Insurance Data

This Jupyter Notebook takes an insurance data set from Kaggle and looks into the relationships between the different parameters given. First, we check for a correlation (a linear relationship) between BMI and insurance charges, and then we see if being a smoker influences what you’re charged by insurance companies using an A/B Test.

import pandas as pd
import numpy as np
import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline
# plt.style.use('ggplot')

# read table
insurance = pd.read_csv('insurance.csv')
insurance.head()
age sex bmi children smoker region charges
0 19 female 27.900 0 yes southwest 16884.92400
1 18 male 33.770 1 no southeast 1725.55230
2 28 male 33.000 3 no southeast 4449.46200
3 33 male 22.705 0 no northwest 21984.47061
4 32 male 28.880 0 no northwest 3866.85520

Data Exploration

The best place to start with any data-oriented project is to figure out how the data look. To this end, we take a look at the distributions of the different data in the insurance DataFrame. The figure created below has a histogram or bar chart for each column to show the counts of the data contained therein.

plt.figure(figsize=[20, 20])

plt.subplot(331)
sns.boxplot(y='age', data=insurance)
plt.title('Boxplot of Ages')
plt.ylabel('Age')

ins_by_sex = insurance.groupby('sex').count()

plt.subplot(332)
sns.barplot(ins_by_sex.index, ins_by_sex['age'])
plt.title('Counts of Sexes')
plt.xlabel('Sex')
plt.ylabel('Count')

plt.subplot(333)
sns.distplot(insurance['bmi'], bins=np.arange(15.5, 53.5, 1))
plt.title('Histogram of BMIs, bin width = 1')
plt.xlabel('BMI')
plt.ylabel('Count')

plt.subplot(334)
sns.distplot(insurance['children'], bins=np.arange(-.5, 6.5, 1), kde=False)
plt.title('Histogram of No. of Children')
plt.xlabel('No. of Children')
plt.ylabel('Count')
plt.xlim([-.5, 5.5])

ins_by_smoker = insurance.groupby('smoker').count()

plt.subplot(335)
sns.barplot(ins_by_smoker.index, ins_by_smoker['age'])
plt.title('Counts of Smokers and Non-Smokers')
plt.xlabel('Smoker?')
plt.ylabel('Count')

ins_by_region = insurance.groupby('region').count()

plt.subplot(336)
sns.barplot(ins_by_region.index, ins_by_region['age'])
plt.title('Counts of Regions')
plt.xlabel('Region')
plt.ylabel('Count')

plt.subplot(338)
sns.distplot(insurance['charges'])
plt.title('Histogram of Charges, bin width = $2,000')
plt.xlabel('Charges ($)')
plt.ylabel('Count')

plt.suptitle('Data Exploration', y=.92, fontsize=24);

png

Is there a correlation between BMI and insurance charges?

Correlation is calculated by taking two data points, putting them in standard units, multiplying the coordinates elementwise, and then finding the mean (all of this is defined in the correlation function below). The value of $r$, heretofore referred to as correlation, ranges from -1 to 1; a value near 1 indicates a positive linear relationship (i.e. a line with a positive slope), near -1 indicates a negative linear relationship, and near 0 indicates little/no linear relationship.

Accompanying the calculation of $r$ is a scatter plot and a joint density plot of the data, with BMI on the $x$-axis and insurance charges on the $y$-axis.

standard_units = lambda x: (x - np.std(x))/np.mean(x)
correlation = lambda x, y: np.mean(standard_units(x) * standard_units(y))
slope = lambda x, y: correlation(x, y) * np.std(y) / np.std(x)
intercept = lambda x, y: np.mean(y) - slope * np.mean(x)
plt.figure(figsize=[12, 5])
plt.suptitle(r'Insurance Charges vs. BMI, $r$ = ' + str(np.round(correlation(insurance['bmi'], insurance['charges']), 3)))
plt.subplots_adjust(top=0.9, wspace=0.3)

plt.subplot(121)
sns.scatterplot('bmi', 'charges', data=insurance)

plt.subplot(122)
sns.kdeplot(insurance['bmi'], insurance['charges']);

png

Based on the fact that the value of $r$ was around 0.1, there might be a small correlation between BMI and insurance charges, but the relationship is not as strong as it could be among other data points. There doesn’t seem to be such a huge correlation between BMI and insurance charges, but it kind of looks like there might be something (very) loosely positive there. In order to check, we see if there is perhaps a more linear relationship between BMI and the insurance charges on a log scale:

insurance['log10_charges'] = np.log10(insurance['charges'])
sns.jointplot('bmi', 'log10_charges', data=insurance, kind='kde')
plt.suptitle(r'Log10 of Insurance Charges vs. BMI, $r$ = ' + str(np.round(correlation(insurance['bmi'], np.log10(insurance['charges'])), 3)))
plt.subplots_adjust(top=0.9);

png

Conclusion

As it turns out, although the correlation is .7, there really doesn’t look to be much there. So while there may be some linear relationship between BMI and insurance charges (or its log), it’s not too apparent.

Does being a smoker affect what you’re charged by the insurance company?

Null Hypothesis: Being a smoker does not affect your charges; any differences in the observed values are due to random chance.

Alternative Hypothesis: Being a smoker does affect what you are charged in insurance premiums.

This question, from a data science perspective, is asking whether or not the charges for the groups smoker and non-smoker come from the same underlying distribution. To find this out, we use an A/B Test, which involves shuffling up the data in question in a sample without replacement, computing a test statistic, and finding the p-value. By convention, if the p-value is less than .05 (meaning less than 5% of the simulated data point in the same direction as the original data set), we lean in the direction of the alternative hypothesis. In this A/B test, the test statistic will be the absolute difference between the mean charges for smokers and non-smokers.

Before doing the actual permutation test, we will run through a single permutation to demonstrate the process that will eventually be done thousands of times to obtain a p-value. The first part of the permutation test is to shuffle up the charges column of the insurance DataFrame and create a new column with the shuffled-up charges. It is from this permuting of the sample that permutation tests get their name.

smoker_and_charges = insurance[['smoker', 'charges']]
charges_shuffled = list(smoker_and_charges.sample(frac=1)['charges'])
shuffled_charges_df = smoker_and_charges.assign(shuffled_charges=charges_shuffled)
shuffled_charges_df.head()
smoker charges shuffled_charges
0 yes 16884.92400 6600.3610
1 no 1725.55230 12231.6136
2 no 4449.46200 3484.3310
3 no 21984.47061 8442.6670
4 no 3866.85520 42983.4585

The next part of the test is to compute the value of the test statistic, which is the absolute difference between the mean charges for smokers and non-smokers. (A high value of this statistic points in the direction of the alternative hypothesis.) To this end, we define the function ts which takes a DataFrame and a column name as its arguments and returns the absolute difference of the mean value of col_name after df has been grouped by the column smoker.

def ts(df, col_name):
    df_grouped = df.groupby('smoker').mean()
    return abs(df_grouped[col_name].iloc[0] - df_grouped[col_name].iloc[1])


# computing the test statistic on the table shuffled_charges_df
test_stat_1 = ts(shuffled_charges_df, 'shuffled_charges')
test_stat_1
668.2917138336725

Finally, we are ready for the permutation test. The function perm_test below taks a DataFrame as its argument and the number of replications, reps, to go through. For each replication, it permutes df as we did above and computes the value of the test statistic, collecting them in the list stats. After collecting these values, it computes the test statistic for the original data, and returns a p-value by taking the percentage of test statistics that are greater than or equal to the observed value.

def perm_test(df, reps):
    stats = []
    for _ in np.arange(reps):
        charges_shuffled = list(df.sample(frac=1)['charges'])
        df = df.assign(shuffled_charges=charges_shuffled)
        stat = ts(df, 'shuffled_charges')
        stats += [stat]
        
    observed_ts = ts(df, 'charges')
    
    return np.count_nonzero(stats >= observed_ts) / len(stats)
# run the permutation test with 100,000 repetitions
perm_test(smoker_and_charges, 100000)
0.0

Conclusion

Because the p-value is 0, we know that none of the shuffled sets were as far or farther in the direction of the alternative hypothesis than was the original data set; this means that in all likelihood, the observed differences are not due to random chance. Thus, we lean in the direction of the alternative hypothesis: that being a smoker affects what you’re charged by insurance companies. Conventional wisdom, I know, but it is still nice to have it proven empirically.