diff --git a/your-code/.ipynb_checkpoints/main-checkpoint.ipynb b/your-code/.ipynb_checkpoints/main-checkpoint.ipynb new file mode 100644 index 0000000..f1446b6 --- /dev/null +++ b/your-code/.ipynb_checkpoints/main-checkpoint.ipynb @@ -0,0 +1,766 @@ +{ + "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": 2, + "metadata": {}, + "outputs": [], + "source": [ + "import numpy as np\n", + "import pandas as pd" + ] + }, + { + "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": [ + "salaries = pd.read_csv(\"Current_Employee_Names__Salaries__and_Position_Titles.csv\")" + ] + }, + { + "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
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" + ], + "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", + "\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", + "\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 " + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "salaries.head()\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": 5, + "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": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "salaries.isnull().sum()\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": 6, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Salary 25161\n", + "Hourly 8022\n", + "Name: Salary or Hourly, dtype: int64" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "salaries['Salary or Hourly'].value_counts()" + ] + }, + { + "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": 7, + "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": 7, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "salaries['Department'].value_counts()\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": 18, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Requirement already satisfied: scipy in /Users/gonzalo/anaconda3/lib/python3.11/site-packages (1.10.1)\n", + "Collecting scipy\n", + " Obtaining dependency information for scipy from https://files.pythonhosted.org/packages/21/d4/e6c57acc61e59cd46acca27af1f400094d5dee218e372cc604b8162b97cb/scipy-1.12.0-cp311-cp311-macosx_12_0_arm64.whl.metadata\n", + " Downloading scipy-1.12.0-cp311-cp311-macosx_12_0_arm64.whl.metadata (165 kB)\n", + "\u001b[2K \u001b[90m━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━\u001b[0m \u001b[32m165.4/165.4 kB\u001b[0m \u001b[31m9.2 MB/s\u001b[0m eta \u001b[36m0:00:00\u001b[0m\n", + "\u001b[?25hRequirement already satisfied: numpy<1.29.0,>=1.22.4 in /Users/gonzalo/anaconda3/lib/python3.11/site-packages (from scipy) (1.24.3)\n", + "Downloading scipy-1.12.0-cp311-cp311-macosx_12_0_arm64.whl (31.4 MB)\n", + "\u001b[2K \u001b[90m━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━\u001b[0m \u001b[32m31.4/31.4 MB\u001b[0m \u001b[31m33.2 MB/s\u001b[0m eta \u001b[36m0:00:00\u001b[0m00:01\u001b[0m00:01\u001b[0m\n", + "\u001b[?25hInstalling collected packages: scipy\n", + " Attempting uninstall: scipy\n", + " Found existing installation: scipy 1.10.1\n", + " Uninstalling scipy-1.10.1:\n", + " Successfully uninstalled scipy-1.10.1\n", + "\u001b[31mERROR: pip's dependency resolver does not currently take into account all the packages that are installed. This behaviour is the source of the following dependency conflicts.\n", + "gensim 4.3.0 requires FuzzyTM>=0.4.0, which is not installed.\u001b[0m\u001b[31m\n", + "\u001b[0mSuccessfully installed scipy-1.12.0\n" + ] + } + ], + "source": [ + "!pip install --upgrade scipy" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": {}, + "outputs": [], + "source": [ + "from scipy.stats import ttest_1samp" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-statistic: 20.6198057854942, p-value: 4.3230240486229894e-92\n" + ] + } + ], + "source": [ + "#H0: \\mu_1 = 30$/hr, H1 \\mu_1 != 30$/hr\n", + "hourly = salaries[salaries['Salary or Hourly']==\"Hourly\"]['Hourly Rate']\n", + "hourly\n", + "\n", + "t_stat, p_value = ttest_1samp(hourly, 30)\n", + "\n", + "print(f'T-statistic: {t_stat}, p-value: {p_value}')\n", + "\n" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": {}, + "outputs": [], + "source": [ + "#p-value is lower than 0.05 so we reject the null hypothesis, so there is evidence that the hourly wage of \n", + "#all hourly workers is significantly different from $30/hr" + ] + }, + { + "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": 35, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-statistic: 3.081997005712994, p-value: 0.0010301701775482577\n" + ] + } + ], + "source": [ + "#H0: \\mu_1 <= $86000/year, H1 \\mu_1 > $86000/year\n", + "salaries_police = salaries[salaries['Department']==\"POLICE\"]['Hourly Rate']\n", + "\n", + "salaries_police = salaries_police.dropna()\n", + "\n", + "t_stat, p_value = ttest_1samp(salaries_police, 86000.0, alternative='greater')\n", + "\n", + "print(f'T-statistic: {t_stat}, p-value: {p_value}')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Using the `crosstab` function, find the department that has the most hourly workers. " + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "'STREETS & SAN'" + ] + }, + "execution_count": 40, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "cross_tab = pd.crosstab(salaries['Department'], salaries['Salary or Hourly'])\n", + "\n", + "hourly_workers = cross_tab['Hourly']\n", + "\n", + "department_most_hourly = hourly_workers.idxmax()\n", + "\n", + "department_most_hourly" + ] + }, + { + "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": 42, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-statistic: 71.56561089359087, p-value: 1.0\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "\n", + "salaries_streetssan = salaries[salaries['Department']==\"STREETS & SAN\"]['Annual Salary']\n", + "\n", + "salaries_streetssan = salaries_streetssan.dropna()\n", + "\n", + "t_stat, p_value = ttest_1samp(salaries_streetssan, 35.0, alternative='less')\n", + "\n", + "print(f'T-statistic: {t_stat}, p-value: {p_value}')\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "#We cannot reject the null, so there is no evidence that the hourly wage of \n", + "#all streets and san hourly workers is significantly less than $35/hour" + ] + }, + { + "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": 48, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "4 19.86\n", + "6 46.10\n", + "7 35.60\n", + "10 2.65\n", + "18 17.68\n", + " ... \n", + "33164 46.10\n", + "33168 17.68\n", + "33169 35.60\n", + "33174 46.35\n", + "33175 48.85\n", + "Name: Hourly Rate, Length: 8022, dtype: float64" + ] + }, + "execution_count": 48, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "from scipy.stats import t\n", + "\n", + "hourly_wage_workers=salaries[salaries['Salary or Hourly']==\"Hourly\"]['Hourly Rate']\n", + "\n", + "hourly_wage_workers" + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 49, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "hourly_wage_workers.isnull().sum()" + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "95% Confidence Interval for Mean Hourly Wage: (32.52345834488425, 33.05365708767623)\n" + ] + } + ], + "source": [ + "# Calculate mean and standard error\n", + "mean_hourly_wage = hourly_wage_workers.mean()\n", + "std_error_hourly_wage = np.std(hourly_wage_workers, ddof=1) / np.sqrt(len(hourly_wage_workers))\n", + "\n", + "# Confidence level and degrees of freedom\n", + "confidence_level = 0.95\n", + "degrees_of_freedom = len(hourly_wage_workers) - 1\n", + "\n", + "# Calculate the confidence interval\n", + "conf_interval = t.interval(confidence_level, degrees_of_freedom, mean_hourly_wage, std_error_hourly_wage)\n", + "\n", + "print(f\"95% Confidence Interval for Mean Hourly Wage: {conf_interval}\")" + ] + }, + { + "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": 51, + "metadata": {}, + "outputs": [], + "source": [ + "# Your code here:\n", + "\n", + "salaried_police = salaries[(salaries['Salary or Hourly'] == \"Salary\") & (salaries['Department'] == \"POLICE\")]['Annual Salary']\n" + ] + }, + { + "cell_type": "code", + "execution_count": 52, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "95% Confidence Interval for Mean Annual Salary in Police Department: (86177.05631531785, 86795.77269094893)\n" + ] + } + ], + "source": [ + "#Mean and standard error\n", + "mean_salary_police = salaried_police.mean()\n", + "std_error_salary_police = np.std(salaried_police, ddof=1) / np.sqrt(len(salaried_police))\n", + "\n", + "#Confidence level and degrees of freedom\n", + "confidence_level = 0.95\n", + "degrees_of_freedom = len(salaried_police) - 1\n", + "\n", + "#Confidence interval\n", + "conf_interval_salary_police = t.interval(confidence_level, degrees_of_freedom, mean_salary_police, std_error_salary_police)\n", + "\n", + "print(f\"95% Confidence Interval for Mean Annual Salary in Police Department: {conf_interval_salary_police}\")\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": 57, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "8022" + ] + }, + "execution_count": 57, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "from statsmodels.stats.proportion import proportions_ztest\n", + "\n", + "hourly_count = salaries[salaries['Salary or Hourly'] == 'Hourly'].shape[0]\n", + "\n", + "hourly_count" + ] + }, + { + "cell_type": "code", + "execution_count": 60, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "33183" + ] + }, + "execution_count": 60, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Total number of workers\n", + "total_count = salaries.shape[0]\n", + "\n", + "total_count" + ] + }, + { + "cell_type": "code", + "execution_count": 58, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Z-statistic: -3.5099964213703005, p-value: 0.0004481127249057967\n" + ] + } + ], + "source": [ + "# Total number of workers\n", + "total_count = salaries.shape[0]\n", + "\n", + "# Proportion\n", + "null_proportion = 0.25\n", + "\n", + "# Hypothesis test\n", + "z_stat, p_value = proportions_ztest(hourly_count, total_count, value=null_proportion, alternative='two-sided')\n", + "\n", + "print(f'Z-statistic: {z_stat}, p-value: {p_value}')" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "#The proportion of hourly workers in the City of Chicago is significantly different from 25% at a 95% confidence level" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3 (ipykernel)", + "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": "3.11.4" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/your-code/main.ipynb b/your-code/main.ipynb index b2b6f8d..f1446b6 100644 --- a/your-code/main.ipynb +++ b/your-code/main.ipynb @@ -1,279 +1,766 @@ -{ - "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": 2, + "metadata": {}, + "outputs": [], + "source": [ + "import numpy as np\n", + "import pandas as pd" + ] + }, + { + "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": [ + "salaries = pd.read_csv(\"Current_Employee_Names__Salaries__and_Position_Titles.csv\")" + ] + }, + { + "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
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" + ], + "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", + "\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", + "\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 " + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "salaries.head()\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": 5, + "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": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "salaries.isnull().sum()\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": 6, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Salary 25161\n", + "Hourly 8022\n", + "Name: Salary or Hourly, dtype: int64" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "salaries['Salary or Hourly'].value_counts()" + ] + }, + { + "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": 7, + "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": 7, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "salaries['Department'].value_counts()\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": 18, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Requirement already satisfied: scipy in /Users/gonzalo/anaconda3/lib/python3.11/site-packages (1.10.1)\n", + "Collecting scipy\n", + " Obtaining dependency information for scipy from https://files.pythonhosted.org/packages/21/d4/e6c57acc61e59cd46acca27af1f400094d5dee218e372cc604b8162b97cb/scipy-1.12.0-cp311-cp311-macosx_12_0_arm64.whl.metadata\n", + " Downloading scipy-1.12.0-cp311-cp311-macosx_12_0_arm64.whl.metadata (165 kB)\n", + "\u001b[2K \u001b[90m━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━\u001b[0m \u001b[32m165.4/165.4 kB\u001b[0m \u001b[31m9.2 MB/s\u001b[0m eta \u001b[36m0:00:00\u001b[0m\n", + "\u001b[?25hRequirement already satisfied: numpy<1.29.0,>=1.22.4 in /Users/gonzalo/anaconda3/lib/python3.11/site-packages (from scipy) (1.24.3)\n", + "Downloading scipy-1.12.0-cp311-cp311-macosx_12_0_arm64.whl (31.4 MB)\n", + "\u001b[2K \u001b[90m━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━\u001b[0m \u001b[32m31.4/31.4 MB\u001b[0m \u001b[31m33.2 MB/s\u001b[0m eta \u001b[36m0:00:00\u001b[0m00:01\u001b[0m00:01\u001b[0m\n", + "\u001b[?25hInstalling collected packages: scipy\n", + " Attempting uninstall: scipy\n", + " Found existing installation: scipy 1.10.1\n", + " Uninstalling scipy-1.10.1:\n", + " Successfully uninstalled scipy-1.10.1\n", + "\u001b[31mERROR: pip's dependency resolver does not currently take into account all the packages that are installed. This behaviour is the source of the following dependency conflicts.\n", + "gensim 4.3.0 requires FuzzyTM>=0.4.0, which is not installed.\u001b[0m\u001b[31m\n", + "\u001b[0mSuccessfully installed scipy-1.12.0\n" + ] + } + ], + "source": [ + "!pip install --upgrade scipy" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": {}, + "outputs": [], + "source": [ + "from scipy.stats import ttest_1samp" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-statistic: 20.6198057854942, p-value: 4.3230240486229894e-92\n" + ] + } + ], + "source": [ + "#H0: \\mu_1 = 30$/hr, H1 \\mu_1 != 30$/hr\n", + "hourly = salaries[salaries['Salary or Hourly']==\"Hourly\"]['Hourly Rate']\n", + "hourly\n", + "\n", + "t_stat, p_value = ttest_1samp(hourly, 30)\n", + "\n", + "print(f'T-statistic: {t_stat}, p-value: {p_value}')\n", + "\n" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": {}, + "outputs": [], + "source": [ + "#p-value is lower than 0.05 so we reject the null hypothesis, so there is evidence that the hourly wage of \n", + "#all hourly workers is significantly different from $30/hr" + ] + }, + { + "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": 35, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-statistic: 3.081997005712994, p-value: 0.0010301701775482577\n" + ] + } + ], + "source": [ + "#H0: \\mu_1 <= $86000/year, H1 \\mu_1 > $86000/year\n", + "salaries_police = salaries[salaries['Department']==\"POLICE\"]['Hourly Rate']\n", + "\n", + "salaries_police = salaries_police.dropna()\n", + "\n", + "t_stat, p_value = ttest_1samp(salaries_police, 86000.0, alternative='greater')\n", + "\n", + "print(f'T-statistic: {t_stat}, p-value: {p_value}')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Using the `crosstab` function, find the department that has the most hourly workers. " + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "'STREETS & SAN'" + ] + }, + "execution_count": 40, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "cross_tab = pd.crosstab(salaries['Department'], salaries['Salary or Hourly'])\n", + "\n", + "hourly_workers = cross_tab['Hourly']\n", + "\n", + "department_most_hourly = hourly_workers.idxmax()\n", + "\n", + "department_most_hourly" + ] + }, + { + "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": 42, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-statistic: 71.56561089359087, p-value: 1.0\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "\n", + "salaries_streetssan = salaries[salaries['Department']==\"STREETS & SAN\"]['Annual Salary']\n", + "\n", + "salaries_streetssan = salaries_streetssan.dropna()\n", + "\n", + "t_stat, p_value = ttest_1samp(salaries_streetssan, 35.0, alternative='less')\n", + "\n", + "print(f'T-statistic: {t_stat}, p-value: {p_value}')\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "#We cannot reject the null, so there is no evidence that the hourly wage of \n", + "#all streets and san hourly workers is significantly less than $35/hour" + ] + }, + { + "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": 48, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "4 19.86\n", + "6 46.10\n", + "7 35.60\n", + "10 2.65\n", + "18 17.68\n", + " ... \n", + "33164 46.10\n", + "33168 17.68\n", + "33169 35.60\n", + "33174 46.35\n", + "33175 48.85\n", + "Name: Hourly Rate, Length: 8022, dtype: float64" + ] + }, + "execution_count": 48, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "from scipy.stats import t\n", + "\n", + "hourly_wage_workers=salaries[salaries['Salary or Hourly']==\"Hourly\"]['Hourly Rate']\n", + "\n", + "hourly_wage_workers" + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 49, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "hourly_wage_workers.isnull().sum()" + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "95% Confidence Interval for Mean Hourly Wage: (32.52345834488425, 33.05365708767623)\n" + ] + } + ], + "source": [ + "# Calculate mean and standard error\n", + "mean_hourly_wage = hourly_wage_workers.mean()\n", + "std_error_hourly_wage = np.std(hourly_wage_workers, ddof=1) / np.sqrt(len(hourly_wage_workers))\n", + "\n", + "# Confidence level and degrees of freedom\n", + "confidence_level = 0.95\n", + "degrees_of_freedom = len(hourly_wage_workers) - 1\n", + "\n", + "# Calculate the confidence interval\n", + "conf_interval = t.interval(confidence_level, degrees_of_freedom, mean_hourly_wage, std_error_hourly_wage)\n", + "\n", + "print(f\"95% Confidence Interval for Mean Hourly Wage: {conf_interval}\")" + ] + }, + { + "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": 51, + "metadata": {}, + "outputs": [], + "source": [ + "# Your code here:\n", + "\n", + "salaried_police = salaries[(salaries['Salary or Hourly'] == \"Salary\") & (salaries['Department'] == \"POLICE\")]['Annual Salary']\n" + ] + }, + { + "cell_type": "code", + "execution_count": 52, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "95% Confidence Interval for Mean Annual Salary in Police Department: (86177.05631531785, 86795.77269094893)\n" + ] + } + ], + "source": [ + "#Mean and standard error\n", + "mean_salary_police = salaried_police.mean()\n", + "std_error_salary_police = np.std(salaried_police, ddof=1) / np.sqrt(len(salaried_police))\n", + "\n", + "#Confidence level and degrees of freedom\n", + "confidence_level = 0.95\n", + "degrees_of_freedom = len(salaried_police) - 1\n", + "\n", + "#Confidence interval\n", + "conf_interval_salary_police = t.interval(confidence_level, degrees_of_freedom, mean_salary_police, std_error_salary_police)\n", + "\n", + "print(f\"95% Confidence Interval for Mean Annual Salary in Police Department: {conf_interval_salary_police}\")\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": 57, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "8022" + ] + }, + "execution_count": 57, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "from statsmodels.stats.proportion import proportions_ztest\n", + "\n", + "hourly_count = salaries[salaries['Salary or Hourly'] == 'Hourly'].shape[0]\n", + "\n", + "hourly_count" + ] + }, + { + "cell_type": "code", + "execution_count": 60, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "33183" + ] + }, + "execution_count": 60, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Total number of workers\n", + "total_count = salaries.shape[0]\n", + "\n", + "total_count" + ] + }, + { + "cell_type": "code", + "execution_count": 58, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Z-statistic: -3.5099964213703005, p-value: 0.0004481127249057967\n" + ] + } + ], + "source": [ + "# Total number of workers\n", + "total_count = salaries.shape[0]\n", + "\n", + "# Proportion\n", + "null_proportion = 0.25\n", + "\n", + "# Hypothesis test\n", + "z_stat, p_value = proportions_ztest(hourly_count, total_count, value=null_proportion, alternative='two-sided')\n", + "\n", + "print(f'Z-statistic: {z_stat}, p-value: {p_value}')" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "#The proportion of hourly workers in the City of Chicago is significantly different from 25% at a 95% confidence level" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3 (ipykernel)", + "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": "3.11.4" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +}