Clustering Analysis of Universities: Schools in Alabama, California, and Florida According to SAT Averages and Admission Rates

Thank you for checking out my post on Clustering earlier last week. I wanted to expanded on the method with an exercise on clustering universities in the states of Alabama, California, and Florida. I hope you have fun with this one. I know I did!

Let’s begin by loading packages, setting our directory and loading our data set. We will be using the “colleges” data set created by the US Department of Education, and cleaned by Fred Nwanganga and Mike Chapple.

As always, if you wish to follow along, you can download the data set on the Google Drive by clicking on the "Data Sets" tab on the homepage. And like all of my other posts, the code that you can replicate is highlighted in BLUE.

library(tidyverse)

## -- Attaching packages --------------------------------------- tidyverse 1.3.1 --

## v ggplot2 3.3.5     v purrr   0.3.4
## v tibble  3.1.4     v dplyr   1.0.7
## v tidyr   1.1.3     v stringr 1.4.0
## v readr   2.0.1     v forcats 0.5.1

## Warning: package 'stringr' was built under R version 4.1.2

## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()

setwd("C:/Users/firstnamelastinitial/OneDrive/Documents/R")

college <- read_csv("college.csv", col_types = "nccfffffnnnnnnnnn")

Let’s take the time to do some fundamental exploratory data analysis:

dim(college)

## [1] 1270   17

str(college)

## spec_tbl_df [1,270 x 17] (S3: spec_tbl_df/tbl_df/tbl/data.frame)
##  $ id                : num [1:1270] 102669 101648 100830 101879 100858 ...
##  $ name              : chr [1:1270] "Alaska Pacific University" "Marion Military Institute" "Auburn University at Montgomery" "University of North Alabama" ...
##  $ city              : chr [1:1270] "Anchorage" "Marion" "Montgomery" "Florence" ...
##  $ state             : Factor w/ 51 levels "AK","AL","AR",..: 1 2 2 2 2 2 2 2 2 2 ...
##  $ region            : Factor w/ 4 levels "West","South",..: 1 2 2 2 2 2 2 2 2 2 ...
##  $ highest_degree    : Factor w/ 4 levels "Graduate","Associate",..: 1 2 1 1 1 1 1 1 1 1 ...
##  $ control           : Factor w/ 2 levels "Private","Public": 1 2 2 2 2 2 2 1 2 2 ...
##  $ gender            : Factor w/ 3 levels "CoEd","Women",..: 1 1 1 1 1 1 1 1 1 1 ...
##  $ admission_rate    : num [1:1270] 0.421 0.614 0.802 0.679 0.835 ...
##  $ sat_avg           : num [1:1270] 1054 1055 1009 1029 1215 ...
##  $ undergrads        : num [1:1270] 275 433 4304 5485 20514 ...
##  $ tuition           : num [1:1270] 19610 8778 9080 7412 10200 ...
##  $ faculty_salary_avg: num [1:1270] 5804 5916 7255 7424 9487 ...
##  $ loan_default_rate : num [1:1270] 0.077 0.136 0.106 0.111 0.045 0.062 0.096 0.007 0.103 0.063 ...
##  $ median_debt       : num [1:1270] 23250 11500 21335 21500 21831 ...
##  $ lon               : num [1:1270] -149.9 -87.3 -86.3 -87.7 -85.5 ...
##  $ lat               : num [1:1270] 61.2 32.6 32.4 34.8 32.6 ...
##  - attr(*, "spec")=
##   .. cols(
##   ..   id = col_number(),
##   ..   name = col_character(),
##   ..   city = col_character(),
##   ..   state = col_factor(levels = NULL, ordered = FALSE, include_na = FALSE),
##   ..   region = col_factor(levels = NULL, ordered = FALSE, include_na = FALSE),
##   ..   highest_degree = col_factor(levels = NULL, ordered = FALSE, include_na = FALSE),
##   ..   control = col_factor(levels = NULL, ordered = FALSE, include_na = FALSE),
##   ..   gender = col_factor(levels = NULL, ordered = FALSE, include_na = FALSE),
##   ..   admission_rate = col_number(),
##   ..   sat_avg = col_number(),
##   ..   undergrads = col_number(),
##   ..   tuition = col_number(),
##   ..   faculty_salary_avg = col_number(),
##   ..   loan_default_rate = col_number(),
##   ..   median_debt = col_number(),
##   ..   lon = col_number(),
##   ..   lat = col_number()
##   .. )
##  - attr(*, "problems")=<externalptr>

colnames(college)

##  [1] "id"                 "name"               "city"             
##  [4] "state"              "region"             "highest_degree"   
##  [7] "control"            "gender"             "admission_rate"   
## [10] "sat_avg"            "undergrads"         "tuition"          
## [13] "faculty_salary_avg" "loan_default_rate"  "median_debt"      
## [16] "lon"                "lat"

There are a total of 1270 rows, and 17 columns. We can confirm the data types with this function. There are 17 variables.

This time, we will do our clustering analysis on the state of Alabama. So let’s create a new data set with ONLY schools from this state.

alabama_schools <- college %>%
  filter(state == "AL") %>%
  column_to_rownames(var = "name")

After we pass the state code “MD” in the filter() function, we use the column_to_rowname() function which allows us to see the name of the school for our observations. Let’s view the new data set:

View(alabama_schools)

We now have 24 observations, or schools, and 16 variables in our new data set with only schools from Maryland. Now, let’s use the select() function to specify which features we are interested in. 

For this case, we are interested in admission_rate and sat_avg. Sorry for making that decision for you!! 

alabama_schools %>%
  select(admission_rate, sat_avg) %>%
  summary()

##  admission_rate      sat_avg   
##  Min.   :0.4414   Min.   : 811 
##  1st Qu.:0.5309   1st Qu.: 969 
##  Median :0.5927   Median :1035 
##  Mean   :0.6523   Mean   :1033 
##  3rd Qu.:0.8064   3rd Qu.:1109 
##  Max.   :1.0000   Max.   :1219

Before moving on, I would like to observe that we are using the pipe (%>%) operator which allows us to compose cleaner code. Let’s move on:

As expected, we can notice that the range of values between both variables have vastly different ranges. As a reminder, we MUST normalize our variables BEFORE building our model.

alabama_schools_scaled <- alabama_schools %>%
  select(admission_rate, sat_avg) %>%
  scale()

We create a new object, using our alabama_schools data set, that utilizes the scale() function to “normalize” our data to z-scores. This creates an optimal scenario to enable us to use a clustering model.

Let’s see the new values for each variable by piping the summary() function into our new data object.

alabama_schools_scaled %>%
  summary()

##  admission_rate       sat_avg       
##  Min.   :-1.3758   Min.   :-1.92418 
##  1st Qu.:-0.7922   1st Qu.:-0.55343 
##  Median :-0.3884   Median : 0.01916 
##  Mean   : 0.0000   Mean   : 0.00000 
##  3rd Qu.: 1.0051   3rd Qu.: 0.66332 
##  Max.   : 2.2685   Max.   : 1.61547

Our variables have been standardized to z-scores and our data is officially normalized. Remember, by normalizing our data we don’t need to worry about extreme differences between our variables. Let’s load the stats package so we can run the algorithm.

Let’s set the seed so we can have reproducible results.

library(stats)

set.seed(1234)

Notice that we have set our seed to randomize the data. Simply remember the number that you use in the argument so that you can reproduce the results every time you run this algorithm.

Reminder:

The kmeans() functions takes three arguments. Our scaled data set, which recall is alabama_schools_scaled, is the first argument. Centers will determine how many cluster centers we will have. This is essentially our k-value. We are going to set our centers (k-value) to 3, because we want three clusters. The nstart will be set to 25, and this will be the total number of configuration attempts. Let’s run it!

k_3 <- kmeans(alabama_schools_scaled, centers = 3, nstart = 25)

Great! Let’s start to explore this cluster. Let’s see how many observations are in each of the three clusters.

k_3$size

## [1]  8  6 10

This output tells us that one cluster has 8 observations, another has 6 observations, and the third cluster has the remaining 10 observations.

We can also get a sense of the values of the three cluster centers by using the centers attribute.

k_3$centers

##   admission_rate    sat_avg
## 1     -0.6824578  0.5743977
## 2     -0.8282133 -1.2431427
## 3      1.0428942  0.2863675

Let’s break this down per cluster:

The first cluster center has a value of roughly -.68 for admission_rate and .57 for sat_avg.

The second cluster center has a value of roughly -.82 for admission_rate and -1.24 for sat_avg.

The third cluster center has a value of roughly 1.04 for admission_rate and approximately .28 for sat_avg.

THESE VALUES ARE NORMALIZED VALUES OF THE ORIGINAL DATASET!!!

With this in mind, we have lost some ability to interpret this part of the data. This is fine as our objective is to group these schools as accurately as possible.

What good is clustering if we can’t have a visualization of our model. Let’s load another package called factoextra.

library(factoextra)

## Warning: package 'factoextra' was built under R version 4.1.2

## Welcome! Want to learn more? See two factoextra-related books at https://goo.gl/ve3WBa

Next, will pass three arguments into a very powerful function called the fviz_cluster() function. The first argument will be an object where our model is stored, k_3. The second component in the function will be the data set, which is the scaled data set, alabama_schools_scaled. The last element of the function will contain a specification that will help organize our labels.

fviz_cluster(k_3, data = alabama_schools_scaled, repel = TRUE)

## Warning: ggrepel: 1 unlabeled data points (too many overlaps). Consider
## increasing max.overlaps

There is a lot to unpack here, but for brevity, I will describe the clusters as simply as possible.

Cluster 1 is labeled in red and as you can notice it contains schools that have the highest SAT average and lowest admission rates. I would argue that the 8 schools in this cluster are the most elite schools in Alabama as observed by the two metrics that we have used.

Cluster 2 is labeled in green. Universities in this cluster have the lowest SAT average, and are low in admission rates.

Cluster 3 is labeled in blue. Schools in this cluster have high admission rate and have high sat averages; this cluster is similar to cluster 1 with regards to high SAT scores.

I suppose all school clusters will have similar arrangements on two dimensional space if they are measured by same features, sat_avg and admission_rate. 

Let’s see the how the three clusters looks for the state of California, and Florida. Just to test out my hypothesis.

california_schools <- college %>%
  filter(state == "CA") %>%
  column_to_rownames(var = "name")


california_schools %>%
  select(admission_rate, sat_avg) %>%
  summary()

##  admission_rate      sat_avg     
##  Min.   :0.0509   Min.   : 871.0 
##  1st Qu.:0.4051   1st Qu.: 993.5 
##  Median :0.5800   Median :1086.0 
##  Mean   :0.5477   Mean   :1113.7 
##  3rd Qu.:0.7425   3rd Qu.:1222.0 
##  Max.   :0.8750   Max.   :1545.0

california_schools_scaled <- california_schools %>%
  select(admission_rate, sat_avg) %>%
  scale()


set.seed(1234)


k_3 <- kmeans(california_schools_scaled, centers = 3, nstart = 25)


fviz_cluster(k_3, data = california_schools_scaled, repel = TRUE)

## Warning: ggrepel: 58 unlabeled data points (too many overlaps). Consider
## increasing max.overlaps

Now for Florida:

florida_schools <- college %>%
  filter(state == "FL") %>%
  column_to_rownames(var = "name")


florida_schools %>%
  select(admission_rate, sat_avg) %>%
  summary()

##  admission_rate      sat_avg   
##  Min.   :0.2286   Min.   : 803 
##  1st Qu.:0.4481   1st Qu.: 948 
##  Median :0.5188   Median :1065 
##  Mean   :0.5436   Mean   :1057 
##  3rd Qu.:0.6157   3rd Qu.:1153 
##  Max.   :1.0000   Max.   :1330

florida_schools_scaled <- florida_schools %>%
  select(admission_rate, sat_avg) %>%
  scale()


set.seed(1234)


k_3 <- kmeans(florida_schools_scaled, centers = 3, nstart = 25)


fviz_cluster(k_3, data = florida_schools_scaled, repel = TRUE)

## Warning: ggrepel: 19 unlabeled data points (too many overlaps). Consider
## increasing max.overlaps

An observation that I would like to make is that the clusters are labeled different from graphic to graphic. For example, Alabama’s elite schools are labeled red, California’s elite schools are labeled color green, and Florida’s top schools are labeled blue.

Yet, regardless of how the cluster is labeled by color, we can deduce that elite level schools are positioned in the upper left portion of the quadrant.

THANK YOU!

Thank you so much reading this different take on Clustering. This machine learning method is used widely in a Marketing context, namely when managers are looking to categorize, or "segment," groups of costumers.

I am proud of all of you! I know that the steps that you are taking towards a safer, healthier life, are tedious, and at times cumbersome. 

But you know what? 

Your dedication to ridding yourselves of a toxic profession will pay off. You will live happier lives, filled with new found love from family and friends, and even acquaintances who are drawn to your hard work and professional ethic. 

Thank you for allowing me to be part of your journey to becoming data analysts, and data entrepreneurs. 

Read More

Regression Analysis Using Small Data Sets (No CSV File)

 WELCOME

Hello Ladies! I am thrilled that you are coming back for more! Here is a homework assignment that I did for class. I hope that this sample work can show you by example, and not by explanation. If you get frustrated, read through it again, so that you can assimilate the concepts more efficiently. Enjoy!

PROBLEM 1:

 

In this problem, we are given three data sets. The following are the sets that we are given, along with a description of the data sets:

 

Percentage in Poverty: This dataset gives us proportional data about the poverty rates of the 50 states. Each observation in the set represents the poverty rate. For example: the first observation is Alabama, and it has a 20.1 percent poverty rate.

 

Birth Rate 15-16: Similarly, this data set gives us the proportion of 15-16 year olds in each state that have children. More specifically, the first observation will give us 31.5 percent per every 1000 citizens in Alabama give birth to children.

 

Birth Rate 17-18: This data set is nearly identical to the previous data set. The only difference is the age group. Each observation gives us a proportion for every 1000 citizens, in each of the 50 states.

 

This problem will test our assumptions whether there is a correlation between poverty and differing age groups. Is there a relationship? Does poverty (our explanatory variable) impact the Birth Rate per 1000 citizens? 

Can we use a regression line to make predictions for Birth Rates within the various age groups? Do higher amounts of poverty within a given state also explain the differences between the various age groups? Is the difference significant?

 

These are questions that I will attempt to answer in this analysis.

 

Let begin by loading tidyverse (because I love this package) and setting our working directory.

 

library(tidyverse)

## -- Attaching packages --------------------------------------- tidyverse 1.3.1 --

## v ggplot2 3.3.5     v purrr   0.3.4
## v tibble  3.1.4     v dplyr   1.0.7
## v tidyr   1.1.3     v stringr 1.4.0
## v readr   2.0.1     v forcats 0.5.1

## Warning: package 'stringr' was built under R version 4.1.2

## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()

setwd("C:/Users/firstnamelastinitial/Documents/R")

 

Next, I will store three vectors in each appropriate variable.

 

I will call the Percent in Poverty variable: pec_pov

 

I will call the Birth Rate 15-16 variable: br1

 

I will call the Birth Rate 17-18 variable: br2

 

pec_pov <- c(20.1, 7.1, 16.1, 14.9, 16.7, 8.8, 9.7, 10.3, 22, 16.2, 12.1, 10.3, 14.5, 12.4, 9.6, 12.2, 10.8, 14.7, 19.7, 11.2, 10.1, 11, 12.2, 9.2, 23.5, 9.4, 15.3, 9.6, 11.1, 5.3, 7.8, 25.3, 16.5, 12.6, 12, 11.5, 17.1, 11.2, 12.2, 10.6, 19.9, 14.5, 15.5, 17.4, 8.4, 10.3, 10.2, 12.5, 16.7, 8.5, 12.2)

br1 <- c(31.5, 18.9, 35, 31.6, 22.6, 26.2, 14.1, 24.7, 44.8, 23.2, 31.4, 17.7, 18.4, 23.4, 22.6, 16.4, 21.4, 26.5, 31.7, 11.9, 20, 12.5, 18, 14.2, 37.6, 22.2, 17.8, 18.3, 28, 8.1, 14.7, 37.8, 15.7, 28.6, 11.7, 20.1, 30.1, 18.2, 17.2, 19.6, 29.2, 17.3, 28.2, 38.2, 17.8, 10.4, 19, 16.8, 21.5, 15.9, 17.7)

br2 <- c(88.7, 73.7, 102.5, 101.7, 69.1, 79.1, 45.1, 77.8, 101.5, 78.4, 92.8, 66.4, 69.1, 70.5, 78.5, 55.4, 74.2, 84.8, 96.1, 45.2, 59.6, 39.6, 60.8, 47.3, 103.3, 76.6, 63.3, 64.2, 96.7, 39, 46.1, 99.5, 50.1, 89.3, 48.7, 69.4, 97.6, 64.8, 53.7, 59, 87.2, 67.8, 94.2, 104.3, 62.4, 44.4, 66, 57.6, 80.7, 57.1, 72.1)

 

I shall make two scatter plots, comparing pec_pov and br1, and pec_pov and br2. I don’t have enough information at this point to determine if regression analysis is feasible or even appropriate, so I will see if there is a pattern in the data. I will also add titles to the plots, as well as a regression line to see if regression is appropriate.

 

plot(pec_pov, br1,
     main = "Does Poverty Explain Birth rates for 15-16 Year Olds in America",
     xlab = "Poverty Percentage in Each State",
     ylab = "Birth Rate for 15-16 Year Old per 1,000")
abline(lm(br1~pec_pov))

This first plot appears to be linear. I need to do further investigating to be certain of this. I will plot the second scatter plot, with the variables pec_pov and br2. As I did previously, I will label each axis and generate a regression line.

 

plot(pec_pov, br2,
     main = "Does Poverty Explain Birth rates for 17-18 Year Olds in America",
     xlab = "Poverty Percentage in Each State",
     ylab = "Birth Rate for 17-18 Year Olds per 1,000")
abline(lm(br2~pec_pov))

This plot also appears to follow a linear pattern. Yet, I am still not sure if regression is appropriate for this data set. I need to investigate further.

 

POVERTY VS BIRTH RATES 15-16 ANALYSIS AND DISCUSSION

 

As part of my analysis, I will program a model for both graphs to see if I can calculate an appropriate R value. This will give me more confidence that regression can in fact be used or not.

 

The model for pec_pov and br1 will look like such:

 

mod1 <- lm(br1~pec_pov)

 

Now I can call up the summary of the model:

 

summary(mod1)

##
## Call:
## lm(formula = br1 ~ pec_pov)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -11.2275  -3.6554  -0.0407   2.4972  10.5152
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)   
## (Intercept)   4.2673     2.5297   1.687    0.098 . 
## pec_pov       1.3733     0.1835   7.483 1.19e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.551 on 49 degrees of freedom
## Multiple R-squared:  0.5333, Adjusted R-squared:  0.5238
## F-statistic:    56 on 1 and 49 DF,  p-value: 1.188e-09

 

I notice in the Residuals section of the summary that our Adjusted R-Squared Score is .5238. So, in order to obtain the R value, I need to perform the square root of this value.

 

sqrt(.5238)

## [1] 0.7237403

 

Our R value is .7237, which tells us that our R value has a strong positive correlation. This means that there is a high correlation between the explanatory variable (pec_pov) and the response variable br1. 

We can be confident in assuming that as poverty increases, the birth rates for ages 15-16 will increase as well. Likewise, we can also be enthusiastic that regression analysis is appropriate for this type of data.

 

Although, it is worth mentioning that Poverty only explains roughly 52.4% of birth rates from ages 15-16. Conversely, the use of other variables might make our model more predictive, like income, ethnicity, socioeconomic status, etc. We can be confident that 47.6% of the model is NOT being explained by the two variables that we have used in our model, called mod1.

 

Our equation for this model looks like this:

 

Birth Rate for 15-16 ^ = 4.2673 + 1.3733*(Percentage in Poverty)

 

So, if I wanted to predict what the birth rate would be if the poverty rate was 20, then I would use the following equation:

 

y <- 4.2673 +  1.3733*(20)
print(y)

## [1] 31.7333

This means that if the poverty rate was 20% then the birth rate of 15-16 year olds would be 31.73% per 1000 citizens in this age group.

 

In summation, the model that we have created for these variables is strong, BUT it can be stronger, if we added more variables to the regression equation (as in multivariate regression). 

Maybe there is a variable that explains birth rates much better than poverty. Does the LOCATION of the state, or even CLIMATE have any predictive power in a regression model? We need more data and more features to ensure that this model has a better predictive influence.

 

I will create a residual plot now to see if any of my analysis held merit.

 

plot(mod1)

 

The residual plot supports that regression analysis was appropriate, as the data points in the plots do NOT conform to a specific or obvious pattern. But, I would like to add that other variables should be used to explore a better model that can explain more of the response variable. However, this is a good start.

 

I will now conduct the same analysis for the Birth Rate 17-18 and Percentage in Poverty and conclude this report with Managerial Implications for Laypersons involved in policy making.

 

POVERTY VS BIRTH RATES 17-18 ANALYSIS AND DISCUSSION

 

Now, I will run a regression model with Birth Rates for 17-18 year olds per 1000 citizens.

 

mod2 <- lm(br2~pec_pov)
summary(mod2)

##
## Call:
## lm(formula = br2 ~ pec_pov)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -31.668 -10.541  -1.611  11.381  30.496
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)   
## (Intercept)  34.2124     6.6414   5.151 4.59e-06 ***
## pec_pov       2.8822     0.4818   5.982 2.50e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 14.57 on 49 degrees of freedom
## Multiple R-squared:  0.4221, Adjusted R-squared:  0.4103
## F-statistic: 35.78 on 1 and 49 DF,  p-value: 2.495e-07

 

Immediately, I notice that our Adjusted R^2 Score is less than the previous age group. So, I am assuming that our R value will be less as well. Let me confirm this:

 

sqrt(0.4103)

## [1] 0.6405466

 

My assumption was correct! The R value is much lower, at roughly (rounded up) to .641. Despite this lower number, our R value still represents a positive moderate correlation. Meaning that the percentage of poverty in the state DOES impact Birth rates per 1000 citizens of 17-18 year olds.

 

We can further deduce that we ABSOLUTELY REQUIRE more variables to improve the predictive power of our model. Unless we find an explanatory variable that has a higher correlation than poverty, we will need multiple variables to explain the Birth Rates in this age group.

 

Our equation for these two variables looks like this:

 

Birth Rates 17-18^ = 34.2124 + 2.8822*(Percentage in Poverty)

 

I am now going to create a residual plot to see if my regression analysis holds some merit.

 

plot(mod2)

I am on the fence here! I can see that the plot titled, “Residuals vs. Fitted,” has a subtle pattern. IT IS NOT OBVIOUS, but I can see that there are more observations to the left of the value 80, and less observations passed the value of 80. At this point, I can conclude that I would need to go beyond a regression model or make use of other variables.

 

I have also neglected to mention that I have not removed any outliers or influence points from the vectors that I created in the beginning of this analysis. The main reason for this is because our sample size is only 50 (n=50). 

Since our sample set is relatively smaller, I didn’t want to take away any information that could be valuable to me in this analysis. Yet, one could also argue that the removal of three, perhaps 4, outliers and influence points, might enable our model’s predictive power to be higher.

 

If our R^2 score was higher in this case, would that necessarily translate well in a real world setting, in which human lives are so vital in the model’s predictive outcome? I lack the experience and knowledge to answer this question at this time.

 

Now for the Managerial Implications.

 

MANAGERIAL IMPLICATIONS FOR POLICY MAKERS (HYPOTHETICAL LECTURE)

 

To all of the policy makers who have gathered today at this TedTalk, I beg you all to hear me out on some key points that I have discovered when it comes to birth rates in the US.

 

Policy makers who create birth control programs will need to be receptive to that fact that each state has different poverty levels. Additionally, we can be confident that the Percentage in Poverty does explain birth rates in both ages groups, from 15-16 and 17-18 years of age.

 

HOWEVER, and this is important, Poverty is NOT the only variable that can explain Birth Rates amongst these age groups. We need to collect more data, and consider more, and perhaps, use multiple variables to create regression models that you can build your programs around.

 

Policy makers can be confident in using birth control programs to target those living in states, where income levels are 20% or less for the 15-16 year age group. Although, there are no guarantees and certainties in statistics, we can be confident of this fact.

 

hist(br1,
     main = "Distribution of Birth Rates per 1000",
     xlab = "Proportion of Birthday Rates per 1000 Citizens")

The histogram below shows the distribution of Birth Rates for 17-18 Years Old. You might note that these proportions are higher for this age group than for 15-16 year olds. I am going to make a hypothesis that the main reason for the disparity between groups is because 17-18 year olds are of legal age to drive, and have driver’s licenses. You might consider having flyers or promotional birth control documents and solutions at DMVs, and even hotels–these kids need some place private to “get it on.”

 

hist(br2)

Thank you very much for attending my Ted Talk today. I hope my advice is enough to persuade you to take action!

THANK YOU!

Thank you so very much for coming to my blog to learn data analytics and machine learning. I know that you are making a tremendous effort to learn this material, and that some of it may be foreign to you. 

This fine! Take your time with it. Go at a pace that is comfortable to you, and enjoy the ride!

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