Question 1(3.75 points)

Consider the following Python commands:

import scipy.stats as st

st.norm.interval(0.99, 0.50, 0.05),

What does the 0.99 represent?

Question 1 options:

a)

proportion

b)

standard error

c)

confidence level

d)

mean

Question 2(3.75 points)

Which of the following Python functions is used to calculate a confidence interval based on Student's t-distribution?

Note: st is from the import command import scipy.stats as st

Question 2 options:

a)

st.t.normal

b)

st.t.confidence_interval

c)

st.t.interval

d)

st.norm.confidence_interval

Question 3(3.75 points)

If n = 100 and mean = 219 and sample standard deviation = 35, which of the following Python lines outputs the 95% confidence interval?

Question 3 options:

a)

import scipy.stats as st

n = 100

df = n - 1

mean = 219

stderror = 35.0/(n**0.5)

print(st.t.interval(0.90, df, mean, stderror))

b)

import scipy.stats as st

n = 100

df = n - 1

mean = 219

stderror = 35.0/(n**0.5)

print(st.t.interval(0.95, df, mean, stderror))

c)

import scipy.stats as st

n = 100

df = n - 1

mean = 219

stderror = 0.5/(n**35)

print(st.t.interval(0.95, df, mean, stderror))

d)

import scipy.stats as st

n = 219

df = n - 1

mean = 100

stderror = 35.0/(n**0.5)

print(st.t.interval(0.95, df, mean, stderror))

Question 4(3.75 points)

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))

Question 5(3.75 points)

If n = 99 and proportion (p) = 0.75, which of the following Python lines outputs the 99% confidence interval?

Question 5 options:

a)

import scipy.stats as st

n = 99

p = 0.25

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

print(st.norm.interval(0.99, 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 = 99

p = 0.75

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

print(st.norm.interval(0.98, 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))

Question 6(3.75 points)

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))

Question 7(3.75 points)

Which of the following Python functions is used to calculate a confidence interval based on normal distribution?

Note: st is from the import command import scipy.stats as st

Question 7 options:

a)

st.norm.normal

b)

st.norm.confidence_interval

c)

st.t.confidence_interval

d)

st.norm.interval

Question 8(3.75 points)

Which of the following Python lines calculates the 95% confidence interval for the mean of variable Exam1 from a CSV file named ExamScores?

Question 8 options:

a)

import pandas as pd

import scipy.stats as st

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

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

df = n - 1

mean = scores[['Exam1']].std()

stdev = scores[['Exam1']].mean()

stderror = stdev/(n**0.5)

print(st.t.interval(0.95, df, mean, stderror))

b)

import pandas as pd

import scipy.stats as st

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

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

mean = scores[['Exam1']].mean()

stdev = scores[['Exam1']].std()

stderror = stdev/(n**0.5)

print(st.t.interval(0.95, df, mean, stderror))

c)

import pandas as pd

import scipy.stats as st

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

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

df = n - 1

mean = scores[['Exam1']].mean()

stdev = scores[['Exam1']].std()

stderror = stdev/(n**0.5)

print(st.t.interval(0.99, df, mean, stderror))

d)

import pandas as pd

import scipy.stats as st

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

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

df = n - 1

mean = scores[['Exam1']].mean()

stdev = scores[['Exam1']].std()

stderror = stdev/(n**0.5)

print(st.t.interval(0.95, df, mean, stderror))