diff --git a/your-code/maiin.ipynb b/your-code/maiin.ipynb new file mode 100644 index 0000000..19775ad --- /dev/null +++ b/your-code/maiin.ipynb @@ -0,0 +1,744 @@ +{ + "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": 38, + "metadata": {}, + "outputs": [], + "source": [ + "# import numpy and pandas\n", + "\n", + "import pandas as pd\n", + "import numpy as np\n", + "import scipy.stats as st\n", + "import matplotlib.pyplot as plt\n", + "import seaborn as sns\n", + "from scipy.stats import ttest_1samp" + ] + }, + { + "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": 39, + "metadata": {}, + "outputs": [ + { + "data": { + "text/html": [ + "
\n", + "\n", + "\n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + "
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
...........................
33178ZYLINSKA, KATARZYNAPOLICE OFFICERPOLICEFSalaryNaN72510.0NaN
33179ZYMANTAS, LAURA CPOLICE OFFICERPOLICEFSalaryNaN48078.0NaN
33180ZYMANTAS, MARK EPOLICE OFFICERPOLICEFSalaryNaN90024.0NaN
33181ZYRKOWSKI, CARLO EPOLICE OFFICERPOLICEFSalaryNaN93354.0NaN
33182ZYSKOWSKI, DARIUSZCHIEF DATA BASE ANALYSTDoITFSalaryNaN115932.0NaN
\n", + "

33183 rows × 8 columns

\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", + "... ... ... \n", + "33178 ZYLINSKA, KATARZYNA POLICE OFFICER \n", + "33179 ZYMANTAS, LAURA C POLICE OFFICER \n", + "33180 ZYMANTAS, MARK E POLICE OFFICER \n", + "33181 ZYRKOWSKI, CARLO E POLICE OFFICER \n", + "33182 ZYSKOWSKI, DARIUSZ CHIEF DATA BASE ANALYST \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", + "33178 POLICE F Salary NaN \n", + "33179 POLICE F Salary NaN \n", + "33180 POLICE F Salary NaN \n", + "33181 POLICE F Salary NaN \n", + "33182 DoIT 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", + "... ... ... \n", + "33178 72510.0 NaN \n", + "33179 48078.0 NaN \n", + "33180 90024.0 NaN \n", + "33181 93354.0 NaN \n", + "33182 115932.0 NaN \n", + "\n", + "[33183 rows x 8 columns]" + ] + }, + "execution_count": 39, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "salaries = pd.read_csv(r\"C:\\Users\\dulce\\OneDrive\\Documentos\\Ironhack git\\Labs\\Labs week 5\\lab-hypothesis-testing-1\\your-code\\Current_Employee_Names__Salaries__and_Position_Titles.csv\")\n", + "salaries" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Examine the `salaries` dataset using the `head` function below." + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": {}, + "outputs": [ + { + "data": { + "text/html": [ + "
\n", + "\n", + "\n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + "
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
\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", + "\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": 40, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "salaries.head()" + ] + }, + { + "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": 41, + "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": 41, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "salaries.isnull().sum()" + ] + }, + { + "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": 42, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "33183" + ] + }, + "execution_count": 42, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "salaries['Salary or Hourly'].count()" + ] + }, + { + "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": 43, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Name\n", + "Job Titles\n", + "Department\n", + "Full or Part-Time\n", + "Salary or Hourly\n", + "Typical Hours\n", + "Annual Salary\n", + "Hourly Rate\n" + ] + } + ], + "source": [ + "for Department in salaries:\n", + " print(Department)" + ] + }, + { + "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": 87, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "20.6198057854942\n", + "Reject the null hypothesis.\n" + ] + } + ], + "source": [ + "alpha = 0.05\n", + "\n", + "t_statistic, p_value = ttest_1samp(hourly_rate, 30)\n", + "\n", + "print(t_statistic)\n", + "\n", + "if p_value < alpha:\n", + " print(\"Reject the null hypothesis.\")\n", + "else:\n", + " print(\"Fail to reject the null hypothesis.\")" + ] + }, + { + "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": 44, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reject the null hypothesis: Police salaries are significantly higher.\n" + ] + } + ], + "source": [ + "alpha = 0.05 \n", + "\n", + "police_salaries = salaries[(salaries['Department'] == 'POLICE') & (salaries['Salary or Hourly'] == 'Salary')]\n", + "\n", + "police_salaries = police_salaries['Annual Salary']\n", + "\n", + "result = ttest_1samp(police_salaries, 86000, alternative='greater')\n", + "\n", + "p_value = result.pvalue / 2\n", + "\n", + "if p_value < alpha:\n", + " print(\"Reject the null hypothesis: Police salaries are significantly higher.\")\n", + "else:\n", + " print(\"Fail to reject the null hypothesis.\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Using the `crosstab` function, find the department that has the most hourly workers. " + ] + }, + { + "cell_type": "code", + "execution_count": 45, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "'STREETS & SAN'" + ] + }, + "execution_count": 45, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "cross_table = pd.crosstab(salaries['Department'], salaries['Salary or Hourly'])\n", + "\n", + "most_hourly_department = contingency_table['Hourly'].idxmax()\n", + "\n", + "most_hourly_department" + ] + }, + { + "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": 89, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "-16.352363851001144\n", + "Reject the null hypothesis.\n" + ] + } + ], + "source": [ + "alpha = 0.05\n", + "\n", + "t_statistic, p_value = stats.ttest_1samp(hourly_rate, 35)\n", + "\n", + "print(t_statistic)\n", + "\n", + "if p_value < alpha:\n", + " print(\"Reject the null hypothesis.\")\n", + "else:\n", + " print(\"Fail to reject the null hypothesis.\")" + ] + }, + { + "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": 90, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "(32.52345834488425, 33.05365708767623)" + ] + }, + "execution_count": 90, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "mean = hourly.mean()\n", + "std_error = stats.sem(hourly)\n", + "\n", + "confidence_interval = stats.t.interval(0.95, len(hourly_rate)-1, mean, std_error)\n", + "\n", + "confidence_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": null, + "metadata": {}, + "outputs": [], + "source": [ + "hourly = salaries[salaries['Salary or Hourly'] == 'Hourly']\n", + "\n", + "hourly_rate = hourly['Hourly Rate']\n", + "\n", + "mean = hourly_rate.mean()\n", + "std_error = stats.sem(hourly_rate)\n", + "\n", + "confidence_interval = stats.t.interval(0.95, len(hourly_rate)-1, mean, std_error)\n", + "\n", + "confidence_interval" + ] + }, + { + "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 (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": "3.10.9" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +}