From 5148035d51df2d52ffa734981762c8c0ccc494c4 Mon Sep 17 00:00:00 2001 From: RocioARM Date: Mon, 8 Jan 2024 18:35:20 +0100 Subject: [PATCH] Update --- .../.ipynb_checkpoints/main-checkpoint.ipynb | 779 ++++++++++++ your-code/main.ipynb | 1058 ++++++++++++----- 2 files changed, 1558 insertions(+), 279 deletions(-) create mode 100644 your-code/.ipynb_checkpoints/main-checkpoint.ipynb diff --git a/your-code/.ipynb_checkpoints/main-checkpoint.ipynb b/your-code/.ipynb_checkpoints/main-checkpoint.ipynb new file mode 100644 index 0000000..5363c0d --- /dev/null +++ b/your-code/.ipynb_checkpoints/main-checkpoint.ipynb @@ -0,0 +1,779 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Before your start:\n", + "- Read the README.md file\n", + "- Comment as much as you can and use the resources (README.md file)\n", + "- Happy learning!" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [], + "source": [ + "# import numpy and pandas\n", + "import numpy as np\n", + "import pandas as pd\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Challenge 1 - Exploring the Data\n", + "\n", + "In this challenge, we will examine all salaries of employees of the City of Chicago. We will start by loading the dataset and examining its contents." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": {}, + "outputs": [], + "source": [ + "# Your code here:\n", + "data = pd.read_csv('Current_Employee_Names__Salaries__and_Position_Titles.csv')\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Examine the `salaries` dataset using the `head` function below." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": {}, + "outputs": [ + { + "data": { + "text/html": [ + "
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NameJob TitlesDepartmentFull or Part-TimeSalary or HourlyTypical HoursAnnual SalaryHourly Rate
0AARON, JEFFERY MSERGEANTPOLICEFSalaryNaN101442.0NaN
1AARON, KARINAPOLICE OFFICER (ASSIGNED AS DETECTIVE)POLICEFSalaryNaN94122.0NaN
2AARON, KIMBERLEI RCHIEF CONTRACT EXPEDITERGENERAL SERVICESFSalaryNaN101592.0NaN
3ABAD JR, VICENTE MCIVIL ENGINEER IVWATER MGMNTFSalaryNaN110064.0NaN
4ABASCAL, REECE ETRAFFIC CONTROL AIDE-HOURLYOEMCPHourly20.0NaN19.86
5ABBASI, CHRISTOPHERSTAFF ASST TO THE ALDERMANCITY COUNCILFSalaryNaN50436.0NaN
6ABBATACOLA, ROBERT JELECTRICAL MECHANICAVIATIONFHourly40.0NaN46.10
7ABBATE, JOSEPH LPOOL MOTOR TRUCK DRIVERSTREETS & SANFHourly40.0NaN35.60
8ABBATEMARCO, JAMES JFIRE ENGINEER-EMTFIREFSalaryNaN103350.0NaN
9ABBATE, TERRY MPOLICE OFFICERPOLICEFSalaryNaN93354.0NaN
\n", + "
" + ], + "text/plain": [ + " Name Job Titles \\\n", + "0 AARON, JEFFERY M SERGEANT \n", + "1 AARON, KARINA POLICE OFFICER (ASSIGNED AS DETECTIVE) \n", + "2 AARON, KIMBERLEI R CHIEF CONTRACT EXPEDITER \n", + "3 ABAD JR, VICENTE M CIVIL ENGINEER IV \n", + "4 ABASCAL, REECE E TRAFFIC CONTROL AIDE-HOURLY \n", + "5 ABBASI, CHRISTOPHER STAFF ASST TO THE ALDERMAN \n", + "6 ABBATACOLA, ROBERT J ELECTRICAL MECHANIC \n", + "7 ABBATE, JOSEPH L POOL MOTOR TRUCK DRIVER \n", + "8 ABBATEMARCO, JAMES J FIRE ENGINEER-EMT \n", + "9 ABBATE, TERRY M POLICE OFFICER \n", + "\n", + " Department Full or Part-Time Salary or Hourly Typical Hours \\\n", + "0 POLICE F Salary NaN \n", + "1 POLICE F Salary NaN \n", + "2 GENERAL SERVICES F Salary NaN \n", + "3 WATER MGMNT F Salary NaN \n", + "4 OEMC P Hourly 20.0 \n", + "5 CITY COUNCIL F Salary NaN \n", + "6 AVIATION F Hourly 40.0 \n", + "7 STREETS & SAN F Hourly 40.0 \n", + "8 FIRE F Salary NaN \n", + "9 POLICE F Salary NaN \n", + "\n", + " Annual Salary Hourly Rate \n", + "0 101442.0 NaN \n", + "1 94122.0 NaN \n", + "2 101592.0 NaN \n", + "3 110064.0 NaN \n", + "4 NaN 19.86 \n", + "5 50436.0 NaN \n", + "6 NaN 46.10 \n", + "7 NaN 35.60 \n", + "8 103350.0 NaN \n", + "9 93354.0 NaN " + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "data.head(10)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We see from looking at the `head` function that there is quite a bit of missing data. Let's examine how much missing data is in each column. Produce this output in the cell below" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Name 0\n", + "Job Titles 0\n", + "Department 0\n", + "Full or Part-Time 0\n", + "Salary or Hourly 0\n", + "Typical Hours 25161\n", + "Annual Salary 8022\n", + "Hourly Rate 25161\n", + "dtype: int64" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "null_values = data.isnull().sum()\n", + "null_values" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Let's also look at the count of hourly vs. salaried employees. Write the code in the cell below" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Salary 25161\n", + "Hourly 8022\n", + "Name: Salary or Hourly, dtype: int64" + ] + }, + "execution_count": 8, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "data['Salary or Hourly'].value_counts()\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "What this information indicates is that the table contains information about two types of employees - salaried and hourly. Some columns apply only to one type of employee while other columns only apply to another kind. This is why there are so many missing values. Therefore, we will not do anything to handle the missing values." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "There are different departments in the city. List all departments and the count of employees in each department." + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "POLICE 13414\n", + "FIRE 4641\n", + "STREETS & SAN 2198\n", + "OEMC 2102\n", + "WATER MGMNT 1879\n", + "AVIATION 1629\n", + "TRANSPORTN 1140\n", + "PUBLIC LIBRARY 1015\n", + "GENERAL SERVICES 980\n", + "FAMILY & SUPPORT 615\n", + "FINANCE 560\n", + "HEALTH 488\n", + "CITY COUNCIL 411\n", + "LAW 407\n", + "BUILDINGS 269\n", + "COMMUNITY DEVELOPMENT 207\n", + "BUSINESS AFFAIRS 171\n", + "COPA 116\n", + "BOARD OF ELECTION 107\n", + "DoIT 99\n", + "PROCUREMENT 92\n", + "INSPECTOR GEN 87\n", + "MAYOR'S OFFICE 85\n", + "CITY CLERK 84\n", + "ANIMAL CONTRL 81\n", + "HUMAN RESOURCES 79\n", + "CULTURAL AFFAIRS 65\n", + "BUDGET & MGMT 46\n", + "ADMIN HEARNG 39\n", + "DISABILITIES 28\n", + "TREASURER 22\n", + "HUMAN RELATIONS 16\n", + "BOARD OF ETHICS 8\n", + "POLICE BOARD 2\n", + "LICENSE APPL COMM 1\n", + "Name: Department, dtype: int64" + ] + }, + "execution_count": 11, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "data['Department'].value_counts()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Challenge 2 - Hypothesis Tests\n", + "\n", + "In this section of the lab, we will test whether the hourly wage of all hourly workers is significantly different from $30/hr. Import the correct one sample test function from scipy and perform the hypothesis test for a 95% two sided confidence interval." + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-Statistic: 20.6198057854942\n", + "P-Value: 4.3230240486229894e-92\n", + "Rechazamos la hipótesis nula.Hay dif. signifiticativas respecto al valor $30/hora\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "from scipy.stats import ttest_1samp\n", + "hyp_0 = 30\n", + "hourly_wage = data[data['Salary or Hourly'] == 'Hourly']['Hourly Rate'].dropna()\n", + "\n", + "t_statistic, p_value = ttest_1samp(hourly_wage, hyp_0)\n", + "print(f\"T-Statistic: {t_statistic}\")\n", + "print(f\"P-Value: {p_value}\")\n", + "\n", + "alpha = 0.05\n", + "if p_value < alpha:\n", + " print(\"Rechazamos la hipótesis nula.Hay dif. signifiticativas respecto al valor $30/hora\")\n", + "else:\n", + " print(\"No hay evidencia para rechazar la hipótesis nula.\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We are also curious about salaries in the police force. The chief of police in Chicago claimed in a press briefing that salaries this year are higher than last year's mean of $86000/year a year for all salaried employees. Test this one sided hypothesis using a 95% confidence interval.\n", + "\n", + "Hint: A one tailed test has a p-value that is half of the two tailed p-value. If our hypothesis is greater than, then to reject, the test statistic must also be positive." + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-Statistic: 3.081997005712994\n", + "P-Value: 0.0010301701775482569\n", + "Hipotesis nula <=86.000$\n", + "Rechazamos la hipótesis nula. La media de los salarios si es superior a 86.000$\n" + ] + }, + { + "data": { + "text/plain": [ + "86486.41450313339" + ] + }, + "execution_count": 24, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "hyp_0_police = 86000\n", + "police_salary = data[(data['Department'] == 'POLICE') & (data['Salary or Hourly'] == 'Salary')]['Annual Salary'].dropna()\n", + "\n", + "t_statistic, p_value = ttest_1samp(police_salary, hyp_0_police, alternative='greater')\n", + "\n", + "\n", + "\n", + "print(f\"T-Statistic: {t_statistic}\")\n", + "print(f\"P-Value: {p_value}\")\n", + "\n", + "alpha = 0.05\n", + "if p_value < alpha:\n", + " print('Hipotesis nula <=86.000$')\n", + " print(\"Rechazamos la hipótesis nula. La media de los salarios si es superior a 86.000$\")\n", + "else:\n", + " print('Hipotesis nula <=86.000$')\n", + " print(\"No hay evidencia para rechazar la hipótesis nula\")\n", + "\n", + "police_salary.mean()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Using the `crosstab` function, find the department that has the most hourly workers. " + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "STREETS & SAN\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "cross_table = pd.crosstab(data['Department'], data['Salary or Hourly'])\n", + "\n", + "print(cross_table['Hourly'].idxmax())\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The workers from the department with the most hourly workers have complained that their hourly wage is less than $35/hour. Using a one sample t-test, test this one-sided hypothesis at the 95% confidence level." + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-Statistic: -9.567447887848152\n", + "P-Value: 1.6689265282353859e-21\n", + "Hipotesis nula >=35$/hora\n", + "Rechazamos la hipótesis nula. El precio medio la hora si es inferior a 35$\n" + ] + }, + { + "data": { + "text/plain": [ + "33.728378088076845" + ] + }, + "execution_count": 31, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "hyp_0_streets = 35\n", + "streets_hourly_rate = data[data['Department']=='STREETS & SAN']['Hourly Rate'].dropna()\n", + "t_statistic, p_value = ttest_1samp(streets_hourly_rate, hyp_0_streets, alternative='less')\n", + "print(f\"T-Statistic: {t_statistic}\")\n", + "print(f\"P-Value: {p_value}\")\n", + "\n", + "alpha = 0.05\n", + "if p_value < alpha:\n", + " print('Hipotesis nula >=35$/hora')\n", + " print(\"Rechazamos la hipótesis nula. El precio medio la hora si es inferior a 35$\")\n", + "else:\n", + " print('Hipotesis nula >=35$/hora')\n", + " print(\"No hay evidencia para rechazar la hipótesis nula\")\n", + "\n", + "streets_hourly_rate.mean()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Challenge 3: To practice - Constructing Confidence Intervals\n", + "\n", + "While testing our hypothesis is a great way to gather empirical evidence for accepting or rejecting the hypothesis, another way to gather evidence is by creating a confidence interval. A confidence interval gives us information about the true mean of the population. So for a 95% confidence interval, we are 95% sure that the mean of the population is within the confidence interval. \n", + ").\n", + "\n", + "To read more about confidence intervals, click [here](https://en.wikipedia.org/wiki/Confidence_interval).\n", + "\n", + "\n", + "In the cell below, we will construct a 95% confidence interval for the mean hourly wage of all hourly workers. \n", + "\n", + "The confidence interval is computed in SciPy using the `t.interval` function. You can read more about this function [here](https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.stats.t.html).\n", + "\n", + "To compute the confidence interval of the hourly wage, use the 0.95 for the confidence level, number of rows - 1 for degrees of freedom, the mean of the sample for the location parameter and the standard error for the scale. The standard error can be computed using [this](https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.sem.html) function in SciPy." + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "95% Confidence Interval for Mean Hourly Wage: (32.52345834488529, 33.05365708767727)\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "from scipy.stats import t\n", + "hourly_workers = data['Hourly Rate'].dropna()\n", + "mean_hourly_wage = np.mean(hourly_workers)\n", + "std_dev_hourly_wage = np.std(hourly_workers, ddof=1)\n", + "confidence_level = 0.95\n", + "# Calculate the margin of error\n", + "margin_of_error = t.ppf((1 + confidence_level) / 2, len(hourly_workers) - 1) * (std_dev_hourly_wage / np.sqrt(len(hourly_workers)))\n", + "\n", + "# Calculate the confidence interval\n", + "confidence_interval = (mean_hourly_wage - margin_of_error, mean_hourly_wage + margin_of_error)\n", + "\n", + "# Print the results\n", + "print(f\"95% Confidence Interval for Mean Hourly Wage: {confidence_interval}\")\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Now construct the 95% confidence interval for all salaried employeed in the police in the cell below." + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "95% Confidence Interval for Mean Salary Wage: (86526.99656774198, 87047.00301256108)\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "\n", + "salary_workers = data['Annual Salary'].dropna()\n", + "mean_salary_wage = np.mean(salary_workers)\n", + "std_dev_wage = np.std(salary_workers, ddof=1)\n", + "confidence_level = 0.95\n", + "# Calculate the margin of error\n", + "margin_of_error = t.ppf((1 + confidence_level) / 2, len(salary_workers) - 1) * (std_dev_wage / np.sqrt(len(salary_workers)))\n", + "\n", + "# Calculate the confidence interval\n", + "confidence_interval = (mean_salary_wage - margin_of_error, mean_salary_wage + margin_of_error)\n", + "\n", + "# Print the results\n", + "print(f\"95% Confidence Interval for Mean Salary Wage: {confidence_interval}\")\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Bonus Challenge - Hypothesis Tests of Proportions\n", + "\n", + "Another type of one sample test is a hypothesis test of proportions. In this test, we examine whether the proportion of a group in our sample is significantly different than a fraction. \n", + "\n", + "You can read more about one sample proportion tests [here](http://sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/SAS/SAS6-CategoricalData/SAS6-CategoricalData2.html).\n", + "\n", + "In the cell below, use the `proportions_ztest` function from `statsmodels` to perform a hypothesis test that will determine whether the number of hourly workers in the City of Chicago is significantly different from 25% at the 95% confidence level." + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Z-Statistic: -3.5099964213703005\n", + "P-Value: 0.0004481127249057967\n", + "Hipotesis nula proporcion de trabajadores por hora del 25%\n", + "Rechazamos hipotesis nula.\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "from statsmodels.stats.proportion import proportions_ztest\n", + "\n", + "# Hourly workers and total workers:\n", + "observed_hourly_workers = len(data[data['Salary or Hourly']=='Hourly'])\n", + "total_workers = len(data)\n", + "\n", + "# Hypothesized proportion (25%)\n", + "hyp_proportion = 0.25\n", + "\n", + "# Perform proportions z-test\n", + "z_statistic, p_value = proportions_ztest(observed_hourly_workers, total_workers, value=hyp_proportion, alternative='two-sided')\n", + "\n", + "# Print the results\n", + "print(f\"Z-Statistic: {z_statistic}\")\n", + "print(f\"P-Value: {p_value}\")\n", + "\n", + "# Compare p-value to significance level (e.g., 0.05)\n", + "alpha = 0.05\n", + "if p_value < alpha:\n", + " print('Hipotesis nula proporcion de trabajadores por hora del 25%')\n", + " print(\"Rechazamos hipotesis nula.\")\n", + "else:\n", + " print(\"No hay evidencias para rechazar la hipotesis.\")\n" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Salary 25161\n", + "Hourly 8022\n", + "Name: Salary or Hourly, dtype: int64" + ] + }, + "execution_count": 41, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "data['Salary or Hourly'].value_counts()\n" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0.24175029382515142" + ] + }, + "execution_count": 42, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "8022/(8022+25161)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/your-code/main.ipynb b/your-code/main.ipynb index b2b6f8d..5363c0d 100644 --- a/your-code/main.ipynb +++ b/your-code/main.ipynb @@ -1,279 +1,779 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Before your start:\n", - "- Read the README.md file\n", - "- Comment as much as you can and use the resources (README.md file)\n", - "- Happy learning!" - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# import numpy and pandas\n", - "\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Challenge 1 - Exploring the Data\n", - "\n", - "In this challenge, we will examine all salaries of employees of the City of Chicago. We will start by loading the dataset and examining its contents." - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Examine the `salaries` dataset using the `head` function below." - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n", - "\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "We see from looking at the `head` function that there is quite a bit of missing data. Let's examine how much missing data is in each column. Produce this output in the cell below" - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n", - "\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Let's also look at the count of hourly vs. salaried employees. Write the code in the cell below" - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n", - "\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "What this information indicates is that the table contains information about two types of employees - salaried and hourly. Some columns apply only to one type of employee while other columns only apply to another kind. This is why there are so many missing values. Therefore, we will not do anything to handle the missing values." - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "There are different departments in the city. List all departments and the count of employees in each department." - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n", - "\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Challenge 2 - Hypothesis Tests\n", - "\n", - "In this section of the lab, we will test whether the hourly wage of all hourly workers is significantly different from $30/hr. Import the correct one sample test function from scipy and perform the hypothesis test for a 95% two sided confidence interval." - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n", - "\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "We are also curious about salaries in the police force. The chief of police in Chicago claimed in a press briefing that salaries this year are higher than last year's mean of $86000/year a year for all salaried employees. Test this one sided hypothesis using a 95% confidence interval.\n", - "\n", - "Hint: A one tailed test has a p-value that is half of the two tailed p-value. If our hypothesis is greater than, then to reject, the test statistic must also be positive." - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n", - "\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Using the `crosstab` function, find the department that has the most hourly workers. " - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n", - "\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "The workers from the department with the most hourly workers have complained that their hourly wage is less than $35/hour. Using a one sample t-test, test this one-sided hypothesis at the 95% confidence level." - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n", - "\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Challenge 3: To practice - Constructing Confidence Intervals\n", - "\n", - "While testing our hypothesis is a great way to gather empirical evidence for accepting or rejecting the hypothesis, another way to gather evidence is by creating a confidence interval. A confidence interval gives us information about the true mean of the population. So for a 95% confidence interval, we are 95% sure that the mean of the population is within the confidence interval. \n", - ").\n", - "\n", - "To read more about confidence intervals, click [here](https://en.wikipedia.org/wiki/Confidence_interval).\n", - "\n", - "\n", - "In the cell below, we will construct a 95% confidence interval for the mean hourly wage of all hourly workers. \n", - "\n", - "The confidence interval is computed in SciPy using the `t.interval` function. You can read more about this function [here](https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.stats.t.html).\n", - "\n", - "To compute the confidence interval of the hourly wage, use the 0.95 for the confidence level, number of rows - 1 for degrees of freedom, the mean of the sample for the location parameter and the standard error for the scale. The standard error can be computed using [this](https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.sem.html) function in SciPy." - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n", - "\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Now construct the 95% confidence interval for all salaried employeed in the police in the cell below." - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n", - "\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Bonus Challenge - Hypothesis Tests of Proportions\n", - "\n", - "Another type of one sample test is a hypothesis test of proportions. In this test, we examine whether the proportion of a group in our sample is significantly different than a fraction. \n", - "\n", - "You can read more about one sample proportion tests [here](http://sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/SAS/SAS6-CategoricalData/SAS6-CategoricalData2.html).\n", - "\n", - "In the cell below, use the `proportions_ztest` function from `statsmodels` to perform a hypothesis test that will determine whether the number of hourly workers in the City of Chicago is significantly different from 25% at the 95% confidence level." - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n", - "\n" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 3", - "language": "python", - "name": "python3" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 3 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython3", - "version": "3.7.3" - } - }, - "nbformat": 4, - "nbformat_minor": 2 -} +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Before your start:\n", + "- Read the README.md file\n", + "- Comment as much as you can and use the resources (README.md file)\n", + "- Happy learning!" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [], + "source": [ + "# import numpy and pandas\n", + "import numpy as np\n", + "import pandas as pd\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Challenge 1 - Exploring the Data\n", + "\n", + "In this challenge, we will examine all salaries of employees of the City of Chicago. We will start by loading the dataset and examining its contents." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": {}, + "outputs": [], + "source": [ + "# Your code here:\n", + "data = pd.read_csv('Current_Employee_Names__Salaries__and_Position_Titles.csv')\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Examine the `salaries` dataset using the `head` function below." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": {}, + "outputs": [ + { + "data": { + "text/html": [ + "
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NameJob TitlesDepartmentFull or Part-TimeSalary or HourlyTypical HoursAnnual SalaryHourly Rate
0AARON, JEFFERY MSERGEANTPOLICEFSalaryNaN101442.0NaN
1AARON, KARINAPOLICE OFFICER (ASSIGNED AS DETECTIVE)POLICEFSalaryNaN94122.0NaN
2AARON, KIMBERLEI RCHIEF CONTRACT EXPEDITERGENERAL SERVICESFSalaryNaN101592.0NaN
3ABAD JR, VICENTE MCIVIL ENGINEER IVWATER MGMNTFSalaryNaN110064.0NaN
4ABASCAL, REECE ETRAFFIC CONTROL AIDE-HOURLYOEMCPHourly20.0NaN19.86
5ABBASI, CHRISTOPHERSTAFF ASST TO THE ALDERMANCITY COUNCILFSalaryNaN50436.0NaN
6ABBATACOLA, ROBERT JELECTRICAL MECHANICAVIATIONFHourly40.0NaN46.10
7ABBATE, JOSEPH LPOOL MOTOR TRUCK DRIVERSTREETS & SANFHourly40.0NaN35.60
8ABBATEMARCO, JAMES JFIRE ENGINEER-EMTFIREFSalaryNaN103350.0NaN
9ABBATE, TERRY MPOLICE OFFICERPOLICEFSalaryNaN93354.0NaN
\n", + "
" + ], + "text/plain": [ + " Name Job Titles \\\n", + "0 AARON, JEFFERY M SERGEANT \n", + "1 AARON, KARINA POLICE OFFICER (ASSIGNED AS DETECTIVE) \n", + "2 AARON, KIMBERLEI R CHIEF CONTRACT EXPEDITER \n", + "3 ABAD JR, VICENTE M CIVIL ENGINEER IV \n", + "4 ABASCAL, REECE E TRAFFIC CONTROL AIDE-HOURLY \n", + "5 ABBASI, CHRISTOPHER STAFF ASST TO THE ALDERMAN \n", + "6 ABBATACOLA, ROBERT J ELECTRICAL MECHANIC \n", + "7 ABBATE, JOSEPH L POOL MOTOR TRUCK DRIVER \n", + "8 ABBATEMARCO, JAMES J FIRE ENGINEER-EMT \n", + "9 ABBATE, TERRY M POLICE OFFICER \n", + "\n", + " Department Full or Part-Time Salary or Hourly Typical Hours \\\n", + "0 POLICE F Salary NaN \n", + "1 POLICE F Salary NaN \n", + "2 GENERAL SERVICES F Salary NaN \n", + "3 WATER MGMNT F Salary NaN \n", + "4 OEMC P Hourly 20.0 \n", + "5 CITY COUNCIL F Salary NaN \n", + "6 AVIATION F Hourly 40.0 \n", + "7 STREETS & SAN F Hourly 40.0 \n", + "8 FIRE F Salary NaN \n", + "9 POLICE F Salary NaN \n", + "\n", + " Annual Salary Hourly Rate \n", + "0 101442.0 NaN \n", + "1 94122.0 NaN \n", + "2 101592.0 NaN \n", + "3 110064.0 NaN \n", + "4 NaN 19.86 \n", + "5 50436.0 NaN \n", + "6 NaN 46.10 \n", + "7 NaN 35.60 \n", + "8 103350.0 NaN \n", + "9 93354.0 NaN " + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "data.head(10)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We see from looking at the `head` function that there is quite a bit of missing data. Let's examine how much missing data is in each column. Produce this output in the cell below" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Name 0\n", + "Job Titles 0\n", + "Department 0\n", + "Full or Part-Time 0\n", + "Salary or Hourly 0\n", + "Typical Hours 25161\n", + "Annual Salary 8022\n", + "Hourly Rate 25161\n", + "dtype: int64" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "null_values = data.isnull().sum()\n", + "null_values" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Let's also look at the count of hourly vs. salaried employees. Write the code in the cell below" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Salary 25161\n", + "Hourly 8022\n", + "Name: Salary or Hourly, dtype: int64" + ] + }, + "execution_count": 8, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "data['Salary or Hourly'].value_counts()\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "What this information indicates is that the table contains information about two types of employees - salaried and hourly. Some columns apply only to one type of employee while other columns only apply to another kind. This is why there are so many missing values. Therefore, we will not do anything to handle the missing values." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "There are different departments in the city. List all departments and the count of employees in each department." + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "POLICE 13414\n", + "FIRE 4641\n", + "STREETS & SAN 2198\n", + "OEMC 2102\n", + "WATER MGMNT 1879\n", + "AVIATION 1629\n", + "TRANSPORTN 1140\n", + "PUBLIC LIBRARY 1015\n", + "GENERAL SERVICES 980\n", + "FAMILY & SUPPORT 615\n", + "FINANCE 560\n", + "HEALTH 488\n", + "CITY COUNCIL 411\n", + "LAW 407\n", + "BUILDINGS 269\n", + "COMMUNITY DEVELOPMENT 207\n", + "BUSINESS AFFAIRS 171\n", + "COPA 116\n", + "BOARD OF ELECTION 107\n", + "DoIT 99\n", + "PROCUREMENT 92\n", + "INSPECTOR GEN 87\n", + "MAYOR'S OFFICE 85\n", + "CITY CLERK 84\n", + "ANIMAL CONTRL 81\n", + "HUMAN RESOURCES 79\n", + "CULTURAL AFFAIRS 65\n", + "BUDGET & MGMT 46\n", + "ADMIN HEARNG 39\n", + "DISABILITIES 28\n", + "TREASURER 22\n", + "HUMAN RELATIONS 16\n", + "BOARD OF ETHICS 8\n", + "POLICE BOARD 2\n", + "LICENSE APPL COMM 1\n", + "Name: Department, dtype: int64" + ] + }, + "execution_count": 11, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "data['Department'].value_counts()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Challenge 2 - Hypothesis Tests\n", + "\n", + "In this section of the lab, we will test whether the hourly wage of all hourly workers is significantly different from $30/hr. Import the correct one sample test function from scipy and perform the hypothesis test for a 95% two sided confidence interval." + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-Statistic: 20.6198057854942\n", + "P-Value: 4.3230240486229894e-92\n", + "Rechazamos la hipótesis nula.Hay dif. signifiticativas respecto al valor $30/hora\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "from scipy.stats import ttest_1samp\n", + "hyp_0 = 30\n", + "hourly_wage = data[data['Salary or Hourly'] == 'Hourly']['Hourly Rate'].dropna()\n", + "\n", + "t_statistic, p_value = ttest_1samp(hourly_wage, hyp_0)\n", + "print(f\"T-Statistic: {t_statistic}\")\n", + "print(f\"P-Value: {p_value}\")\n", + "\n", + "alpha = 0.05\n", + "if p_value < alpha:\n", + " print(\"Rechazamos la hipótesis nula.Hay dif. signifiticativas respecto al valor $30/hora\")\n", + "else:\n", + " print(\"No hay evidencia para rechazar la hipótesis nula.\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We are also curious about salaries in the police force. The chief of police in Chicago claimed in a press briefing that salaries this year are higher than last year's mean of $86000/year a year for all salaried employees. Test this one sided hypothesis using a 95% confidence interval.\n", + "\n", + "Hint: A one tailed test has a p-value that is half of the two tailed p-value. If our hypothesis is greater than, then to reject, the test statistic must also be positive." + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-Statistic: 3.081997005712994\n", + "P-Value: 0.0010301701775482569\n", + "Hipotesis nula <=86.000$\n", + "Rechazamos la hipótesis nula. La media de los salarios si es superior a 86.000$\n" + ] + }, + { + "data": { + "text/plain": [ + "86486.41450313339" + ] + }, + "execution_count": 24, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "hyp_0_police = 86000\n", + "police_salary = data[(data['Department'] == 'POLICE') & (data['Salary or Hourly'] == 'Salary')]['Annual Salary'].dropna()\n", + "\n", + "t_statistic, p_value = ttest_1samp(police_salary, hyp_0_police, alternative='greater')\n", + "\n", + "\n", + "\n", + "print(f\"T-Statistic: {t_statistic}\")\n", + "print(f\"P-Value: {p_value}\")\n", + "\n", + "alpha = 0.05\n", + "if p_value < alpha:\n", + " print('Hipotesis nula <=86.000$')\n", + " print(\"Rechazamos la hipótesis nula. La media de los salarios si es superior a 86.000$\")\n", + "else:\n", + " print('Hipotesis nula <=86.000$')\n", + " print(\"No hay evidencia para rechazar la hipótesis nula\")\n", + "\n", + "police_salary.mean()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Using the `crosstab` function, find the department that has the most hourly workers. " + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "STREETS & SAN\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "cross_table = pd.crosstab(data['Department'], data['Salary or Hourly'])\n", + "\n", + "print(cross_table['Hourly'].idxmax())\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The workers from the department with the most hourly workers have complained that their hourly wage is less than $35/hour. Using a one sample t-test, test this one-sided hypothesis at the 95% confidence level." + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-Statistic: -9.567447887848152\n", + "P-Value: 1.6689265282353859e-21\n", + "Hipotesis nula >=35$/hora\n", + "Rechazamos la hipótesis nula. El precio medio la hora si es inferior a 35$\n" + ] + }, + { + "data": { + "text/plain": [ + "33.728378088076845" + ] + }, + "execution_count": 31, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "hyp_0_streets = 35\n", + "streets_hourly_rate = data[data['Department']=='STREETS & SAN']['Hourly Rate'].dropna()\n", + "t_statistic, p_value = ttest_1samp(streets_hourly_rate, hyp_0_streets, alternative='less')\n", + "print(f\"T-Statistic: {t_statistic}\")\n", + "print(f\"P-Value: {p_value}\")\n", + "\n", + "alpha = 0.05\n", + "if p_value < alpha:\n", + " print('Hipotesis nula >=35$/hora')\n", + " print(\"Rechazamos la hipótesis nula. El precio medio la hora si es inferior a 35$\")\n", + "else:\n", + " print('Hipotesis nula >=35$/hora')\n", + " print(\"No hay evidencia para rechazar la hipótesis nula\")\n", + "\n", + "streets_hourly_rate.mean()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Challenge 3: To practice - Constructing Confidence Intervals\n", + "\n", + "While testing our hypothesis is a great way to gather empirical evidence for accepting or rejecting the hypothesis, another way to gather evidence is by creating a confidence interval. A confidence interval gives us information about the true mean of the population. So for a 95% confidence interval, we are 95% sure that the mean of the population is within the confidence interval. \n", + ").\n", + "\n", + "To read more about confidence intervals, click [here](https://en.wikipedia.org/wiki/Confidence_interval).\n", + "\n", + "\n", + "In the cell below, we will construct a 95% confidence interval for the mean hourly wage of all hourly workers. \n", + "\n", + "The confidence interval is computed in SciPy using the `t.interval` function. You can read more about this function [here](https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.stats.t.html).\n", + "\n", + "To compute the confidence interval of the hourly wage, use the 0.95 for the confidence level, number of rows - 1 for degrees of freedom, the mean of the sample for the location parameter and the standard error for the scale. The standard error can be computed using [this](https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.sem.html) function in SciPy." + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "95% Confidence Interval for Mean Hourly Wage: (32.52345834488529, 33.05365708767727)\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "from scipy.stats import t\n", + "hourly_workers = data['Hourly Rate'].dropna()\n", + "mean_hourly_wage = np.mean(hourly_workers)\n", + "std_dev_hourly_wage = np.std(hourly_workers, ddof=1)\n", + "confidence_level = 0.95\n", + "# Calculate the margin of error\n", + "margin_of_error = t.ppf((1 + confidence_level) / 2, len(hourly_workers) - 1) * (std_dev_hourly_wage / np.sqrt(len(hourly_workers)))\n", + "\n", + "# Calculate the confidence interval\n", + "confidence_interval = (mean_hourly_wage - margin_of_error, mean_hourly_wage + margin_of_error)\n", + "\n", + "# Print the results\n", + "print(f\"95% Confidence Interval for Mean Hourly Wage: {confidence_interval}\")\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Now construct the 95% confidence interval for all salaried employeed in the police in the cell below." + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "95% Confidence Interval for Mean Salary Wage: (86526.99656774198, 87047.00301256108)\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "\n", + "salary_workers = data['Annual Salary'].dropna()\n", + "mean_salary_wage = np.mean(salary_workers)\n", + "std_dev_wage = np.std(salary_workers, ddof=1)\n", + "confidence_level = 0.95\n", + "# Calculate the margin of error\n", + "margin_of_error = t.ppf((1 + confidence_level) / 2, len(salary_workers) - 1) * (std_dev_wage / np.sqrt(len(salary_workers)))\n", + "\n", + "# Calculate the confidence interval\n", + "confidence_interval = (mean_salary_wage - margin_of_error, mean_salary_wage + margin_of_error)\n", + "\n", + "# Print the results\n", + "print(f\"95% Confidence Interval for Mean Salary Wage: {confidence_interval}\")\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Bonus Challenge - Hypothesis Tests of Proportions\n", + "\n", + "Another type of one sample test is a hypothesis test of proportions. In this test, we examine whether the proportion of a group in our sample is significantly different than a fraction. \n", + "\n", + "You can read more about one sample proportion tests [here](http://sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/SAS/SAS6-CategoricalData/SAS6-CategoricalData2.html).\n", + "\n", + "In the cell below, use the `proportions_ztest` function from `statsmodels` to perform a hypothesis test that will determine whether the number of hourly workers in the City of Chicago is significantly different from 25% at the 95% confidence level." + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Z-Statistic: -3.5099964213703005\n", + "P-Value: 0.0004481127249057967\n", + "Hipotesis nula proporcion de trabajadores por hora del 25%\n", + "Rechazamos hipotesis nula.\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "from statsmodels.stats.proportion import proportions_ztest\n", + "\n", + "# Hourly workers and total workers:\n", + "observed_hourly_workers = len(data[data['Salary or Hourly']=='Hourly'])\n", + "total_workers = len(data)\n", + "\n", + "# Hypothesized proportion (25%)\n", + "hyp_proportion = 0.25\n", + "\n", + "# Perform proportions z-test\n", + "z_statistic, p_value = proportions_ztest(observed_hourly_workers, total_workers, value=hyp_proportion, alternative='two-sided')\n", + "\n", + "# Print the results\n", + "print(f\"Z-Statistic: {z_statistic}\")\n", + "print(f\"P-Value: {p_value}\")\n", + "\n", + "# Compare p-value to significance level (e.g., 0.05)\n", + "alpha = 0.05\n", + "if p_value < alpha:\n", + " print('Hipotesis nula proporcion de trabajadores por hora del 25%')\n", + " print(\"Rechazamos hipotesis nula.\")\n", + "else:\n", + " print(\"No hay evidencias para rechazar la hipotesis.\")\n" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Salary 25161\n", + "Hourly 8022\n", + "Name: Salary or Hourly, dtype: int64" + ] + }, + "execution_count": 41, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "data['Salary or Hourly'].value_counts()\n" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0.24175029382515142" + ] + }, + "execution_count": 42, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "8022/(8022+25161)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +}