def gaussian_case(sig):
sample_size = 200
count_error = math.sqrt(sample_size)
m1 = 1
m2 = 2
s1 = generate_sample(sample_size, m1, sig)
s2 = generate_sample(sample_size, m2, sig)
h_testing = Hedges_d(s1, s2)
h_testing.hedges_d_unbiased() #answer is in self.d
approx_95CI_lower, approx_95CI_upper = h_testing.approx_CI()
bs_95CI_lower, bs_95CI_upper = h_testing.bootstrap_CI(5000)
print(mean_SD(s1), mean_SD(s2))
print("h_testing.d, analytic, correction = ", h_testing.d, (m2 - m1) / sig,
h_testing.correction)
print("lower: approx, bootstrap", approx_95CI_lower, bs_95CI_lower)
print("upper: approx, bootstrap", approx_95CI_upper, bs_95CI_upper)
#bootstrap is similar at high d but gives wider intervals at low d
assert close_enough(approx_95CI_lower, bs_95CI_lower, count_error)
assert close_enough(approx_95CI_upper, bs_95CI_upper, count_error)
assert close_enough(h_testing.d, (m2 - m1) / sig, count_error)
def bootstrap_CI(self, repeats, bCA=False):
# repeats is the number of times that it is repeated.
# bCA: bias-corrected and accelerated - recommended to improve pctile coverage
# bCA not implemented yet
random.seed(1984)
hedges_d_bs = [] #unbiased, values from bootstrap
n1m = len(self.s1) - 1 #we only use these below - simplifies
n2m = len(self.s2) - 1
correction = 1.0 - 3.0 / (4 * (n1m + n2m) - 1)
for n in range(repeats):
# bootstrap # RL 31/03/19 removed call to local bootstrap function (redundant)
#br1 = bootstrap(self.s1)
br1 = random.choices(self.s1, k=len(self.s1))
#br2 = bootstrap(self.s2)
br2 = random.choices(self.s2, k=len(self.s2))
#means and SDs of bootstrap sampled distributions
mbr1, sbr1 = mean_SD(br1)
mbr2, sbr2 = mean_SD(br2)
#need to call a function for Hedges_d that returns a value, not storing it
#current alternative, just include here, it is so simple.
s_pooled = math.sqrt((n2m * sbr2**2 + n1m * sbr1**2) / (n1m + n2m))
#Hedges' g
# RL: quick fix (probably not very meaningful) for a case of a sample with two measurments only
if s_pooled == 0.0:
biased_d = 0.0
else:
biased_d = (mbr2 - mbr1) / s_pooled
# correction has no influence on the bootstrap distribution
# so apply it later to save multiplications
hedges_d_bs.append(biased_d)
hedges_d_bs.sort() # put the values into rank order
self.lower95CI = hedges_d_bs[int(0.025 * repeats)] * correction
self.upper95CI = hedges_d_bs[int(0.975 * repeats)] * correction
return (self.lower95CI, self.upper95CI)
def hedges_d_unbiased(self):
#d_unbiased = d_biased [ 1 - 3/ (4 * ( n1 + n2 - 2)-1)]
#from Nakagawa and Cuthill (2007) Biol. Rev.
#count each time to make formulas easier to read
n1 = len(self.s1)
n2 = len(self.s2)
#means and standard deviations
m1, s1 = mean_SD(self.s1)
m2, s2 = mean_SD(self.s2)
#pooled variance
s_pooled = math.sqrt(
((n2 - 1) * s2**2 + (n1 - 1) * s1**2) / (n1 + n2 - 2))
#Hedges' g
biased_d = (m2 - m1) / s_pooled
self.correction = 1.0 - 3.0 / (4 * (n1 + n2 - 2) - 1)
#store answer
self.d = biased_d * self.correction
def bootstrap_CI(self, repeats, bCA=False):
# repeats is the number of times that it is repeated.
# bCA: bias-corrected and accelerated - recommended to improve pctile coverage
# bCA not implemented yet
hedges_d_bs = [] #unbiased, values from bootstrap
n1m = len(self.s1) - 1 #we only use these below - simplifies
n2m = len(self.s2) - 1
correction = 1.0 - 3.0 / (4 * (n1m + n2m) - 1)
for n in range(repeats):
br1 = bootstrap(self.s1)
br2 = bootstrap(self.s2)
#means and SDs of bootstrap sampled distributions
mbr1, sbr1 = mean_SD(br1)
mbr2, sbr2 = mean_SD(br2)
#need to call a function for Hedges_d that returns a value, not storing it
#current alternative, just include here, it is so simple.
s_pooled = math.sqrt((n2m * sbr2**2 + n1m * sbr1**2) / (n1m + n2m))
#Hedges' g
biased_d = (mbr2 - mbr1) / s_pooled
# correction has no influence on the bootstrap distribution
# so apply it later to save multiplications
hedges_d_bs.append(biased_d)
hedges_d_bs.sort() # put the values into rank order
self.lower95CI = hedges_d_bs[int(0.025 * repeats)] * correction
self.upper95CI = hedges_d_bs[int(0.975 * repeats)] * correction
return (self.lower95CI, self.upper95CI)