If n = 99 and proportion (p) = 0.75, which of the following

Python lines outputs the 95% confidence interval?

Question 6 options:

a)

import scipy.stats as st

n = 99

p = 0.25

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

print(st.norm.interval(0.95, p, stderror))

b)

import scipy.stats as st

n = 99

p = 0.75

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

print(st.norm.interval(0.95, p, stderror))

c)

import scipy.stats as st

n = 100

p = 0.75

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

print(st.norm.interval(0.95, p, stderror))

d)

import scipy.stats as st

n = 99

p = 0.75

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

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