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)	level of significance

b)	proportion

c)	standard error

d)

mean

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Question 2 (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 2 options:

a)	n = 100 df = n - 1 mean = 219 stderror = 0.5/(n**35) print(st.t.interval(0.95, df, mean, stderror))

b)	n = 219 df = n - 1 mean = 100 stderror = 35.0/(n**0.5) print(st.t.interval(0.95, df, mean, stderror))

c)	n = 100 df = n - 1 mean = 219 stderror = 35.0/(n**0.5) print(st.t.interval(0.95, df, mean, stderror))

d)	n = 100 df = n - 1 mean = 219 stderror = 35.0/(n**0.5) print(st.t.interval(0.90, df, mean, stderror))

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Question 3 (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 3 options:

a)	import pandas as pd scores = pd.read_csv('ExamScores.csv') n = scores[['Exam1']].count() x = (scores[['Exam1']] > 90).values.sum() p = x*1.0 stderror = (p * (1 - p))**0.5 print(st.norm.interval(0.90, p, stderror))

b)	import pandas as pd scores = pd.read_csv('ExamScores.csv') n = scores[['Exam1']].count() x = (scores[['Exam1']] > 90).values.sum() p = x*1.0 stderror = (p * (1 - p))**0.5 print(st.norm.interval(0.99, p, stderror))

c)	import pandas as pd scores = pd.read_csv('ExamScores.csv') x = scores[['Exam1']].count() n = (scores[['Exam1']] > 90).values.sum() p = x*1.0 stderror = (p * (1 - p))**0.5 print(st.norm.interval(0.99, p, stderror))

d)	import pandas as pd scores = pd.read_csv('ExamScores.csv') n = scores[['Exam1']].count() x = (scores[['Exam1']] >= 90).values.sum() p = x*1.0 stderror = (p * (1 - p))**0.5 print(st.norm.interval(0.99, p, stderror))

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Question 4 (3.75 points)

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

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

Question 4 options:

a)	st.t.normal

b)	st.t.interval

c)	st.t.confidence_interval

d)	st.norm.confidence_interval

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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 = 0.75 stderror = (p * (1 - p))**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))**0.5 print(st.norm.interval(0.99, p, stderror))

c)	import scipy.stats as st n = 99 p = 0.75 stderror = (p * (1 - p))**0.5 print(st.norm.interval(0.98, p, stderror))

d)	import scipy.stats as st n = 99 p = 0.25 stderror = (p * (1 - p))**0.5 print(st.norm.interval(0.99, p, stderror))

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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 = 0.75 stderror = (p * (1 - p))**0.5 print(st.norm.interval(0.95, p, stderror))

b)	import scipy.stats as st n = 100 p = 0.75 stderror = (p * (1 - p))**0.5 print(st.norm.interval(0.95, p, stderror))

c)	import scipy.stats as st n = 99 p = 0.25 stderror = (p * (1 - p))**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))**0.5 print(st.norm.interval(0.99, p, stderror))

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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.t.confidence_interval

b)	st.norm.confidence_interval

c)	st.norm.normal

d)	st.norm.interval

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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 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))

b)	import pandas as pd 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))

c)	import pandas as pd 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))

d)	import pandas as pd 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))

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