# lecture 7, quantitative analysis with Python

from scipy.stats import f_oneway

#enter exam scores for each group
group1 = [85, 86, 88, 75, 78, 94, 98, 79, 71, 80]
group2 = [91, 92, 93, 85, 87, 84, 82, 88, 95, 96]
group3 = [49, 28, 48, 34, 52, 65, 33, 25, 42, 51]

#perform one-way ANOVA
result = f_oneway(group1, group2, group3)

# 0.05 is a common "error margin" or alpha value that is
# considered "statistically significant" in MOST CASES
# if pValue is less than 0.05 => GROUPS ARE SIMILAR
# if pValue is more than 0.05 => GROUPS ARE DIFFERENT
print(result)

# NEW FILE -> ANCOVA testing

import numpy as np
import pandas as pd

from pingouin import ancova


#create data
df = pd.DataFrame({'technique': np.repeat(['Traditional', 'Video lessons', 'Self-study'], 5),
                   'current_grade': [97, 78, 75, 77, 85,
                                     92, 69, 77, 74, 88,
                                     96, 91, 88, 82, 80],
                   'exam_score': [77, 89, 82, 84, 89,
                                  78, 88, 93, 94, 90,
                                  85, 81, 83, 88, 89]})



#perform ANCOVA
result = ancova(data=df, dv='exam_score', covar='current_grade', between='technique')

# p-unc, if value is less than 0.05 => different study techniques affect the grade scores
# otherwise, the study technique does not matter!
print(result)

# NEW FILE -> CHI SQUARE
# TWO DIFFERENT TESTS -> INDEPENDENCE, OR GOODNESS OF FIT

import pandas as pd
import numpy as np
from scipy.stats import chi2_contingency

df = pd.DataFrame({'Gender' : ['M', 'M', 'M', 'F', 'F'] * 10,
                   'isSmoker' : ['Smoker', 'Smoker', 'Non-Smoker', 'Non-Smoker', 'Smoker'] * 10
                  })


contigency = pd.crosstab(df['Gender'], df['isSmoker'])

contigency_pct = pd.crosstab(df['Gender'], df['isSmoker'], normalize='index')


# Chi-square test of independence.
c, p, dof, expected = chi2_contingency(contigency)

# shop owner claims this
expected = [50, 50, 50, 50, 50]

# the researcher counted these
observed = [50, 60, 40, 47, 53]

import scipy.stats as stats

#perform Chi-Square Goodness of Fit Test
result = stats.chisquare(f_obs=observed, f_exp=expected)
print(result)

# NEW FILE
# k-means demo

import matplotlib.pyplot as plt
import numpy as np
from sklearn.cluster import KMeans

X = np.array([[5,3],
     [10,15],
     [15,12],
     [24,10],
     [30,45],
     [35,43],
     [45,42],
     [85,70],
     [71,80],
     [60,78],
     [55,52],
     [80,91],])

plt.clf()
plt.scatter(X[:,0],X[:,1], label='True Position')
plt.figure()

kmeans = KMeans(n_clusters=3)
kmeans.fit(X)

plt.clf()
plt.scatter(X[:,0],X[:,1], c=kmeans.labels_, cmap='rainbow')
plt.figure()

# test if the clusters (groups) work with test values
print(kmeans.predict([[10,10], [42, 45], [82,87]]))

# NEW FILE
# Student's t-test, individual

import pandas as pd
import numpy as np
import random
from numpy.random import seed
from numpy.random import randn
from scipy.stats import ttest_ind
from scipy.stats import t

seed(1)

df=pd.DataFrame({"female":np.random.randint(10, 100, size=10),"male":np.random.randint(10, 140, size=10)})

t_stat, p = ttest_ind(df['male'], df['female'])

# scipy makes a paired t-test by default
# to get the individual t-test p-value:
p = p/2

print(f't={t_stat}, p={p}')