Which of the following Python lines calculates the 99%

confidence interval for proportion of Exam1 scores with scores

greater than 90 from a CSV file named ExamScores?

Question 4 options:

a)

import pandas as pd

import scipy.stats as st

scores = pd.read_csv('ExamScores.csv')

x = scores[['Exam1']].count()

n = (scores[['Exam1']] > 90).values.sum()

p = x/n*1.0

stderror = (p * (1 - p)/n)**0.5

print(st.norm.interval(0.99, p, stderror))

b)

import pandas as pd

import scipy.stats as st

scores = pd.read_csv('ExamScores.csv')

n = scores[['Exam1']].count()

x = (scores[['Exam1']] >= 90).values.sum()

p = x/n*1.0

stderror = (p * (1 - p)/n)**0.5

print(st.norm.interval(0.99, p, stderror))

c)

import pandas as pd

import scipy.stats as st

scores = pd.read_csv('ExamScores.csv')

n = scores[['Exam1']].count()

x = (scores[['Exam1']] > 90).values.sum()

p = x/n*1.0

stderror = (p * (1 - p)/n)**0.5

print(st.norm.interval(0.99, p, stderror))

d)

import pandas as pd

import scipy.stats as st

scores = pd.read_csv('ExamScores.csv')

n = scores[['Exam1']].count()

x = (scores[['Exam1']] > 90).values.sum()

p = x/n*1.0

stderror = (p * (1 - p)/n)**0.5

print(st.norm.interval(0.90, p, stderror))