★모평균에 대한 가설 검증★모분산 모를땐 t-분포★신뢰구간 구하기★기초통계학-[연습문제 01 - 10]

1. 어느 회사에서 제조되는 1.5V 소형 건전지의 평균 수명을 알아보기 위하여 15개를 임의로 조사한 결과, 평균 71.5시간 , 표준편차 3.8시간으로 측정되었다. 이 회사에서 제조되는 소형 건전지의 평균 수명에 대한 95% 신뢰구간을 구하라.

 

|X = 71.5

s = 3.8

n = 15

 

==> 모분산 모른다 ==> t-분포 활용

 

X = np.arange(-5,5 , .01)

fig = plt.figure(figsize=(15,8))




# A = [3.1 , 1.9 , 2.4 , 2.8 , 2.9 , 3.0 , 2.8 , 2.3, 2.2 , 2.6]

MEANS = 71.5
STDS = 3.8




n = 15 #표본개수
dof_2 = [n-1] #자유도c

ax = sns.lineplot(x = X , y=scipy.stats.t(dof_2).pdf(X) )
trust = 95 #신뢰도
trust = round( (1- trust/100)/2 , 4)
t_r =  scipy.stats.t(dof_2).ppf(1- trust)
print(t_r)
t_l = scipy.stats.t(dof_2).ppf(trust)
print(t_l)

E = round(float(t_r * STDS / math.sqrt(n)),4)



# =========================================================
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X>=t_r) | (X<=t_l) , facecolor = 'skyblue') # x값 , y값 , 0 , X조건 인곳 , 색깔
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X<t_r) & (X>t_l) , facecolor = 'orange') # x값 , y값 , 0 , X조건 인곳 , 색깔
area = round(float(scipy.stats.t(dof_2).cdf(t_r) - scipy.stats.t(dof_2).cdf(t_l)),4)


plt.annotate('' , xy=(0, .2), xytext=(-2.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(-4.6 , .27, f'평균(MEANS) = {MEANS}\n alpha = {round(1-area,4)}\n' + r'$t_{\dfrac{\alpha}{2}} = {%.4f}$' % t_r +f'\n n = {n} \n 표준편차(s) = {STDS}\n' +r'오차한계 $e_{95\% } = t_{\dfrac{\alpha}{2}}*\dfrac{s}{\sqrt{n}}$'+f'= {E}',fontsize=15)

plt.annotate('' , xy=(0, .25), xytext=(1.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(1.6 , .25, r'$P(t_{%.3f}<T<t_{%.3f})$' % (trust , 1-trust) + f'= {area}\n' + r'신뢰구간 = (MEANS -$e_{\alpha}$ , MEANS + $e_{\alpha}$)' +f'\n' + r' = $({%.4f} - {%.4f} , {%.4f} + {%.4f})$' % (MEANS, E , MEANS , E)  +f'\n' +r'$ = ({%.4f} , {%.4f})$' % (MEANS-E , MEANS+E)  ,fontsize=15)

ax.vlines(x = t_r ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_r) , colors = 'black')
ax.vlines(x = t_l ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_l) , colors = 'black')

plt.annotate('' , xy=(3.0, .007), xytext=(2.5 , .1)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(-3.0, .007), xytext=(-3.5 , .1)  , arrowprops = dict(facecolor = 'black'))

ax.text(1.71 , .13, r'$P(T>t_{%.3f})$' % trust + f'= {round(float(1- scipy.stats.t(dof_2).cdf(t_r)),3)}',fontsize=15)
ax.text(-3.71 , .13, r'$P(T<t_{%.3f})$' % trust + f'= {round(float(scipy.stats.t(dof_2).cdf(t_l)),3)}',fontsize=15)

ax.text(t_r - 1 , 0.02 , r'$t_r$' + f'= {t_r}'  , fontsize = 13)
ax.text(t_l + .2 , 0.02 , r'$t_l$' + f'= {t_l}'  , fontsize = 13)





b = ['t-(n={})'.format(i) for i in dof_2]
plt.legend(b , fontsize = 15)

t-분포 신뢰구간

신뢰구간 : (69.3956 , 73.6044)

 

2. 어느 회사에서는 직원들의 후생 복지를 지원하기 위하여 먼저 직원들이 여가 시간에 자기 계발을 위하여 하루 동안 투자하는 시간을 조사하였고, 그 결과는 다음과 같았다. 물음에 답하라.

A = "40 30 70 60 50 60 60 30 40 50 90 60 50 30 30"
A = list(map(int , A.split(' ')))

A = [40, 30, 70, 60, 50, 60, 60, 30, 40, 50, 90, 60, 50, 30, 30]

 

1> 전 직원이 자기 계발을 위하여 투자하는 평균 시간에 대한 95% 신뢰구간을 구하라.

 

X = np.arange(-5,5 , .01)

fig = plt.figure(figsize=(15,8))




A = "40 30 70 60 50 60 60 30 40 50 90 60 50 30 30"
A = list(map(int , A.split(' ')))
print(A)
MEANS = np.mean(A)
STDS = np.std(A , ddof=1)




n = len(A) #표본개수
dof_2 = [n-1] #자유도c

ax = sns.lineplot(x = X , y=scipy.stats.t(dof_2).pdf(X) )
trust = 95 #신뢰도
trust = round( (1- trust/100)/2 , 4)
t_r =  scipy.stats.t(dof_2).ppf(1- trust)
print(t_r)
t_l = scipy.stats.t(dof_2).ppf(trust)
print(t_l)

E = round(float(t_r * STDS / math.sqrt(n)),4)



# =========================================================
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X>=t_r) | (X<=t_l) , facecolor = 'skyblue') # x값 , y값 , 0 , X조건 인곳 , 색깔
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X<t_r) & (X>t_l) , facecolor = 'orange') # x값 , y값 , 0 , X조건 인곳 , 색깔
area = round(float(scipy.stats.t(dof_2).cdf(t_r) - scipy.stats.t(dof_2).cdf(t_l)),4)


plt.annotate('' , xy=(0, .2), xytext=(-2.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(-4.6 , .27, f'평균(MEANS) = {MEANS}\n alpha = {round(1-area,4)}\n' + r'$t_{\dfrac{\alpha}{2}} = {%.4f}$' % t_r +f'\n n = {n} \n 표준편차(s) = {STDS}\n' +r'오차한계 $e_{95\% } = t_{\dfrac{\alpha}{2}}*\dfrac{s}{\sqrt{n}}$'+f'= {E}',fontsize=15)

plt.annotate('' , xy=(0, .25), xytext=(1.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(1.6 , .25, r'$P(t_{%.3f}<T<t_{%.3f})$' % (trust , 1-trust) + f'= {area}\n' + r'신뢰구간 = (MEANS -$e_{\alpha}$ , MEANS + $e_{\alpha}$)' +f'\n' + r' = $({%.4f} - {%.4f} , {%.4f} + {%.4f})$' % (MEANS, E , MEANS , E)  +f'\n' +r'$ = ({%.4f} , {%.4f})$' % (MEANS-E , MEANS+E)  ,fontsize=15)

ax.vlines(x = t_r ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_r) , colors = 'black')
ax.vlines(x = t_l ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_l) , colors = 'black')

plt.annotate('' , xy=(3.0, .007), xytext=(2.5 , .1)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(-3.0, .007), xytext=(-3.5 , .1)  , arrowprops = dict(facecolor = 'black'))

ax.text(1.71 , .13, r'$P(T>t_{%.3f})$' % trust + f'= {round(float(1- scipy.stats.t(dof_2).cdf(t_r)),3)}',fontsize=15)
ax.text(-3.71 , .13, r'$P(T<t_{%.3f})$' % trust + f'= {round(float(scipy.stats.t(dof_2).cdf(t_l)),3)}',fontsize=15)

ax.text(t_r - 1 , 0.02 , r'$t_r$' + f'= {t_r}'  , fontsize = 13)
ax.text(t_l + .2 , 0.02 , r'$t_l$' + f'= {t_l}'  , fontsize = 13)





b = ['t-(n={})'.format(i) for i in dof_2]
plt.legend(b , fontsize = 15)

t-분포 신뢰구간

 

2> 직원들의 자기 계발을 위한 평균 투자 시간이 1시간에 미달하는지 유의수준 5%에서 조사하라.

 

H_0 : m>= 1 (하단측 검정)

 

X = np.arange(-5,5 , .01)

fig = plt.figure(figsize=(15,8))


#
# A = "1073 1067 1103 1122 1057 1096 1057 1053 1089 1102 1100 1091 1053 1138 1063 1120 1077 1091"
# A = list(map(int, A.split(' ')))


A = "40 30 70 60 50 60 60 30 40 50 90 60 50 30 30"
A = list(map(int , A.split(' ')))
print(A)
MEANS = np.mean(A)
STDS = np.std(A , ddof=1)




n = len(A) #표본개수
dof_2 = [n-1] #자유도c

MO_MEAN = 60

ax = sns.lineplot(x = X , y=scipy.stats.t(dof_2).pdf(X) )
trust = 95 #신뢰도
trust = round( (1- trust/100) , 4)
t_r =  scipy.stats.t(dof_2).ppf(1- trust)
print(t_r)
t_l = scipy.stats.t(dof_2).ppf(trust)
print(t_l)

E = round(float(t_r * STDS / math.sqrt(n)),4) #오차한계


ax.set_title('하단측검정' , fontsize = 15)
# =========================================================

ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X>=-t_r)  , facecolor = 'orange') # x값 , y값 , 0 , X조건 인곳 , 색깔
area = round(float(1- scipy.stats.t(dof_2).cdf(t_l)),4)


plt.annotate('' , xy=(0, .2), xytext=(-2.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(-4.6 , .32, f'평균(MEANS) = {MEANS}\n'  +f' n = {n} \n 표준편차(s) = {STDS}\n' +r'오차한계 $e_{%d} = t_{{\alpha}}*\dfrac{s}{\sqrt{n}}$' % ((1-  trust)*100 ) +f'= {E}' ,fontsize=15)

plt.annotate('' , xy=(0, .25), xytext=(1.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(1.6 , .25, r'$P(t_{%.3f}<T)$' % (trust) + f'= {area}\n' + r'신뢰구간 = (MEANS -$e_{\alpha} , \infty$)' +f'\n' + r' = $({%.4f} - {%.4f} , \infty)$' % (MEANS, E)  +f'\n' +r'$ = ({%.4f} , \infty)$' % (MEANS-E)  ,fontsize=15)

# ax.vlines(x = t_r ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_r) , colors = 'black')
ax.vlines(x = t_l ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_l) , colors = 'black')




# plt.annotate('' , xy=(t_r, .007), xytext=(2.5 , .1)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(t_l, .007), xytext=(-3.5 , .1)  , arrowprops = dict(facecolor = 'black'))
# ax.text(1.71 , .13, r'$t_{\dfrac{\alpha}{2}} = {%.4f}$' % t_r + '\n' +r'$\dfrac{\alpha}{2}$ =' + f'{round(float(1- scipy.stats.t(dof_2).cdf(t_r)),3)}',fontsize=15)
ax.text(-3.71 , .13, r'$t_{{\alpha}} = {%.4f}$' % t_l + '\n' +r'${\alpha}$ =' +f'{round(float(scipy.stats.t(dof_2).cdf(t_l)),3)}',fontsize=15)

# ax.text(t_r - 1 , 0.02 , r'$t_r$' + f'= {t_r}'  , fontsize = 13)
ax.text(t_l + .2 , 0.02 , r'$t_l$' + f'= {t_l}'  , fontsize = 13)




#==================================== 가설 검정 ==========================================



t_1 = round((MEANS - MO_MEAN)/ (STDS / math.sqrt(n)),4)

print(t_1)
t_1 = abs(t_1)
area = round(float(scipy.stats.t(dof_2).cdf(-t_1) ),4)
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where =  (X<=-t_1) , facecolor = 'skyblue') # x값 , y값 , 0 , X조건 인곳 , 색깔
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X<=t_l) , facecolor = 'red') # x값 , y값 , 0 , X조건 인곳 , 색깔

# ax.vlines(x= t_1, ymin= 0 , ymax= stats.t(dof_2).pdf(t_1) , color = 'green' , linestyle ='solid' , label ='{}'.format(2))
ax.vlines(x= -t_1, ymin= 0 , ymax= stats.t(dof_2).pdf(-t_1) , color = 'green' , linestyle ='solid' , label ='{}'.format(2))

annotate_len = stats.t(dof_2).pdf(t_1) /2
# plt.annotate('' , xy=(t_1, annotate_len), xytext=(-t_1/2 , annotate_len)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(-t_1, annotate_len), xytext=(t_1/2 , annotate_len)  , arrowprops = dict(facecolor = 'black'))
ax.text(0.7, annotate_len+0.03 , f'P-value : \nP(T<={-t_1}) \n = {area}',fontsize=15)

mo = '모평균'

ax.text(-4.6 , .22, r'T = $\dfrac{\overline{X} - {\mu}}{\dfrac{s}{\sqrt{n}}}$' + f'= { round((MEANS - MO_MEAN)/(STDS / math.sqrt(n)),4) }' ,fontsize=15)






b = ['t-(n={})'.format(i) for i in dof_2]
plt.legend(b , fontsize = 15)

t-분포 하단측검정

 

H_0 : 평균시간 >= 60(하단측 검정)

 

p-value : 0.0211

 

alpha = 0.05

 

p-value < alpha ==> 0.0211 < 0.05 ==> H_0: 평균시간 >= 1 기각 ==> 평균시간은 1시간에 미달한다.

 

 

3. 건강에 관심이 많은 어느 사회단체는 건강한 성인이 하루에 소비하는 물의 양은 2L 이상이라고 하였다. 이것을 확인하기 위하여 12명의 건강한 성인을 임의로 선정하여 하루에 소비하는 물의 양을 다음과 같이 조사하였다. 건강한 성인이 하루에 평균 2L 이상 소비하는 지 유의수준 1%에서 조사하라.

 

H_0 : m>= 2 (하단측 검정)

 

X = np.arange(-5,5 , .01)

fig = plt.figure(figsize=(15,8))


#
# A = "1073 1067 1103 1122 1057 1096 1057 1053 1089 1102 1100 1091 1053 1138 1063 1120 1077 1091"
# A = list(map(int, A.split(' ')))


A = "2.1 2.2 1.5 1.7 2.0 1.6 1.7 1.5 2.4 1.6 2.5 1.9"
A = list(map(float , A.split(' ')))
print(A)
MEANS = np.mean(A)
STDS = np.std(A , ddof=1)




n = len(A) #표본개수
dof_2 = [n-1] #자유도c

MO_MEAN = 2

ax = sns.lineplot(x = X , y=scipy.stats.t(dof_2).pdf(X) )
trust = 99 #신뢰도
trust = round( (1- trust/100) , 4)
t_r =  scipy.stats.t(dof_2).ppf(1- trust)
print(t_r)
t_l = scipy.stats.t(dof_2).ppf(trust)
print(t_l)

E = round(float(t_r * STDS / math.sqrt(n)),4) #오차한계


ax.set_title('하단측검정' , fontsize = 15)
# =========================================================

ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X>=-t_r)  , facecolor = 'orange') # x값 , y값 , 0 , X조건 인곳 , 색깔
area = round(float(1- scipy.stats.t(dof_2).cdf(t_l)),4)


plt.annotate('' , xy=(0, .2), xytext=(-2.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(-4.6 , .32, f'평균(MEANS) = {MEANS}\n'  +f' n = {n} \n 표준편차(s) = {STDS}\n' +r'오차한계 $e_{%d} = t_{{\alpha}}*\dfrac{s}{\sqrt{n}}$' % ((1-  trust)*100 ) +f'= {E}' ,fontsize=15)

plt.annotate('' , xy=(0, .25), xytext=(1.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(1.6 , .25, r'$P(t_{%.3f}<T)$' % (trust) + f'= {area}\n' + r'신뢰구간 = (MEANS -$e_{\alpha} , \infty}$)' +f'\n' + r' = $({%.4f} - {%.4f}, \infty)$' % (MEANS, E )  +f'\n' +r'$ = ({%.4f} , \infty)$' % (MEANS-E )  ,fontsize=15)

# ax.vlines(x = t_r ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_r) , colors = 'black')
ax.vlines(x = t_l ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_l) , colors = 'black')




# plt.annotate('' , xy=(t_r, .007), xytext=(2.5 , .1)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(t_l, .007), xytext=(-3.5 , .1)  , arrowprops = dict(facecolor = 'black'))
# ax.text(1.71 , .13, r'$t_{\dfrac{\alpha}{2}} = {%.4f}$' % t_r + '\n' +r'$\dfrac{\alpha}{2}$ =' + f'{round(float(1- scipy.stats.t(dof_2).cdf(t_r)),3)}',fontsize=15)
ax.text(-3.71 , .13, r'$t_{{\alpha}} = {%.4f}$' % t_l + '\n' +r'${\alpha}$ =' +f'{round(float(scipy.stats.t(dof_2).cdf(t_l)),3)}',fontsize=15)

# ax.text(t_r - 1 , 0.02 , r'$t_r$' + f'= {t_r}'  , fontsize = 13)
ax.text(t_l + .2 , 0.02 , r'$t_l$' + f'= {t_l}'  , fontsize = 13)




#==================================== 가설 검정 ==========================================



t_1 = round((MEANS - MO_MEAN)/ (STDS / math.sqrt(n)),4)

print(t_1)
t_1 = abs(t_1)
area = round(float(scipy.stats.t(dof_2).cdf(-t_1) ),4)
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where =  (X<=-t_1) , facecolor = 'skyblue') # x값 , y값 , 0 , X조건 인곳 , 색깔
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X<=t_l) , facecolor = 'red') # x값 , y값 , 0 , X조건 인곳 , 색깔

# ax.vlines(x= t_1, ymin= 0 , ymax= stats.t(dof_2).pdf(t_1) , color = 'green' , linestyle ='solid' , label ='{}'.format(2))
ax.vlines(x= -t_1, ymin= 0 , ymax= stats.t(dof_2).pdf(-t_1) , color = 'green' , linestyle ='solid' , label ='{}'.format(2))

annotate_len = stats.t(dof_2).pdf(t_1) /2
# plt.annotate('' , xy=(t_1, annotate_len), xytext=(-t_1/2 , annotate_len)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(-t_1, annotate_len), xytext=(t_1/2 , annotate_len)  , arrowprops = dict(facecolor = 'black'))
ax.text(0.7, annotate_len+0.03 , f'P-value : \nP(T<={-t_1}) \n = {area}',fontsize=15)

mo = '모평균'

ax.text(-4.6 , .22, r'T = $\dfrac{\overline{X} - {\mu}}{\dfrac{s}{\sqrt{n}}}$' + f'= { round((MEANS - MO_MEAN)/(STDS / math.sqrt(n)),4) }' ,fontsize=15)






b = ['t-(n={})'.format(i) for i in dof_2]
plt.legend(b , fontsize = 15)

t-분포 하단측검정

H_0 : m>= 2 (하단측 검정)

 

p-value : 0.1517

 

alpha = 0.01

 

p-value > alpha ==> 0.1517 > 0.01 ==> H_0: m>= 2 채택 ==> 유의수준 1%에서 건강한 성인이 하루에 평균 2L이상 소비한다.

 

 

4. 정규모집단의 모평균을 알아보기 위하여 크기 10인 표본을 조사하여 |x = 24.04 , s = 1.2를 얻었다.

 

n = 10

|x = 24.04

s = 1.2

 

==> 모분산 모른다! ==> t-분포 활용

 

1> 신뢰도 95%인 모평균에 대한 신뢰구간

 

X = np.arange(-5,5 , .01)

fig = plt.figure(figsize=(15,8))



# A = "1073 1067 1103 1122 1057 1096 1057 1053 1089 1102 1100 1091 1053 1138 1063 1120 1077 1091"
# A = list(map(int, A.split(' ')))


MEANS = 24.04
STDS = 1.2
MO_MEAN = 25



n = 10 #표본개수
dof_2 = [n-1] #자유도c

ax = sns.lineplot(x = X , y=scipy.stats.t(dof_2).pdf(X) )
trust = 95 #신뢰도
trust = round( (1- trust/100)/2 , 4)
t_r =  scipy.stats.t(dof_2).ppf(1- trust)
print(t_r)
t_l = scipy.stats.t(dof_2).ppf(trust)
print(t_l)

E = round(float(t_r * STDS / math.sqrt(n)),4)



# =========================================================

ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X<t_r) & (X>t_l) , facecolor = 'orange') # x값 , y값 , 0 , X조건 인곳 , 색깔
area = round(float(scipy.stats.t(dof_2).cdf(t_r) - scipy.stats.t(dof_2).cdf(t_l)),4)


plt.annotate('' , xy=(0, .2), xytext=(-2.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(-4.6 , .32, f'평균(MEANS) = {MEANS}\n'  +f' n = {n} \n 표준편차(s) = {STDS}\n' +r'오차한계 $e_{95\% } = t_{\dfrac{\alpha}{2}}*\dfrac{s}{\sqrt{n}}$'+f'= {E}',fontsize=15)

plt.annotate('' , xy=(0, .25), xytext=(1.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(1.6 , .25, r'$P(t_{%.3f}<T<t_{%.3f})$' % (trust , 1-trust) + f'= {area}\n' + r'신뢰구간 = (MEANS -$e_{\alpha}$ , MEANS + $e_{\alpha}$)' +f'\n' + r' = $({%.4f} - {%.4f} , {%.4f} + {%.4f})$' % (MEANS, E , MEANS , E)  +f'\n' +r'$ = ({%.4f} , {%.4f})$' % (MEANS-E , MEANS+E)  ,fontsize=15)

ax.vlines(x = t_r ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_r) , colors = 'black')
ax.vlines(x = t_l ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_l) , colors = 'black')




plt.annotate('' , xy=(t_r, .007), xytext=(2.5 , .1)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(t_l, .007), xytext=(-3.5 , .1)  , arrowprops = dict(facecolor = 'black'))
ax.text(1.71 , .13, r'$t_{\dfrac{\alpha}{2}} = {%.4f}$' % t_r + '\n' +r'$\dfrac{\alpha}{2}$ =' + f'{round(float(1- scipy.stats.t(dof_2).cdf(t_r)),3)}',fontsize=15)
ax.text(-3.71 , .13, r'$t_{\dfrac{\alpha}{2}} = {%.4f}$' % t_l + '\n' +r'$\dfrac{\alpha}{2}$ =' +f'{round(float(scipy.stats.t(dof_2).cdf(t_l)),3)}',fontsize=15)

ax.text(t_r - 1 , 0.02 , r'$t_r$' + f'= {t_r}'  , fontsize = 13)
ax.text(t_l + .2 , 0.02 , r'$t_l$' + f'= {t_l}'  , fontsize = 13)




#==================================== 가설 검정 ==========================================



t_1 = round((MEANS - MO_MEAN)/ (STDS / math.sqrt(n)),4)

print(t_1)
t_1 = abs(t_1)
area = round(float(scipy.stats.t(dof_2).cdf(-t_1) + 1 - (scipy.stats.t(dof_2).cdf(t_1))),4)
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X>=t_1) | (X<=-t_1) , facecolor = 'skyblue') # x값 , y값 , 0 , X조건 인곳 , 색깔
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X>=t_r) | (X<=-t_r) , facecolor = 'red') # x값 , y값 , 0 , X조건 인곳 , 색깔

ax.vlines(x= t_1, ymin= 0 , ymax= stats.t(dof_2).pdf(t_1) , color = 'green' , linestyle ='solid' , label ='{}'.format(2))
ax.vlines(x= -t_1, ymin= 0 , ymax= stats.t(dof_2).pdf(-t_1) , color = 'green' , linestyle ='solid' , label ='{}'.format(2))

annotate_len = stats.t(dof_2).pdf(t_1) /2
plt.annotate('' , xy=(t_1, annotate_len), xytext=(-t_1/2 , annotate_len)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(-t_1, annotate_len), xytext=(t_1/2 , annotate_len)  , arrowprops = dict(facecolor = 'black'))
ax.text(-1.5 , annotate_len+0.03 , f'P-value : \nP(T<={-t_1}) + P(T>={t_1}) \n = {area}',fontsize=15)

mo = '모평균'

ax.text(-4.6 , .22, r'T = $\dfrac{\overline{X} - {\mu}}{\dfrac{s}{\sqrt{n}}}$' + f'= { round((MEANS - MO_MEAN)/(STDS / math.sqrt(n)),4) }' ,fontsize=15)






b = ['t-(n={})'.format(i) for i in dof_2]
plt.legend(b , fontsize = 15)

t-분포 신뢰구간

신뢰구간 : (23.1816 , 24.8984)

 

2> 유의수준 alpha = 0.01에서 H_0 : m = 25를 조사하라.(양측검정)

 

X = np.arange(-5,5 , .01)

fig = plt.figure(figsize=(15,8))



# A = "1073 1067 1103 1122 1057 1096 1057 1053 1089 1102 1100 1091 1053 1138 1063 1120 1077 1091"
# A = list(map(int, A.split(' ')))


MEANS = 24.04
STDS = 1.2
MO_MEAN = 25



n = 10 #표본개수
dof_2 = [n-1] #자유도c

ax = sns.lineplot(x = X , y=scipy.stats.t(dof_2).pdf(X) )
trust = 99 #신뢰도
trust = round( (1- trust/100)/2 , 4)
t_r =  scipy.stats.t(dof_2).ppf(1- trust)
print(t_r)
t_l = scipy.stats.t(dof_2).ppf(trust)
print(t_l)

E = round(float(t_r * STDS / math.sqrt(n)),4)



# =========================================================

ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X<t_r) & (X>t_l) , facecolor = 'orange') # x값 , y값 , 0 , X조건 인곳 , 색깔
area = round(float(scipy.stats.t(dof_2).cdf(t_r) - scipy.stats.t(dof_2).cdf(t_l)),4)


plt.annotate('' , xy=(0, .2), xytext=(-2.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(-4.6 , .32, f'평균(MEANS) = {MEANS}\n'  +f' n = {n} \n 표준편차(s) = {STDS}\n' +r'오차한계 $e_{95\% } = t_{\dfrac{\alpha}{2}}*\dfrac{s}{\sqrt{n}}$'+f'= {E}',fontsize=15)

plt.annotate('' , xy=(0, .25), xytext=(1.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(1.6 , .25, r'$P(t_{%.3f}<T<t_{%.3f})$' % (trust , 1-trust) + f'= {area}\n' + r'신뢰구간 = (MEANS -$e_{\alpha}$ , MEANS + $e_{\alpha}$)' +f'\n' + r' = $({%.4f} - {%.4f} , {%.4f} + {%.4f})$' % (MEANS, E , MEANS , E)  +f'\n' +r'$ = ({%.4f} , {%.4f})$' % (MEANS-E , MEANS+E)  ,fontsize=15)

ax.vlines(x = t_r ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_r) , colors = 'black')
ax.vlines(x = t_l ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_l) , colors = 'black')




plt.annotate('' , xy=(t_r, .007), xytext=(2.5 , .1)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(t_l, .007), xytext=(-3.5 , .1)  , arrowprops = dict(facecolor = 'black'))
ax.text(1.71 , .13, r'$t_{\dfrac{\alpha}{2}} = {%.4f}$' % t_r + '\n' +r'$\dfrac{\alpha}{2}$ =' + f'{round(float(1- scipy.stats.t(dof_2).cdf(t_r)),3)}',fontsize=15)
ax.text(-3.71 , .13, r'$t_{\dfrac{\alpha}{2}} = {%.4f}$' % t_l + '\n' +r'$\dfrac{\alpha}{2}$ =' +f'{round(float(scipy.stats.t(dof_2).cdf(t_l)),3)}',fontsize=15)

ax.text(t_r - 1 , 0.02 , r'$t_r$' + f'= {t_r}'  , fontsize = 13)
ax.text(t_l + .2 , 0.02 , r'$t_l$' + f'= {t_l}'  , fontsize = 13)




#==================================== 가설 검정 ==========================================



t_1 = round((MEANS - MO_MEAN)/ (STDS / math.sqrt(n)),4)

print(t_1)
t_1 = abs(t_1)
area = round(float(scipy.stats.t(dof_2).cdf(-t_1) + 1 - (scipy.stats.t(dof_2).cdf(t_1))),4)
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X>=t_1) | (X<=-t_1) , facecolor = 'skyblue') # x값 , y값 , 0 , X조건 인곳 , 색깔
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X>=t_r) | (X<=-t_r) , facecolor = 'red') # x값 , y값 , 0 , X조건 인곳 , 색깔

ax.vlines(x= t_1, ymin= 0 , ymax= stats.t(dof_2).pdf(t_1) , color = 'green' , linestyle ='solid' , label ='{}'.format(2))
ax.vlines(x= -t_1, ymin= 0 , ymax= stats.t(dof_2).pdf(-t_1) , color = 'green' , linestyle ='solid' , label ='{}'.format(2))

annotate_len = stats.t(dof_2).pdf(t_1) /2
plt.annotate('' , xy=(t_1, annotate_len), xytext=(-t_1/2 , annotate_len)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(-t_1, annotate_len), xytext=(t_1/2 , annotate_len)  , arrowprops = dict(facecolor = 'black'))
ax.text(-1.5 , annotate_len+0.03 , f'P-value : \nP(T<={-t_1}) + P(T>={t_1}) \n = {area}',fontsize=15)

mo = '모평균'

ax.text(-4.6 , .22, r'T = $\dfrac{\overline{X} - {\mu}}{\dfrac{s}{\sqrt{n}}}$' + f'= { round((MEANS - MO_MEAN)/(STDS / math.sqrt(n)),4) }' ,fontsize=15)






b = ['t-(n={})'.format(i) for i in dof_2]
plt.legend(b , fontsize = 15)

t-분포 양측검정

H_0 : m= 25 (양측 검정)

 

p-value : 0.0322

 

alpha = 0.01

 

p-value > alpha ==> 0.0322 > 0.01 ==> H_0: m= 25 채택 ==> 유의수준 1%에서 m = 25

 

3> 유의수준 alpha = 0.05에서 H_0 : m = 25를 조사하라.(양측검정)

 

X = np.arange(-5,5 , .01)

fig = plt.figure(figsize=(15,8))



# A = "1073 1067 1103 1122 1057 1096 1057 1053 1089 1102 1100 1091 1053 1138 1063 1120 1077 1091"
# A = list(map(int, A.split(' ')))


MEANS = 24.04
STDS = 1.2
MO_MEAN = 25



n = 10 #표본개수
dof_2 = [n-1] #자유도c

ax = sns.lineplot(x = X , y=scipy.stats.t(dof_2).pdf(X) )
trust = 95 #신뢰도
trust = round( (1- trust/100)/2 , 4)
t_r =  scipy.stats.t(dof_2).ppf(1- trust)
print(t_r)
t_l = scipy.stats.t(dof_2).ppf(trust)
print(t_l)

E = round(float(t_r * STDS / math.sqrt(n)),4)



# =========================================================

ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X<t_r) & (X>t_l) , facecolor = 'orange') # x값 , y값 , 0 , X조건 인곳 , 색깔
area = round(float(scipy.stats.t(dof_2).cdf(t_r) - scipy.stats.t(dof_2).cdf(t_l)),4)


plt.annotate('' , xy=(0, .2), xytext=(-2.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(-4.6 , .32, f'평균(MEANS) = {MEANS}\n'  +f' n = {n} \n 표준편차(s) = {STDS}\n' +r'오차한계 $e_{95\% } = t_{\dfrac{\alpha}{2}}*\dfrac{s}{\sqrt{n}}$'+f'= {E}',fontsize=15)

plt.annotate('' , xy=(0, .25), xytext=(1.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(1.6 , .25, r'$P(t_{%.3f}<T<t_{%.3f})$' % (trust , 1-trust) + f'= {area}\n' + r'신뢰구간 = (MEANS -$e_{\alpha}$ , MEANS + $e_{\alpha}$)' +f'\n' + r' = $({%.4f} - {%.4f} , {%.4f} + {%.4f})$' % (MEANS, E , MEANS , E)  +f'\n' +r'$ = ({%.4f} , {%.4f})$' % (MEANS-E , MEANS+E)  ,fontsize=15)

ax.vlines(x = t_r ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_r) , colors = 'black')
ax.vlines(x = t_l ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_l) , colors = 'black')




plt.annotate('' , xy=(t_r, .007), xytext=(2.5 , .1)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(t_l, .007), xytext=(-3.5 , .1)  , arrowprops = dict(facecolor = 'black'))
ax.text(1.71 , .13, r'$t_{\dfrac{\alpha}{2}} = {%.4f}$' % t_r + '\n' +r'$\dfrac{\alpha}{2}$ =' + f'{round(float(1- scipy.stats.t(dof_2).cdf(t_r)),3)}',fontsize=15)
ax.text(-3.71 , .13, r'$t_{\dfrac{\alpha}{2}} = {%.4f}$' % t_l + '\n' +r'$\dfrac{\alpha}{2}$ =' +f'{round(float(scipy.stats.t(dof_2).cdf(t_l)),3)}',fontsize=15)

ax.text(t_r - 1 , 0.02 , r'$t_r$' + f'= {t_r}'  , fontsize = 13)
ax.text(t_l + .2 , 0.02 , r'$t_l$' + f'= {t_l}'  , fontsize = 13)




#==================================== 가설 검정 ==========================================



t_1 = round((MEANS - MO_MEAN)/ (STDS / math.sqrt(n)),4)

print(t_1)
t_1 = abs(t_1)
area = round(float(scipy.stats.t(dof_2).cdf(-t_1) + 1 - (scipy.stats.t(dof_2).cdf(t_1))),4)
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X>=t_1) | (X<=-t_1) , facecolor = 'skyblue') # x값 , y값 , 0 , X조건 인곳 , 색깔
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X>=t_r) | (X<=-t_r) , facecolor = 'red') # x값 , y값 , 0 , X조건 인곳 , 색깔

ax.vlines(x= t_1, ymin= 0 , ymax= stats.t(dof_2).pdf(t_1) , color = 'green' , linestyle ='solid' , label ='{}'.format(2))
ax.vlines(x= -t_1, ymin= 0 , ymax= stats.t(dof_2).pdf(-t_1) , color = 'green' , linestyle ='solid' , label ='{}'.format(2))

annotate_len = stats.t(dof_2).pdf(t_1) /2
plt.annotate('' , xy=(t_1, annotate_len), xytext=(-t_1/2 , annotate_len)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(-t_1, annotate_len), xytext=(t_1/2 , annotate_len)  , arrowprops = dict(facecolor = 'black'))
ax.text(-1.5 , annotate_len+0.03 , f'P-value : \nP(T<={-t_1}) + P(T>={t_1}) \n = {area}',fontsize=15)

mo = '모평균'

ax.text(-4.6 , .22, r'T = $\dfrac{\overline{X} - {\mu}}{\dfrac{s}{\sqrt{n}}}$' + f'= { round((MEANS - MO_MEAN)/(STDS / math.sqrt(n)),4) }' ,fontsize=15)






b = ['t-(n={})'.format(i) for i in dof_2]
plt.legend(b , fontsize = 15)

t-분포 양측검정

H_0 : m= 25 (양측 검정)

 

p-value : 0.0322

 

alpha = 0.05

 

p-value > alpha ==> 0.0322 < 0.05 ==> H_0: m= 25 기각 ==> 유의수준 1%에서 m = 25의 주장은 기각한다.

 

 

5. 어느 공업 지역 부근을 흐르는 하천 물의 평균 pH농도가 7이라고 한다. 이것을 알아보기 위하여 임의로 21곳의 물을 선정하여 조사한 결과, 평균 7.2 , 표준편차 0.32였다. 이 하천의 pH농도는 정규분포를 따른다고 할 때, 평균 pH농도가 7인지 유의수준 5%에서 조사하라.

 

 

|X = 7.2

s = 0.32

 

H_0 : m = 7 (양측검정)

 

X = np.arange(-5,5 , .01)

fig = plt.figure(figsize=(15,8))



# A = "1073 1067 1103 1122 1057 1096 1057 1053 1089 1102 1100 1091 1053 1138 1063 1120 1077 1091"
# A = list(map(int, A.split(' ')))


MEANS = 7.2
STDS = 0.32
MO_MEAN = 7



n = 21 #표본개수
dof_2 = [n-1] #자유도c

ax = sns.lineplot(x = X , y=scipy.stats.t(dof_2).pdf(X) )
trust = 95 #신뢰도
trust = round( (1- trust/100)/2 , 4)
t_r =  scipy.stats.t(dof_2).ppf(1- trust)
print(t_r)
t_l = scipy.stats.t(dof_2).ppf(trust)
print(t_l)

E = round(float(t_r * STDS / math.sqrt(n)),4)


ax.set_title('양측 검정' , fontsize = 15)
# =========================================================

ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X<t_r) & (X>t_l) , facecolor = 'orange') # x값 , y값 , 0 , X조건 인곳 , 색깔
area = round(float(scipy.stats.t(dof_2).cdf(t_r) - scipy.stats.t(dof_2).cdf(t_l)),4)


plt.annotate('' , xy=(0, .2), xytext=(-2.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(-4.6 , .32, f'평균(MEANS) = {MEANS}\n'  +f' n = {n} \n 표준편차(s) = {STDS}\n' +r'오차한계 $e_{95\% } = t_{\dfrac{\alpha}{2}}*\dfrac{s}{\sqrt{n}}$'+f'= {E}',fontsize=15)

plt.annotate('' , xy=(0, .25), xytext=(1.5 , .25)  , arrowprops = dict(facecolor = 'black'))
ax.text(1.6 , .25, r'$P(t_{%.3f}<T<t_{%.3f})$' % (trust , 1-trust) + f'= {area}\n' + r'신뢰구간 = (MEANS -$e_{\alpha}$ , MEANS + $e_{\alpha}$)' +f'\n' + r' = $({%.4f} - {%.4f} , {%.4f} + {%.4f})$' % (MEANS, E , MEANS , E)  +f'\n' +r'$ = ({%.4f} , {%.4f})$' % (MEANS-E , MEANS+E)  ,fontsize=15)

ax.vlines(x = t_r ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_r) , colors = 'black')
ax.vlines(x = t_l ,ymin=0 , ymax= scipy.stats.t(dof_2).pdf(t_l) , colors = 'black')




plt.annotate('' , xy=(t_r, .007), xytext=(2.5 , .1)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(t_l, .007), xytext=(-3.5 , .1)  , arrowprops = dict(facecolor = 'black'))
ax.text(1.71 , .13, r'$t_{\dfrac{\alpha}{2}} = {%.4f}$' % t_r + '\n' +r'$\dfrac{\alpha}{2}$ =' + f'{round(float(1- scipy.stats.t(dof_2).cdf(t_r)),3)}',fontsize=15)
ax.text(-3.71 , .13, r'$t_{\dfrac{\alpha}{2}} = {%.4f}$' % t_l + '\n' +r'$\dfrac{\alpha}{2}$ =' +f'{round(float(scipy.stats.t(dof_2).cdf(t_l)),3)}',fontsize=15)

ax.text(t_r - 1 , 0.02 , r'$t_r$' + f'= {t_r}'  , fontsize = 13)
ax.text(t_l + .2 , 0.02 , r'$t_l$' + f'= {t_l}'  , fontsize = 13)




#==================================== 가설 검정 ==========================================



t_1 = round((MEANS - MO_MEAN)/ (STDS / math.sqrt(n)),4)

print(t_1)
t_1 = abs(t_1)
area = round(float(scipy.stats.t(dof_2).cdf(-t_1) + 1 - (scipy.stats.t(dof_2).cdf(t_1))),4)
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X>=t_1) | (X<=-t_1) , facecolor = 'skyblue') # x값 , y값 , 0 , X조건 인곳 , 색깔
ax.fill_between(X, scipy.stats.t(dof_2).pdf(X) , 0 , where = (X>=t_r) | (X<=-t_r) , facecolor = 'red') # x값 , y값 , 0 , X조건 인곳 , 색깔

ax.vlines(x= t_1, ymin= 0 , ymax= stats.t(dof_2).pdf(t_1) , color = 'green' , linestyle ='solid' , label ='{}'.format(2))
ax.vlines(x= -t_1, ymin= 0 , ymax= stats.t(dof_2).pdf(-t_1) , color = 'green' , linestyle ='solid' , label ='{}'.format(2))

annotate_len = stats.t(dof_2).pdf(t_1) /2
plt.annotate('' , xy=(t_1, annotate_len), xytext=(-t_1/2 , annotate_len)  , arrowprops = dict(facecolor = 'black'))
plt.annotate('' , xy=(-t_1, annotate_len), xytext=(t_1/2 , annotate_len)  , arrowprops = dict(facecolor = 'black'))
ax.text(-1.5 , annotate_len+0.03 , f'P-value : \nP(T<={-t_1}) + P(T>={t_1}) \n = {area}',fontsize=15)

mo = '모평균'

ax.text(-4.6 , .22, r'T = $\dfrac{\overline{X} - {\mu}}{\dfrac{s}{\sqrt{n}}}$' + f'= { round((MEANS - MO_MEAN)/(STDS / math.sqrt(n)),4) }' ,fontsize=15)






b = ['t-(n={})'.format(i) for i in dof_2]
plt.legend(b , fontsize = 15)

t-분포 양측검정

H_0 : m= 7 (양측 검정)

 

p-value : 0.0096

 

alpha = 0.05

 

p-value < alpha ==> 0.0096 < 0.05 ==> H_0: m= 7 기각 ==> 유의수준 5%에서 pH농도가  7이라는  주장은 기각한다.