From 73565508b4bdd3a1e3e7d71c8fbbc34ac23293bd Mon Sep 17 00:00:00 2001 From: Yuliya Lavrenyuk Date: Sun, 18 Aug 2024 21:19:27 +0200 Subject: [PATCH] yuliya_lavrenyuk --- your-code/main.ipynb | 1157 ++++++++++++++++++++++++++++++++---------- 1 file changed, 878 insertions(+), 279 deletions(-) diff --git a/your-code/main.ipynb b/your-code/main.ipynb index b2b6f8d..d9a7e2b 100644 --- a/your-code/main.ipynb +++ b/your-code/main.ipynb @@ -1,279 +1,878 @@ -{ - "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": 76, + "metadata": {}, + "outputs": [], + "source": [ + "# import numpy and pandas\n", + "import pandas as pd\n", + "import numpy as np\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": 77, + "metadata": {}, + "outputs": [], + "source": [ + "# Your code here:\n", + "df = 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": 78, + "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": 78, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "df.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": 79, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + "RangeIndex: 33183 entries, 0 to 33182\n", + "Data columns (total 8 columns):\n", + " # Column Non-Null Count Dtype \n", + "--- ------ -------------- ----- \n", + " 0 Name 33183 non-null object \n", + " 1 Job Titles 33183 non-null object \n", + " 2 Department 33183 non-null object \n", + " 3 Full or Part-Time 33183 non-null object \n", + " 4 Salary or Hourly 33183 non-null object \n", + " 5 Typical Hours 8022 non-null float64\n", + " 6 Annual Salary 25161 non-null float64\n", + " 7 Hourly Rate 8022 non-null float64\n", + "dtypes: float64(3), object(5)\n", + "memory usage: 2.0+ MB\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "df.info()\n" + ] + }, + { + "cell_type": "code", + "execution_count": 80, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Index(['Name', 'Job Titles', 'Department', 'Full or Part-Time',\n", + " 'Salary or Hourly', 'Typical Hours', 'Annual Salary', 'Hourly Rate'],\n", + " dtype='object')" + ] + }, + "execution_count": 80, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "columns = df.columns\n", + "columns" + ] + }, + { + "cell_type": "code", + "execution_count": 81, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "column \"Name\" contains 0 missing values\n", + "column \"Job Titles\" contains 0 missing values\n", + "column \"Department\" contains 0 missing values\n", + "column \"Full or Part-Time\" contains 0 missing values\n", + "column \"Salary or Hourly\" contains 0 missing values\n", + "column \"Typical Hours\" contains 25161 missing values\n", + "column \"Annual Salary\" contains 8022 missing values\n", + "column \"Hourly Rate\" contains 25161 missing values\n" + ] + } + ], + "source": [ + "for column in columns:\n", + " nulls = df[column].isnull()\n", + " print(f'column \"{column}\" contains {nulls.sum()} missing 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": 82, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Salary or Hourly\n", + "Salary 25161\n", + "Hourly 8022\n", + "Name: count, dtype: int64" + ] + }, + "execution_count": 82, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "df['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": 83, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Department\n", + "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: count, dtype: int64" + ] + }, + "execution_count": 83, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "df['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": 84, + "metadata": {}, + "outputs": [], + "source": [ + "# Null Hypothesis: The mean hourly wage is $30/hr.\n", + "# Alternative Hypothesis: The mean hourly wage is not $30/hr." + ] + }, + { + "cell_type": "code", + "execution_count": 85, + "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": 85, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "hourly_rate = df[df[\"Salary or Hourly\"] =='Hourly']['Hourly Rate']\n", + "\n", + "hourly_rate" + ] + }, + { + "cell_type": "code", + "execution_count": 86, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-Statistic: 20.6198057854942\n", + "P-Value: 4.3230240486229894e-92\n", + "We reject the null hypothesis. The mean hourly wage is significantly different from $30/hr.\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "import scipy.stats as st\n", + "\n", + "t_statistic, p_value = st.ttest_1samp(df[df[\"Salary or Hourly\"] =='Hourly']['Hourly Rate'], 30)\n", + "\n", + "print(f\"T-Statistic: {t_statistic}\")\n", + "print(f\"P-Value: {p_value}\")\n", + "\n", + "# Check if we reject the null hypothesis at the 95% confidence level\n", + "alpha = 0.05\n", + "if p_value < alpha:\n", + " print(\"We reject the null hypothesis. The mean hourly wage is significantly different from $30/hr.\")\n", + "else:\n", + " print(\"We fail to reject the null hypothesis. There is no significant difference between the mean hourly wage and $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": 97, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-Statistic: 3.081997005712994\n", + "Two-Tailed P-Value: 0.0020603403550965137\n", + "One-Tailed P-Value: 0.0010301701775482569\n", + "We reject the null hypothesis. The mean salary this year is significantly greater than $86,000/year.\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "police_salaries = df[(df['Department'] == 'POLICE') & (df['Salary or Hourly'] == 'Salary')]['Annual Salary']\n", + "\n", + "last_year_mean = 86000\n", + "\n", + "# Perform the one-sample t-test\n", + "t_statistic, two_tailed_p_value = st.ttest_1samp(police_salaries, last_year_mean)\n", + "\n", + "# Convert the two-tailed p-value to a one-tailed p-value\n", + "one_tailed_p_value = two_tailed_p_value / 2\n", + "\n", + "print(f\"T-Statistic: {t_statistic}\")\n", + "print(f\"Two-Tailed P-Value: {two_tailed_p_value}\")\n", + "print(f\"One-Tailed P-Value: {one_tailed_p_value}\")\n", + "\n", + "# Check if we reject the null hypothesis at the 95% confidence level\n", + "alpha = 0.05\n", + "if t_statistic > 0 and one_tailed_p_value < alpha:\n", + " print(\"We reject the null hypothesis. The mean salary this year is significantly greater than $86,000/year.\")\n", + "else:\n", + " print(\"We fail to reject the null hypothesis. There is no significant evidence that the mean salary this year is greater than $86,000/year.\")\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Using the `crosstab` function, find the department that has the most hourly workers. " + ] + }, + { + "cell_type": "code", + "execution_count": 104, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "STREETS & SAN 1862\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "hourly_workers = df[df['Salary or Hourly'] == 'Hourly']\n", + "crosstab = pd.crosstab(hourly_workers['Department'], hourly_workers['Salary or Hourly'])\n", + "most_hourly_workers_department = crosstab.idxmax()[0]\n", + "most_hourly_workers_count = crosstab.max()[0]\n", + "\n", + "print(most_hourly_workers_department,most_hourly_workers_count)" + ] + }, + { + "cell_type": "code", + "execution_count": 102, + "metadata": {}, + "outputs": [ + { + "data": { + "text/html": [ + "
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POLICE10
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STREETS & SAN1862
TRANSPORTN725
WATER MGMNT1513
\n", + "
" + ], + "text/plain": [ + "Salary or Hourly Hourly\n", + "Department \n", + "ANIMAL CONTRL 19\n", + "AVIATION 1082\n", + "BUDGET & MGMT 2\n", + "BUSINESS AFFAIRS 7\n", + "CITY COUNCIL 64\n", + "COMMUNITY DEVELOPMENT 4\n", + "CULTURAL AFFAIRS 7\n", + "FAMILY & SUPPORT 287\n", + "FINANCE 44\n", + "FIRE 2\n", + "GENERAL SERVICES 765\n", + "HEALTH 3\n", + "HUMAN RESOURCES 4\n", + "LAW 40\n", + "MAYOR'S OFFICE 8\n", + "OEMC 1273\n", + "POLICE 10\n", + "PROCUREMENT 2\n", + "PUBLIC LIBRARY 299\n", + "STREETS & SAN 1862\n", + "TRANSPORTN 725\n", + "WATER MGMNT 1513" + ] + }, + "execution_count": 102, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "crosstab" + ] + }, + { + "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": 107, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T-Statistic: -9.567447887848152\n", + "Two-Tailed P-Value: 3.3378530564707717e-21\n", + "One-Tailed P-Value: 1.6689265282353859e-21\n", + "We reject the null hypothesis. The mean hourly wage is significantly less than $35/hour.\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "most_hourly_dep = df[(df['Department'] == 'STREETS & SAN') & (df['Salary or Hourly'] == 'Hourly')]['Hourly Rate']\n", + "\n", + "hourly_wage = 35\n", + "\n", + "# Perform the one-sample t-test\n", + "t_statistic, two_tailed_p_value = st.ttest_1samp(most_hourly_dep, hourly_wage)\n", + "\n", + "# Convert the two-tailed p-value to a one-tailed p-value\n", + "one_tailed_p_value = two_tailed_p_value / 2\n", + "\n", + "print(f\"T-Statistic: {t_statistic}\")\n", + "print(f\"Two-Tailed P-Value: {two_tailed_p_value}\")\n", + "print(f\"One-Tailed P-Value: {one_tailed_p_value}\")\n", + "\n", + "# Check if we reject the null hypothesis at the 95% confidence level\n", + "alpha = 0.05\n", + "if t_statistic < 0 and one_tailed_p_value < alpha:\n", + " print(\"We reject the null hypothesis. The mean hourly wage is significantly less than $35/hour.\")\n", + "else:\n", + " print(\"We fail to reject the null hypothesis. There is no significant evidence that the mean hourly wage is less than $35/hour.\")\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": 108, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Sample Mean: 32.78855771628024\n", + "Standard Error: 0.1352368565101596\n", + "95% Confidence Interval: (32.52345834488425, 33.05365708767623)\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "hourly_workers = df[df['Salary or Hourly'] == 'Hourly']['Hourly Rate'].dropna()\n", + "\n", + "sample_mean = hourly_workers.mean()\n", + "\n", + "standard_error = st.sem(hourly_workers)\n", + "\n", + "degrees_of_freedom = len(hourly_workers) - 1\n", + "\n", + "confidence_level = 0.95\n", + "confidence_interval = st.t.interval(confidence_level, degrees_of_freedom, loc=sample_mean, scale=standard_error)\n", + "\n", + "print(f\"Sample Mean: {sample_mean}\")\n", + "print(f\"Standard Error: {standard_error}\")\n", + "print(f\"95% Confidence Interval: {confidence_interval}\")\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": 109, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Sample Mean: 86486.41450313339\n", + "Standard Error: 157.82445675052244\n", + "95% Confidence Interval: (86177.05631531784, 86795.77269094894)\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "police_salaries = df[(df['Department'] == 'POLICE') & (df['Salary or Hourly'] == 'Salary')]['Annual Salary']\n", + "\n", + "sample_mean = police_salaries.mean()\n", + "\n", + "standard_error = st.sem(police_salaries)\n", + "\n", + "degrees_of_freedom = len(police_salaries) - 1\n", + "\n", + "confidence_level = 0.95\n", + "confidence_interval = st.t.interval(confidence_level, degrees_of_freedom, loc=sample_mean, scale=standard_error)\n", + "\n", + "print(f\"Sample Mean: {sample_mean}\")\n", + "print(f\"Standard Error: {standard_error}\")\n", + "print(f\"95% Confidence Interval: {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": 92, + "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": "3.10.11" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +}