| title | description |
|---|---|
Two sample hypothesis testing |
Insert the chapter description here |
type: NormalExercise
lang: python
xp: 100
skills: 1
key: 7942f6cb9a@hint
@instructions
A survey of National Federation of Independence Business (NFIB) indicates that small businesses intended to increase their hiring as well as their capital expenditures during 2017 as compared with 2016. Suppose that, as part of a follow-up survey by NFIB, 20 small businesses, randomly chosen from the NFIB's list of 2,100 companies, show an average hiring from 2016 equal to 3.2 new employees per firm and a standard deviation of 1.5 hires. A random sample of 30 small businesses taken at the end of 2017 shows an average of 5.1 new hires and a standard deviation of 2.3 hires. At the
@pre_exercise_code
def pdf_plotter(mean,std,label):
range = np.arange(mean-4*std, mean+4*std, 0.001)
plt.plot(range, stats.norm.pdf(range, mean, std), label=label)
plt.xlim(mean-4*std, mean+4*std)
import numpy as np
import pandas as pd
import scipy.stats as stats
import matplotlib.pyplot as plt
@sample_code
sb_2016 = np.random.normal(3.2, 1.5, 20)
sb_2017 = np.random.normal(5.1, 2.3, 30)
pop_2016 = np.random.normal(3.2, 1.5, 2100)
pop_2017 = np.random.normal(5.1, 2.3, 2100)
# plot the two distributions using the 'pdf_plotter' function
pdf_plotter(3.2, 1.5, 'sample')
pdf_plotter(5.1, 2.3, 'population')
plt.legend()
plt.show()
# test whether there is a difference between 2016 and 2017
st, p = stats.ttest_ind(pop_2016, pop_2017)
if (p > alpha):
print('p-value > alpha, thus there is a statistically significant difference.')
else:
print('p-value <= alpha, thus there is no statistically significant difference.')
# test if there is an increase in 2017 (directional)
@solution
@sct
*** =sct
type: NormalExercise
lang: python
xp: 100
skills: 1
key: e94f3cb870*** =instructions
Recall the US flights dataset from Chapter 1. The conclusion was that the p-value is small enough to reject the null hypothesis stating that there is no statistically significant difference between the mean of United Airlines delays and mean delays experienced overall.
a) Using again the same data, evaluate whether there are less or more UA delays compared to overall US airline delays. The United Airlines flights have been preloaded in df_ua and delays in ua_delays. The information on all flights and all delays have been stored in df_flights and all_delays.
b) Compare the delays of two competing airline companies (Delta and United) and decide whether they are experiencing statisticall significant delays. If yes, in what direction. The Delta flights have been preloaded in df_dl and delays in dl_delays.
*** =hint
*** =pre_exercise_code
import numpy as np
import pandas as pd
import scipy.stats as stats
import urllib.request
filename = 'https://s3.amazonaws.com/assets.datacamp.com/production/course_6777/datasets/us_flight_delays.csv.gz.csv'
urllib.request.urlretrieve(filename, "us_flight_delays.csv.gz")
df_flights = pd.read_csv('us_flight_delays.csv.gz')
all_delays = df_flights['DEPARTURE_DELAY'].dropna()
df_ua = df_flights.loc[df_flights['AIRLINE'] == 'UA']
ua_delays = df_ua['DEPARTURE_DELAY'].dropna()
df_dl = df_flights.loc[df_flights['AIRLINE'] == 'DL']
dl_delays = df_dl['DEPARTURE_DELAY'].dropna()
*** =sample_code
print(all_delays.mean())
print(ua_delays.mean())
# one sample t-test; compare population mean to one sample mean
popmean = all_delays.mean()
stats.ttest_1samp(ua_delays, popmean)
# test direction of the effect (less or more delays) - one tailed hypothesis test
# compare two independent samples
st, p = stats.ttest_ind(ua_delays, dl_delays)
*** =solution
*** =sct
The CEO of a company claims that 80 percent of his 1,000,000 customers are very satisfied with the service they receive. To test this claim, the Analytics team surveyed 100 customers, using simple random sampling. Among the sampled customers, 73 percent say they are very satisified. Based on these findings, can we reject the CEO's hypothesis that 80% of the customers are very satisfied? Use a 0.05 level of significance.
type: NormalExercise
lang: python
xp: 100
skills: 1
key: 50d0e9a131*** =instructions
*** =hint
*** =pre_exercise_code
import numpy as np
import scipy.stats as stats
import math
# sample proportion
p = 0.73
# hypothesized proportion
P = 0.8
# sample size
n = 100
# population size
N = 1000000
*** =sample_code
# calculate standard deviation
stddev = math.sqrt(P*(1-P)/n)
# calculate z-score as z = (p - P) / stddev
z_score = (p - P) / stddev
# since this is a two-tailed test, the p-value is the probability that the z-score is less than or greater than its +/- value
p_val = stats.norm.cdf(z_score)*2
*** =solution
*** =sct
--- type:MultipleChoiceExercise lang:python xp:50 skills:2 key:a939fe46dc
Suppose the previous exercise is stated a bit differently. Suppose the CEO claims that at least 80% of the company's 1,000,000 customers are very satisfied. What needs to be changed in our hypothesis test to test this claim, given that all the other information remain unchanged (significance level, sample size, etc.)?
*** =instructions
- We use the t-test for this question.
- We use again the z-test to perform a two-tailed test.
- We use again the z-test to perform a one-tailed test.
- Nothing, the result is the same.
*** =hint
The change in the exercise statement is in the "at least 80%". What test do we use to test that something is of a certain value or higher than that?
*** =pre_exercise_code
*** =sct
test_mc(3)