repeat” the history of Major League Baseball by drawing inter-no-hitter times from an exponential distribution with the τ you found and plot the histogram as an approximation to the PDF. NumPy, pandas, matlotlib.pyplot, and seaborn have been imported for you as np, pd, plt, and sns, respectively.
# Seed random number generator
np.random.seed(42)
# Compute mean no-hitter time: tau
tau = np.mean(nohitter_times)
# Draw out of an exponential distribution with parameter tau: inter_nohitter_time
inter_nohitter_time = np.random.exponential(tau, 100000)
# Plot the PDF and label axes
_ = plt.hist(inter_nohitter_time,
bins=50, normed=True, histtype='step')
_ = plt.xlabel('Games between no-hitters')
_ = plt.ylabel('PDF')
# Show the plot
plt.show()
Do the data follow our story?
We have modeled no-hitters using an Exponential distribution. Create an ECDF of the real data. Overlay the theoretical CDF with the ECDF from the data. This helps us to verify that the Exponential distribution describes the observed data. It may be helpful to remind ourself of the function we created in the previous course to compute the ECDF, as well as the code you wrote to plot it.
# Create an ECDF from real data: x, y
x, y = ecdf(nohitter_times)
# Create a CDF from theoretical samples: x_theor, y_theor
x_theor, y_theor = ecdf(inter_nohitter_time)
# Overlay the plots
plt.plot(x_theor, y_theor, marker='.', linestyle='none')
plt.plot(x, y,marker='.', linestyle='none')
# Margins and axis labels
plt.margins(0.02)
plt.xlabel('Games between no-hitters')
plt.ylabel('CDF')
# Show the plot
plt.show()
How is this parameter optimal?
Now sample out of an exponential distribution with τ being twice as large as the optimal τ. Do it again for τ half as large. Make CDFs of these samples and overlay them with your data. We can see that they do not reproduce the data as well. Thus, the τ we computed from the mean inter-no-hitter times is optimal in that it best reproduces the data.
# Plot the theoretical CDFs
plt.plot(x_theor, y_theor)
plt.plot(x, y, marker='.', linestyle='none')
plt.margins(0.02)
plt.xlabel('Games between no-hitters')
plt.ylabel('CDF')
# Take samples with half tau: samples_half
samples_half = np.random.exponential(tau/2,size=10000)
# Take samples with double tau: samples_double
samples_double = np.random.exponential(2*tau,size=10000)
# Generate CDFs from these samples
x_half, y_half = ecdf(samples_half)
x_double, y_double = ecdf(samples_double)
# Plot these CDFs as lines
_ = plt.plot(x_half, y_half)
_ = plt.plot(x_double, y_double)
# Show the plot
plt.show()
EDA of literacy/fertility data
In the next few exercises, we will look at the correlation between female literacy and fertility (defined as the average number of children born per woman) throughout the world. For ease of analysis and interpretation, we will work with the illiteracy rate. It is always a good idea to do some EDA ahead of our analysis. To this end, plot the fertility versus illiteracy and compute the Pearson correlation coefficient. The Numpy array illiteracy has the illiteracy rate among females for most of the world’s nations. The array fertility has the corresponding fertility data. Here, it may be useful to refer back to the function we wrote in the previous course to compute the Pearson correlation coefficient.
# Plot the illiteracy rate versus fertility
_ = plt.plot(x=illiteracy, y=fertility, marker='.', linestyle='none')
# Set the margins and label axes
plt.margins(0.02)
_ = plt.xlabel('percent illiterate')
_ = plt.ylabel('fertility')
# Show the plot
plt.show()
# Show the Pearson correlation coefficient
print(pearson_r(illiteracy, fertility))
Linear regression
We will assume that fertility is a linear function of the female illiteracy rate. That is, f=ai+b, where a is the slope and b is the intercept. We can think of the intercept as the minimal fertility rate, probably somewhere between one and two. The slope tells us how the fertility rate varies with illiteracy. We can find the best fit line using np.polyfit(). Plot the data and the best fit line. Print out the slope and intercept. (Think: what are their units?)
# Plot the illiteracy rate versus fertility
_ = plt.plot(illiteracy, fertility, marker='.', linestyle='none')
plt.margins(0.02)
_ = plt.xlabel('percent illiterate')
_ = plt.ylabel('fertility')
# Perform a linear regression using np.polyfit(): a, b
a, b = np.polyfit(illiteracy,fertility,1)
# Print the results to the screen
print('slope =', a, 'children per woman / percent illiterate')
print('intercept =', b, 'children per woman')
# Make theoretical line to plot
x = np.array([0,100])
y = a * x + b
# Add regression line to your plot
_ = plt.plot(x,y )
# Draw the plot
plt.show()
How is it optimal?
The function np.polyfit() that we used to get our regression parameters finds the optimal slope and intercept. It is optimizing the sum of the squares of the residuals, also known as RSS (for residual sum of squares). In this exercise, we will plot the function that is being optimized, the RSS, versus the slope parameter a. To do this, fix the intercept to be what we found in the optimization. Then, plot the RSS vs. the slope. Where is it minimal?
# Specify slopes to consider: a_vals
a_vals = np.linspace(0,0.1,200)
# Initialize sum of square of residuals: rss
rss = np.empty_like(a_vals)
# Compute sum of square of residuals for each value of a_vals
for i, a in enumerate(a_vals):
rss[i] = np.sum((fertility - a * illiteracy - b)**2)
# Plot the RSS
plt.plot(a_vals,rss, '-')
plt.xlabel('slope (children per woman / percent illiterate)')
plt.ylabel('sum of square of residuals')
plt.show()
Linear regression on appropriate Anscombe data
For practice, perform a linear regression on the data set from Anscombe’s quartet that is most reasonably interpreted with linear regression.
# Perform linear regression: a, b
a, b = np.polyfit(x,y,1)
# Print the slope and intercept
print(a, b)
# Generate theoretical x and y data: x_theor, y_theor
x_theor = np.array([3, 15])
y_theor = a * x_theor + b
# Plot the Anscombe data and theoretical line
_ = plt.plot(x,y,marker='.',linestyle='none')
_ = plt.plot(x_theor,y_theor,marker='.',linestyle='none')
# Label the axes
plt.xlabel('x')
plt.ylabel('y')
# Show the plot
plt.show()
Linear regression on all Anscombe data
Now, to verify that all four of the Anscombe data sets have the same slope and intercept from a linear regression, we will compute the slope and intercept for each set. The data are stored in lists; anscombe_x = [x1, x2, x3, x4] and anscombe_y = [y1, y2, y3, y4], where, for example, x2 and y2 are the x and y values for the second Anscombe data set.
# Iterate through x,y pairs
for x, y in zip(anscombe_x, anscombe_y):
# Compute the slope and intercept: a, b
a, b = np.polyfit(x,y,1)
# Print the result
print('slope:', a, 'intercept:', b)
Getting tripped up over terminology is a common cause of frustration in students. Unfortunately, you often will read and hear other data scientists using different terminology for bootstrap samples and replicates. This is even more reason why we need everything to be clear and consistent for this course. So, before going forward discussing bootstrapping, let’s get our terminology down. If we have a data set with n repeated measurements, a bootstrap sample is an array of length n that was drawn from the original data with replacement, bootstrap replicate is a single value of a statistic computed from a bootstrap sample.
Visualizing bootstrap samples
We will generate bootstrap samples from the set of annual rainfall data measured at the Sheffield Weather Station in the UK from 1883 to 2015. The data are stored in the NumPy array rainfall in units of millimeters (mm). By graphically displaying the bootstrap samples with an ECDF, we can get a feel for how bootstrap sampling allows probabilistic descriptions of data.
for i in range(50):
# Generate bootstrap sample: bs_sample
bs_sample = np.random.choice(rainfall, size=len(rainfall))
# Compute and plot ECDF from bootstrap sample
x, y = ecdf(bs_sample)
_ = plt.plot(x, y, marker='.', linestyle='none',
color='gray', alpha=0.1)
# Compute and plot ECDF from original data
x, y = ecdf(rainfall)
_ = plt.plot(x, y, marker='.')
# Make margins and label axes
plt.margins(0.02)
_ = plt.xlabel('yearly rainfall (mm)')
_ = plt.ylabel('ECDF')
# Show the plot
plt.show()
Generating many bootstrap replicates
Now we’ll write a function, draw_bs_reps(data, func, size=1), which generates many bootstrap replicates from the data set. This function will come in handy for us again and again as we compute confidence intervals and later when we do hypothesis tests. For your reference, the bootstrap_replicate_1d() function is provided below:
def bootstrap_replicate_1d(data, func):
return func(np.random.choice(data, size=len(data)))
def draw_bs_reps(data, func, size=1):
"""Draw bootstrap replicates."""
# Initialize array of replicates: bs_replicates
bs_replicates = np.empty(size)
# Generate replicates
for i in range(size):
bs_replicates[i] = bootstrap_replicate_1d(data,func)
return bs_replicates
Bootstrap replicates of the mean and the SEM
We will compute a bootstrap estimate of the probability density function of the mean annual rainfall at the Sheffield Weather Station. Remember, we are estimating the mean annual rainfall we would get if the Sheffield Weather Station could repeat all of the measurements from 1883 to 2015 over and over again. This is a probabilistic estimate of the mean. You will plot the PDF as a histogram, and you will see that it is Normal.
In fact, it can be shown theoretically that under not-too-restrictive conditions, the value of the mean will always be Normally distributed. (This does not hold in general, just for the mean and a few other statistics.) The standard deviation of this distribution, called the standard error of the mean, or SEM, is given by the standard deviation of the data divided by the square root of the number of data points. I.e., for a data set, sem = np.std(data) / np.sqrt(len(data)). Using hacker statistics, we get this same result without the need to derive it, but we will verify this result from your bootstrap replicates.
# Take 10,000 bootstrap replicates of the mean: bs_replicates
bs_replicates = draw_bs_reps(rainfall,np.mean,size=10000)
# Compute and print SEM
sem = np.std(rainfall) / np.sqrt(len(rainfall))
print(sem)
# Compute and print standard deviation of bootstrap replicates
bs_std = np.std(bs_replicates)
print(bs_std)
# Make a histogram of the results
_ = plt.hist(bs_replicates, bins=50, normed=True)
_ = plt.xlabel('mean annual rainfall (mm)')
_ = plt.ylabel('PDF')
# Show the plot
plt.show()
Bootstrap replicates of other statistics
We saw in a previous exercise that the mean is Normally distributed. This does not necessarily hold for other statistics, but no worry: as hackers, we can always take bootstrap replicates! In this exercise, we’ll generate bootstrap replicates for the variance of the annual rainfall at the Sheffield Weather Station and plot the histogram of the replicates. Here, we will make use of the draw_bs_reps() function you defined a few exercises ago. It is provided below for your reference:
def draw_bs_reps(data, func, size=1):
return np.array([bootstrap_replicate_1d(data, func) for _ in range(size)])
# Generate 10,000 bootstrap replicates of the variance: bs_replicates
bs_replicates = draw_bs_reps(rainfall,np.var,size=10000)
# Put the variance in units of square centimeters
bs_replicates=bs_replicates/100
# Make a histogram of the results
_ = plt.hist(bs_replicates, bins=50,normed=True)
_ = plt.xlabel('variance of annual rainfall (sq. cm)')
_ = plt.ylabel('PDF')
# Show the plot
plt.show()
Confidence interval on the rate of no-hitters
Consider again the inter-no-hitter intervals for the modern era of baseball. Generate 10,000 bootstrap replicates of the optimal parameter τ. Plot a histogram of your replicates and report a 95% confidence interval.
# Draw bootstrap replicates of the mean no-hitter time (equal to tau): bs_replicates
bs_replicates = draw_bs_reps(nohitter_times,np.mean,size=10000)
# Compute the 95% confidence interval: conf_int
conf_int = np.percentile(bs_replicates,[2.5,97.5])
# Print the confidence interval
print('95% confidence interval =', conf_int, 'games')
# Plot the histogram of the replicates
_ = plt.hist(bs_replicates, bins=50, normed=True)
_ = plt.xlabel(r'$\tau$ (games)')
_ = plt.ylabel('PDF')
# Show the plot
plt.show()
A function to do pairs bootstrap
Pairs bootstrap involves resampling pairs of data. Each collection of pairs fit with a line, in this case using np.polyfit(). We do this again and again, getting bootstrap replicates of the parameter values. To have a useful tool for doing pairs bootstrap, we will write a function to perform pairs bootstrap on a set of x,y data.
def draw_bs_pairs_linreg(x, y, size=1):
"""Perform pairs bootstrap for linear regression."""
# Set up array of indices to sample from: inds
inds = np.arange(len(x))
# Initialize replicates: bs_slope_reps, bs_intercept_reps
bs_slope_reps = np.empty(size)
bs_intercept_reps = np.empty(size)
# Generate replicates
for i in range(size):
bs_inds = np.random.choice(inds, size=len(inds))
bs_x, bs_y = x[bs_inds], y[bs_inds]
bs_slope_reps[i], bs_intercept_reps[i] = np.polyfit(bs_x,bs_y,1)
return bs_slope_reps, bs_intercept_reps
Pairs bootstrap of literacy/fertility data
Using the function we just wrote, perform pairs bootstrap to plot a histogram describing the estimate of the slope from the illiteracy/fertility data. Also report the 95% confidence interval of the slope. The data is available to us in the NumPy arrays illiteracy and fertility. As a reminder, draw_bs_pairs_linreg() has a function signature of draw_bs_pairs_linreg(x, y, size=1), and it returns two values: bs_slope_reps and bs_intercept_reps.
# Generate replicates of slope and intercept using pairs bootstrap
bs_slope_reps, bs_intercept_reps = draw_bs_pairs_linreg(illiteracy,fertility,size=1000)
# Compute and print 95% CI for slope
print(np.percentile(bs_slope_reps, [2.5,97.5]))
# Plot the histogram
_ = plt.hist(bs_slope_reps, bins=50, normed=True)
_ = plt.xlabel('slope')
_ = plt.ylabel('PDF')
plt.show()
Plotting bootstrap regressions
A nice way to visualize the variability we might expect in a linear regression is to plot the line we would get from each bootstrap replicate of the slope and intercept. Do this for the first 100 of your bootstrap replicates of the slope and intercept (stored as bs_slope_reps and bs_intercept_reps).
# Generate array of x-values for bootstrap lines: x
x = np.array([0,100])
# Plot the bootstrap lines
for i in range(100):
_ = plt.plot(x, bs_slope_reps[i]*x +bs_intercept_reps[i],
linewidth=0.5, alpha=0.2, color='red')
# Plot the data
_ = plt.plot(illiteracy,fertility,marker='.',linestyle='none')
# Label axes, set the margins, and show the plot
_ = plt.xlabel('illiteracy')
_ = plt.ylabel('fertility')
plt.margins(0.02)
plt.show()
Generating a permutation sample
We learned that permutation sampling is a great way to simulate the hypothesis that two variables have identical probability distributions. This is often a hypothesis you want to test, so in this exercise, we will write a function to generate a permutation sample from two data sets. Remember, a permutation sample of two arrays having respectively n1 and n2 entries is constructed by concatenating the arrays together, scrambling the contents of the concatenated array, and then taking the first n1 entries as the permutation sample of the first array and the last n2 entries as the permutation sample of the second array.
def permutation_sample(data1, data2):
"""Generate a permutation sample from two data sets."""
# Concatenate the data sets: data
data = np.concatenate((data1,data2))
# Permute the concatenated array: permuted_data
permuted_data = np.random.permutation(data)
# Split the permuted array into two: perm_sample_1, perm_sample_2
perm_sample_1 = permuted_data[:len(data1)]
perm_sample_2 = permuted_data[len(data1):]
return perm_sample_1, perm_sample_2
Visualizing permutation sampling
To help see how permutation sampling works, we will generate permutation samples and look at them graphically. We will use the Sheffield Weather Station data again, this time considering the monthly rainfall in July (a dry month) and November (a wet month). We expect these might be differently distributed, so we will take permutation samples to see how their ECDFs would look if they were identically distributed.
The data are stored in the Numpy arrays rain_june and rain_november.
As a reminder, permutation_sample() has a function signature of permutation_sample(data_1, data_2) with a return value of permuted_data[:len(data_1)], permuted_data[len(data_1):], where permuted_data = np.random.permutation(np.concatenate((data_1, data_2)))
for i in range(50):
# Generate permutation samples
perm_sample_1, perm_sample_2 = permutation_sample(rain_june,rain_november)
# Compute ECDFs
x_1, y_1 = ecdf(perm_sample_1)
x_2, y_2 = ecdf(perm_sample_2)
# Plot ECDFs of permutation sample
_ = plt.plot(x_1, y_1, marker='.', linestyle='none',
color='red', alpha=0.02)
_ = plt.plot(x_2, y_2, marker='.', linestyle='none',
color='blue', alpha=0.02)
# Create and plot ECDFs from original data
x_1, y_1 = ecdf(rain_june)
x_2, y_2 = ecdf(rain_november)
_ = plt.plot(x_1, y_1, marker='.', linestyle='none', color='red')
_ = plt.plot(x_2, y_2, marker='.', linestyle='none', color='blue')
# Label axes, set margin, and show plot
plt.margins(0.02)
_ = plt.xlabel('monthly rainfall (mm)')
_ = plt.ylabel('ECDF')
plt.show()
Generating permutation replicates
A permutation replicate is a single value of a statistic computed from a permutation sample. As the draw_bs_reps() function we wrote is useful for you to generate bootstrap replicates, it is useful to have a similar function, draw_perm_reps(), to generate permutation replicates. The function has call signature draw_perm_reps(data_1, data_2, func, size=1). Importantly, func must be a function that takes two arrays as arguments. In most circumstances, func will be a function you write yourself.
def draw_perm_reps(data_1, data_2, func, size=1):
"""Generate multiple permutation replicates."""
# Initialize array of replicates: perm_replicates
perm_replicates = np.empty(size)
for i in range(size):
# Generate permutation sample
perm_sample_1, perm_sample_2 = permutation_sample(data_1,data_2)
# Compute the test statistic
perm_replicates[i] = func(perm_sample_1,perm_sample_2)
return perm_replicates
Look before you leap: EDA before hypothesis testing
Kleinteich and Gorb (Sci. Rep., 4, 5225, 2014) performed an interesting experiment with South American horned frogs. They held a plate connected to a force transducer, along with a bait fly, in front of them. They then measured the impact force and adhesive force of the frog’s tongue when it struck the target.
Frog A is an adult and Frog B is a juvenile. The researchers measured the impact force of 20 strikes for each frog. In the next exercise, we will test the hypothesis that the two frogs have the same distribution of impact forces. But, remember, it is important to do EDA first! Let’s make a bee swarm plot for the data. They are stored in a Pandas data frame, df, where column ID is the identity of the frog and column impact_force is the impact force in Newtons (N).
# Make bee swarm plot
_ = sns.swarmplot(x='ID',y='impact_force',data=df)
# Label axes
_ = plt.xlabel('frog')
_ = plt.ylabel('impact force (N)')
# Show the plot
plt.show()
Permutation test on frog data
The average strike force of Frog A was 0.71 Newtons (N), and that of Frog B was 0.42 N for a difference of 0.29 N. It is possible the frogs strike with the same force and this observed difference was by chance. We will compute the probability of getting at least a 0.29 N difference in mean strike force under the hypothesis that the distributions of strike forces for the two frogs are identical. We use a permutation test with a test statistic of the difference of means to test this hypothesis.
def diff_of_means(data_1, data_2):
"""Difference in means of two arrays."""
# The difference of means of data_1, data_2: diff
diff = np.mean(data_1)-np.mean(data_2)
return diff
# Compute difference of mean impact force from experiment: empirical_diff_means
empirical_diff_means = diff_of_means(force_a,force_b)
# Draw 10,000 permutation replicates: perm_replicates
perm_replicates = draw_perm_reps(force_a,force_b,
diff_of_means, size=10000)
# Compute p-value: p
p = np.sum(perm_replicates >= empirical_diff_means) / len(perm_replicates)
# Print the result
print('p-value =', p)
A one-sample bootstrap hypothesis test
Another juvenile frog was studied, Frog C, and we want to see if Frog B and Frog C have similar impact forces. Unfortunately, we do not have Frog C’s impact forces available, but yoweu know they have a mean of 0.55 N. Because we don’t have the original data, we cannot do a permutation test, and we cannot assess the hypothesis that the forces from Frog B and Frog C come from the same distribution. We will therefore test another, less restrictive hypothesis: The mean strike force of Frog B is equal to that of Frog C.
To set up the bootstrap hypothesis test, we will take the mean as our test statistic. Remember, your goal is to calculate the probability of getting a mean impact force less than or equal to what was observed for Frog B if the hypothesis that the true mean of Frog B’s impact forces is equal to that of Frog C is true. We first translate all of the data of Frog B such that the mean is 0.55 N. This involves adding the mean force of Frog C and subtracting the mean force of Frog B from each measurement of Frog B. This leaves other properties of Frog B’s distribution, such as the variance, unchanged.
# Make an array of translated impact forces: translated_force_b
translated_force_b = force_b - np.mean(force_b) + 0.55
# Take bootstrap replicates of Frog B's translated impact forces: bs_replicates
bs_replicates = draw_bs_reps(translated_force_b,np.mean , 10000)
# Compute fraction of replicates that are less than the observed Frog B force: p
p = np.sum(bs_replicates <= np.mean(force_b)) / 10000
# Print the p-value
print('p = ', p)
A two-sample bootstrap hypothesis test for difference of means
We now want to test the hypothesis that Frog A and Frog B have the same mean impact force, but not necessarily the same distribution, which is also impossible with a permutation test. To do the two-sample bootstrap test, we shift both arrays to have the same mean, since we are simulating the hypothesis that their means are, in fact, equal. We then draw bootstrap samples out of the shifted arrays and compute the difference in means. This constitutes a bootstrap replicate, and we generate many of them. The p-value is the fraction of replicates with a difference in means greater than or equal to what was observed.
# Compute mean of all forces: mean_force
mean_force = np.mean(forces_concat)
# Generate shifted arrays
force_a_shifted = force_a - np.mean(force_a) + mean_force
force_b_shifted = force_b - np.mean(force_b) + mean_force
# Compute 10,000 bootstrap replicates from shifted arrays
bs_replicates_a = draw_bs_reps(force_a_shifted, np.mean, size=10000)
bs_replicates_b = draw_bs_reps(force_b_shifted, np.mean, size=10000)
# Get replicates of difference of means: bs_replicates
bs_replicates = bs_replicates_a- bs_replicates_b
# Compute and print p-value: p
p = np.sum(bs_replicates>=empirical_diff_means) / len(bs_replicates)
print('p-value =', p)
The vote for the Civil Rights Act in 1964
The Civil Rights Act of 1964 was one of the most important pieces of legislation ever passed in the USA. Excluding “present” and “abstain” votes, 153 House Democrats and 136 Republicans voted yea. However, 91 Democrats and 35 Republicans voted nay. Did party affiliation make a difference in the vote?
To answer this question, we will evaluate the hypothesis that the party of a House member has no bearing on his or her vote. We will use the fraction of Democrats voting in favor as your test statistic and evaluate the probability of observing a fraction of Democrats voting in favor at least as small as the observed fraction of 153/244. (That’s right, at least as small as. In 1964, it was the Democrats who were less progressive on civil rights issues.) To do this, permute the party labels of the House voters and then arbitrarily divide them into “Democrats” and “Republicans” and compute the fraction of Democrats voting yea.
# Construct arrays of data: dems, reps
dems = np.array([True] * 153 + [False] * 91)
reps = np.array([True] * 136 + [False] * 35)
def frac_yea_dems(dems, reps):
"""Compute fraction of Democrat yea votes."""
frac = (np.sum(dems)/len(dems)) /(np.sum(reps)/len(reps))
return frac
# Acquire permutation samples: perm_replicates
perm_replicates = draw_perm_reps(dems, reps, frac_yea_dems, size=10000)
# Compute and print p-value: p
p = np.sum(perm_replicates <= 153/244) / len(perm_replicates)
print('p-value =', p)
A time-on-website analog
It turns out that we already did a hypothesis test analogous to an A/B test where we are interested in how much time is spent on the website before and after an ad campaign. The frog tongue force (a continuous quantity like time on the website) is an analog. “Before” = Frog A and “after” = Frog B. Let’s practice this again with something that actually is a before/after scenario.
We return to the no-hitter data set. In 1920, Major League Baseball implemented important rule changes that ended the so-called dead ball era. Importantly, the pitcher was no longer allowed to spit on or scuff the ball, an activity that greatly favors pitchers. In this problem you will perform an A/B test to determine if these rule changes resulted in a slower rate of no-hitters (i.e., longer average time between no-hitters) using the difference in mean inter-no-hitter time as your test statistic. The inter-no-hitter times for the respective eras are stored in the arrays nht_dead and nht_live, where “nht” is meant to stand for “no-hitter time.”
Since you will be using your draw_perm_reps() function in this exercise, it may be useful to remind yourself of its call signature: draw_perm_reps(d1, d2, func, size=1) or even referring back to the exercise in which you defined it.
# Compute the observed difference in mean inter-no-hitter times: nht_diff_obs
nht_diff_obs = diff_of_means(nht_dead,nht_live)
# Acquire 10,000 permutation replicates of difference in mean no-hitter time: perm_replicates
perm_replicates = draw_perm_reps(nht_dead,nht_live,diff_of_means,size=10000)
# Compute and print the p-value: p
p = np.sum(perm_replicates<=nht_diff_obs)/len(perm_replicates)
print('p-val =',p)
Hypothesis test on Pearson correlation
The observed correlation between female illiteracy and fertility may just be by chance; the fertility of a given country may actually be totally independent of its illiteracy. We will test this hypothesis. To do so, permute the illiteracy values but leave the fertility values fixed. This simulates the hypothesis that they are totally independent of each other. For each permutation, compute the Pearson correlation coefficient and assess how many of your permutation replicates have a Pearson correlation coefficient greater than the observed one.
# Compute observed correlation: r_obs
r_obs = pearson_r(illiteracy,fertility)
# Initialize permutation replicates: perm_replicates
perm_replicates = np.empty(10000)
# Draw replicates
for i in range(10000):
# Permute illiteracy measurments: illiteracy_permuted
illiteracy_permuted = np.random.permutation(illiteracy)
# Compute Pearson correlation
perm_replicates[i] = pearson_r(illiteracy_permuted,fertility)
# Compute p-value: p
p = np.sum(perm_replicates>=r_obs)/len(perm_replicates)
print('p-val =', p)
Do neonicotinoid insecticides have unintended consequences?
As a final exercise in hypothesis testing before we put everything together in our case study in the next chapter, we will investigate the effects of neonicotinoid insecticides on bee reproduction. These insecticides are very widely used in the United States to combat aphids and other pests that damage plants.
In a recent study, Straub, et al. (Proc. Roy. Soc. B, 2016) investigated the effects of neonicotinoids on the sperm of pollinating bees. We will study how the pesticide treatment affected the count of live sperm per half milliliter of semen. First, we will do EDA, as usual. Plot ECDFs of the alive sperm count for untreated bees (stored in the Numpy array control) and bees treated with pesticide (stored in the Numpy array treated).
# Compute x,y values for ECDFs
x_control, y_control = ecdf(control)
x_treated, y_treated = ecdf(treated)
# Plot the ECDFs
plt.plot(x_control, y_control, marker='.', linestyle='none')
plt.plot(x_treated, y_treated, marker='.', linestyle='none')
# Set the margins
plt.margins(0.02)
# Add a legend
plt.legend(('control', 'treated'), loc='lower right')
# Label axes and show plot
plt.xlabel('millions of alive sperm per mL')
plt.ylabel('ECDF')
plt.show()
Bootstrap hypothesis test on bee sperm counts
Now, we will test the following hypothesis: On average, male bees treated with neonicotinoid insecticide have the same number of active sperm per milliliter of semen than do untreated male bees. We will use the difference of means as our test statistic.
# Compute the difference in mean sperm count: diff_means
diff_means = np.mean(control) - np.mean(treated)
# Compute mean of pooled data: mean_count
mean_count = np.mean(np.concatenate((control,treated)))
# Generate shifted data sets
control_shifted = control - np.mean(control) + mean_count
treated_shifted = treated - np.mean(treated) + mean_count
# Generate bootstrap replicates
bs_reps_control = draw_bs_reps(control_shifted,
np.mean, size=10000)
bs_reps_treated = draw_bs_reps(treated_shifted,
np.mean, size=10000)
# Get replicates of difference of means: bs_replicates
bs_replicates = bs_reps_control - bs_reps_treated
# Compute and print p-value: p
p = np.sum(bs_replicates >= np.mean(control) - np.mean(treated)) \
/ len(bs_replicates)
print('p-value =', p)
EDA of beak depths of Darwin’s finches
For our first foray into the Darwin finch data, we will study how the beak depth (the distance, top to bottom, of a closed beak) of the finch species Geospiza scandens has changed over time. The Grants have noticed some changes of beak geometry depending on the types of seeds available on the island, and they also noticed that there was some interbreeding with another major species on Daphne Major, Geospiza fortis. These effects can lead to changes in the species over time.
In the next few problems, we will look at the beak depth of G. scandens on Daphne Major in 1975 and in 2012. To start with, let’s plot all of the beak depth measurements in 1975 and 2012 in a bee swarm plot.
The data are stored in a pandas DataFrame called df with columns ‘year’ and ‘beak_depth’. The units of beak depth are millimeters (mm).
# Create bee swarm plot
_ = sns.swarmplot(x='year',y='beak_depth',data=df)
# Label the axes
_ = plt.xlabel('year')
_ = plt.ylabel('beak depth (mm)')
# Show the plot
plt.show()
ECDFs of beak depths
While bee swarm plots are useful, we found that ECDFs are often even better when doing EDA. Plot the ECDFs for the 1975 and 2012 beak depth measurements on the same plot.
# Compute ECDFs
x_1975, y_1975 = ecdf(bd_1975)
x_2012, y_2012 = ecdf(bd_2012)
# Plot the ECDFs
_ = plt.plot(x_1975, y_1975, marker='.', linestyle='none')
_ = plt.plot(x_2012, y_2012, marker='.', linestyle='none')
# Set margins
plt.margins(0.02)
# Add axis labels and legend
_ = plt.xlabel('beak depth (mm)')
_ = plt.ylabel('ECDF')
_ = plt.legend(('1975', '2012'), loc='lower right')
# Show the plot
plt.show()
Parameter estimates of beak depths
Estimate the difference of the mean beak depth of the G. scandens samples from 1975 and 2012 and report a 95% confidence interval.
# Compute the difference of the sample means: mean_diff
mean_diff = np.mean(bd_2012)-np.mean(bd_1975)
# Get bootstrap replicates of means
bs_replicates_1975 = draw_bs_reps(bd_1975,np.mean,10000)
bs_replicates_2012 = draw_bs_reps(bd_2012,np.mean,10000)
# Compute samples of difference of means: bs_diff_replicates
bs_diff_replicates = bs_replicates_2012-bs_replicates_1975
# Compute 95% confidence interval: conf_int
conf_int = np.percentile(bs_diff_replicates,[2.5,97.5])
# Print the results
print('difference of means =', mean_diff, 'mm')
print('95% confidence interval =', conf_int, 'mm')
Hypothesis test: Are beaks deeper in 2012?
Our plot of the ECDF and determination of the confidence interval make it pretty clear that the beaks of G. scandens on Daphne Major have gotten deeper. But is it possible that this effect is just due to random chance? In other words, what is the probability that we would get the observed difference in mean beak depth if the means were the same?
Be careful! The hypothesis we are testing is not that the beak depths come from the same distribution. For that we could use a permutation test. The hypothesis is that the means are equal. To perform this hypothesis test, we need to shift the two data sets so that they have the same mean and then use bootstrap sampling to compute the difference of means.
# Compute mean of combined data set: combined_mean
combined_mean = np.mean(np.concatenate((bd_1975, bd_2012)))
# Shift the samples
bd_1975_shifted = bd_1975-np.mean(bd_1975)+combined_mean
bd_2012_shifted = bd_2012-np.mean(bd_2012)+combined_mean
# Get bootstrap replicates of shifted data sets
bs_replicates_1975 = draw_bs_reps(bd_1975_shifted,np.mean,10000)
bs_replicates_2012 = draw_bs_reps(bd_2012_shifted,np.mean,10000)
# Compute replicates of difference of means: bs_diff_replicates
bs_diff_replicates = bs_replicates_2012-bs_replicates_1975
# Compute the p-value
p = np.sum(bs_diff_replicates >= mean_diff) / len(bs_diff_replicates)
# Print p-value
print('p =', p)
EDA of beak length and depth
The beak length data are stored as bl_1975 and bl_2012, again with units of millimeters (mm). We still have the beak depth data stored in bd_1975 and bd_2012. Make scatter plots of beak depth (y-axis) versus beak length (x-axis) for the 1975 and 2012 specimens.
# Make scatter plot of 1975 data
_ = plt.plot(bl_1975, bd_1975, marker='.',
linestyle='none', color='blue',alpha=0.5)
# Make scatter plot of 2012 data
_ = plt.plot(bl_2012, bd_2012, marker='.',
linestyle='none',color='red',alpha=0.5)
# Label axes and make legend
_ = plt.xlabel('beak length (mm)')
_ = plt.ylabel('beak depth (mm)')
_ = plt.legend(('1975', '2012'), loc='upper left')
# Show the plot
plt.show()
Linear regressions
Perform a linear regression for both the 1975 and 2012 data. Then, perform pairs bootstrap estimates for the regression parameters. Report 95% confidence intervals on the slope and intercept of the regression line. As a reminder, its call signature is draw_bs_pairs_linreg(x, y, size=1), and it returns bs_slope_reps and bs_intercept_reps.
# Compute the linear regressions
slope_1975, intercept_1975 = np.polyfit(bl_1975, bd_1975, 1)
slope_2012, intercept_2012 = np.polyfit(bl_2012, bd_2012, 1)
# Perform pairs bootstrap for the linear regressions
bs_slope_reps_1975, bs_intercept_reps_1975 = \
draw_bs_pairs_linreg(bl_1975, bd_1975, size=1000)
bs_slope_reps_2012, bs_intercept_reps_2012 = \
draw_bs_pairs_linreg(bl_2012, bd_2012, size=1000)
# Compute confidence intervals of slopes
slope_conf_int_1975 = np.percentile(bs_slope_reps_1975,[2.5,97.5])
slope_conf_int_2012 = np.percentile(bs_slope_reps_2012,[2.5,97.5])
intercept_conf_int_1975 = np.percentile(bs_intercept_reps_1975,[2.5,97.5])
intercept_conf_int_2012 = np.percentile(bs_intercept_reps_2012,[2.5,97.5])
# Print the results
print('1975: slope =', slope_1975,
'conf int =', slope_conf_int_1975)
print('1975: intercept =', intercept_1975,
'conf int =', intercept_conf_int_1975)
print('2012: slope =', slope_2012,
'conf int =', slope_conf_int_2012)
print('2012: intercept =', intercept_2012,
'conf int =', intercept_conf_int_2012)
Displaying the linear regression results
Now, we will display our linear regression results on the scatter plot, the code for which is already pre-written for you from our previous exercise. To do this, take the first 100 bootstrap samples (stored in bs_slope_reps_1975, bs_intercept_reps_1975, bs_slope_reps_2012, and bs_intercept_reps_2012) and plot the lines with alpha=0.2 and linewidth=0.5 keyword arguments to plt.plot().
# Make scatter plot of 1975 data
_ = plt.plot(bl_1975, bd_1975, marker='.',
linestyle='none', color='blue', alpha=0.5)
# Make scatter plot of 2012 data
_ = plt.plot(bl_2012, bd_2012, marker='.',
linestyle='none', color='red', alpha=0.5)
# Label axes and make legend
_ = plt.xlabel('beak length (mm)')
_ = plt.ylabel('beak depth (mm)')
_ = plt.legend(('1975', '2012'), loc='upper left')
# Generate x-values for bootstrap lines: x
x = np.array([10, 17])
# Plot the bootstrap lines
for i in range(100):
plt.plot(x,bs_slope_reps_1975[i] * x + bs_intercept_reps_1975[i],
linewidth=0.5, alpha=0.2, color='blue')
plt.plot(x,bs_slope_reps_2012[i] * x + bs_intercept_reps_2012[i],
linewidth=0.5, alpha=0.2, color='red')
# Draw the plot again
plt.show()
Beak length to depth ratio
The linear regressions showed interesting information about the beak geometry. The slope was the same in 1975 and 2012, suggesting that for every millimeter gained in beak length, the birds gained about half a millimeter in depth in both years. However, if we are interested in the shape of the beak, we want to compare the ratio of beak length to beak depth. Let’s make that comparison.
# Compute length-to-depth ratios
ratio_1975 = bl_1975/bd_1975
ratio_2012 = bl_2012/bd_2012
# Compute means
mean_ratio_1975 = np.mean(ratio_1975)
mean_ratio_2012 = np.mean(ratio_2012)
# Generate bootstrap replicates of the means
bs_replicates_1975 = draw_bs_reps(ratio_1975,np.mean,10000)
bs_replicates_2012 = draw_bs_reps(ratio_2012,np.mean,10000)
# Compute the 99% confidence intervals
conf_int_1975 = np.percentile(bs_replicates_1975,[0.5, 99.5])
conf_int_2012 = np.percentile(bs_replicates_2012,[0.5, 99.5])
# Print the results
print('1975: mean ratio =', mean_ratio_1975,
'conf int =', conf_int_1975)
print('2012: mean ratio =', mean_ratio_2012,
'conf int =', conf_int_2012)
EDA of heritability
The array bd_parent_scandens contains the average beak depth (in mm) of two parents of the species G. scandens. The array bd_offspring_scandens contains the average beak depth of the offspring of the respective parents. The arrays bd_parent_fortis and bd_offspring_fortis contain the same information about measurements from G. fortis birds. Make a scatter plot of the average offspring beak depth (y-axis) versus average parental beak depth (x-axis) for both species. Use the alpha=0.5 keyword argument to help you see overlapping points.
# Make scatter plots
_ = plt.plot(bd_parent_fortis,bd_offspring_fortis,
marker='.', linestyle='none', color='blue', alpha=0.5)
_ = plt.plot(bd_parent_scandens, bd_offspring_scandens,
marker='.', linestyle='none', color='red', alpha=0.5)
# Label axes
_ = plt.xlabel('parental beak depth (mm)')
_ = plt.ylabel('offspring beak depth (mm)')
# Add legend
_ = plt.legend(('G. fortis', 'G. scandens'), loc='lower right')
# Show plot
plt.show()
Correlation of offspring and parental data
In an effort to quantify the correlation between offspring and parent beak depths, we would like to compute statistics, such as the Pearson correlation coefficient, between parents and offspring. To get confidence intervals on this, we need to do a pairs bootstrap. We have already written a function to do pairs bootstrap to get estimates for parameters derived from linear regression. Our task in this exercise is to make a new function with call signature draw_bs_pairs(x, y, func, size=1) that performs pairs bootstrap and computes a single statistic on pairs samples defined. The statistic of interested in computed by calling func(bs_x, bs_y). In the next exercise, we will use pearson_r for func.
def draw_bs_pairs(x, y,func, size=1):
"""Perform pairs bootstrap for linear regression."""
# Set up array of indices to sample from: inds
inds = np.arange(len(x))
# Initialize replicates
bs_slope_reps = np.empty(size)
bs_intercept_reps = np.empty(size)
# Generate replicates
for i in range(size):
bs_inds = np.random.choice(inds, size=len(inds))
bs_x, bs_y =x[bs_inds], y[bs_inds]
bs_replicates[i] = func(bs_x,bs_y)
return bs_replicates
Pearson correlation of offspring and parental data
The Pearson correlation coefficient seems like a useful measure of how strongly the beak depth of parents are inherited by their offspring. Compute the Pearson correlation coefficient between parental and offspring beak depths for G. scandens. Do the same for G. fortis. Then, use the function you wrote in the last exercise to compute a 95% confidence interval using pairs bootstrap.
# Compute the Pearson correlation coefficients
r_scandens= pearson_r(bd_parent_scandens,bd_offspring_scandens)
r_fortis = pearson_r(bd_parent_fortis,bd_offspring_fortis)
# Acquire 1000 bootstrap replicates of Pearson r
bs_replicates_scandens = draw_bs_pairs(bd_parent_scandens,bd_offspring_scandens,pearson_r,1000)
bs_replicates_fortis = draw_bs_pairs(bd_parent_fortis,bd_offspring_fortis,pearson_r,1000)
# Compute 95% confidence intervals
conf_int_scandens = np.percentile(bs_replicates_scandens,[2.5,97.5])
conf_int_fortis = np.percentile(bs_replicates_fortis,[2.5,97.5])
# Print results
print('G. scandens:', r_scandens, conf_int_scandens)
print('G. fortis:', r_fortis, conf_int_fortis)
Measuring heritability
Remember that the Pearson correlation coefficient is the ratio of the covariance to the geometric mean of the variances of the two data sets. This is a measure of the correlation between parents and offspring, but might not be the best estimate of heritability. If we stop and think, it makes more sense to define heritability as the ratio of the covariance between parent and offspring to the variance of the parents alone. We will estimate the heritability and perform a pairs bootstrap calculation to get the 95% confidence interval.
This exercise highlights a very important point. Statistical inference (and data analysis in general) is not a plug-n-chug enterprise. We need to think carefully about the questions you are seeking to answer with our data and analyze them appropriately. If you are interested in how heritable traits are, the quantity we defined as the heritability is more apt than the off-the-shelf statistic, the Pearson correlation coefficient.
def heritability(parents, offspring):
"""Compute the heritability from parent and offspring samples."""
covariance_matrix = np.cov(parents, offspring)
return covariance_matrix[0,1] / covariance_matrix[0,0]
# Compute the heritability
heritability_scandens = heritability(bd_parent_scandens,bd_offspring_scandens)
heritability_fortis = heritability(bd_parent_fortis,bd_offspring_fortis)
# Acquire 1000 bootstrap replicates of heritability
replicates_scandens = draw_bs_pairs(
bd_parent_scandens,bd_offspring_scandens, heritability, size=1000)
replicates_fortis = draw_bs_pairs(
bd_parent_fortis,bd_offspring_fortis, heritability, size=1000)
# Compute 95% confidence intervals
conf_int_scandens = np.percentile(replicates_scandens,[2.5,97.5])
conf_int_fortis = np.percentile(replicates_fortis,[2.5,97.5])
# Print results
print('G. scandens:', heritability_scandens, conf_int_scandens)
print('G. fortis:', heritability_fortis, conf_int_fortis)
Is beak depth heritable at all in G. scandens?
The heritability of beak depth in G. scandens seems low. It could be that this observed heritability was just achieved by chance and beak depth is actually not really heritable in the species. We will test that hypothesis here. To do this, we will do a pairs permutation test.
# Initialize array of replicates: perm_replicates
perm_replicates = np.empty(10000)
# Draw replicates
for i in range(10000):
# Permute parent beak depths
bd_parent_permuted = np.random.permutation(bd_parent_scandens)
perm_replicates[i] = heritability(bd_parent_permuted,bd_offspring_scandens)
# Compute p-value: p
p = np.sum(perm_replicates >= heritability_scandens) / len(perm_replicates)
# Print the p-value
print('p-val =', p)