l10

from statistics import mean, variance, stdev
from math import sqrt, prod
# mean(list) = média simples
# variance(list) = variância
# stdev(list) = desvio padrão

# 1.
# a.
# 1021, 1016, 1012, 1011, 1014, 1018, 1022, 1027, 1008, 1015, 1013, 1013, 1017, 1019, 1007, 1003
l = [1021, 1016, 1012, 1011, 1014, 1018, 1022, 1027, 1008, 1015, 1013, 1013, 1017, 1019, 1007, 1003]
# Σx = sum(l) = 16236
# n = len(l) = 16
n = len(l)

# µ = 16236/16 = 1014.75 = mean(l)
# σ² = 34.3125 = variance(l)
# σ = sqrt(σ²) = 5.8577 = stdev(l)

"""
Z ~ N(0,1)
Z = media(X) - µ / (σ/sqrt(n))

P(-z_a/2 < Z < z_a/2) = 1 - a
P(-z_a/2 < media(X) - µ / (σ/sqrt(n)) < z_a/2) = 1 - a
P((σ/sqrt(n)) * -z_a/2 < media(X) - µ < (σ/sqrt(n)) * z_a/2) = 1 - a
P(- media(X) - z_a/2 * (σ/sqrt(n)) < -µ < - media(X) + z_a/2 * (σ/sqrt(n))) = 1 - a
P(media(X) + (σ/sqrt(n)) * z_a/2 > µ > media(X) - (σ/sqrt(n)) * z_a/2) = 1 - a
P(media(X) - (σ/sqrt(n)) * z_a/2 < µ < media(X) + (σ/sqrt(n)) * z_a/2) = 1 - a

IC(µ; 1-a): media(X) ± z_a/2 * σ/sqrt(n)
"""
"""
t Student
T = media(X) - µ / (s/sqrt(n))

P(media(X) - (s/sqrt(n)) * t_a/2 < µ < media(X) + (s/sqrt(n)) * t_a/2) = 1 - a

IC(θ; 1−a): media(X) ± t_a/2 * s/sqrt(n)
onde s = desvio padrao da amostra
"""

a = 0.05
# v = n-1 = 15
v = n-1
## Da tabela normal:
## a/2 = 0.025
## Area entre 0 e z_a/2 = 0.5 - 0.025 = 0.475
##area_a = 0.5 - a/2
## quantil de z_a/2 (olhando pela tabela) = z_.975 = 1.96
# limites de t_a/2 (olhando pela tabela) = t_.v(a)
# t_.v005 = 2.132
t_v005 = 2.132

# IC(µ; 0.95):
IC1 = mean(l) + t_v005 * stdev(l) / sqrt(n)
IC2 = mean(l) - t_v005 * stdev(l) / sqrt(n)

# b.

a2 = 0.01
## Da tabela normal:
## a/2 = 0.005
## Area entre 0 e z_a/2 = 0.5 - 0.005 = 0.495
##area_a2 = 0.5 - a2/2
## quantil de z_a/2 = z_.995 = 3.27
# limites de t_a/2 = t_.v(a)
# t_.v001_2 = 2.947
t_v001 = 2.947

# IC(µ; 0.99):
IC1b = mean(l) + t_v001 * stdev(l) / sqrt(len(l))
IC2b = mean(l) - t_v001 * stdev(l) / sqrt(len(l))

print(IC1, IC2, IC1b, IC2b)
# Intervalo normal:
# 1017.620273 1011.879727 1019.53866975 1009.96133025
# Intervalo Student:
# 1017.872147 1011.627853 1019.06565109 1010.43434891

# 2.
# 1011, 1015, 1017, 1015, 1021, 1021, 1010, 1007, 1022, 1018, 1016, 1015, 1020, 1022, 1025, 1030

l2 = [1011, 1015, 1017, 1015, 1021, 1021, 1010, 1007, 1022, 1018, 1016, 1015, 1020, 1022, 1025, 1030]

ns = [len(l), len(l2)]
ls = [l, l2]

# Scombined = ( variance(l) * (len(l) - 1) + variance(l2) * (len(l2) - 1) ) / ( (len(l2) - 1) + (len(l) - 1) )
Scombined = 0
vcombined = 0
for it in ls:
    Scombined += variance(it) * (len(it) - 1)
    vcombined += (len(it) - 1)

if vcombined != 0:
    Scombined /= vcombined

# Sdiff = sqrt(Scombined * (1/len(l) + 1/len(l2))
Sdiff = sqrt(Scombined * sum(ns) / prod(ns))

# print("v = ", vcombined)
# v = 30
t_vcombined005 = 2.042

# IC(µ₁-µ₂; 1-a): media(X₁) - media(X₂) ± t_a/2 * Sdiff
ICdiff1 = mean(l) - mean(l2) + t_vcombined005 * Sdiff
ICdiff2 = mean(l) - mean(l2) - t_vcombined005 * Sdiff

print(ICdiff1, ICdiff2)
# 1.0954854815341086 -7.220485481534109

# Ao nivel de 95%, o intervalo da diferença contém o valor 0. Com isso, o volume médio fornecido pelas duas máquinas não diferem significativamente. Logo, pode-se afirmar que as máquinas fornecem o mesmo volume médio.

# 3.
# As pressupções que devem ser válidas são, as variáveis terem distribuição normal, as variâncias das populações são iguais e as amostras retiradas são independentes.