Assinment 6

Regression Analysis

Definitions:

Regression Analysis: Used to find the relationship between two variables, and also to investigate the causation.
Coefficient of Determination (r2): shows how much X explains Y/ ranges from 0(no strength) to 1(very strong)

Regression Equation:

• a= the constant: the point where the best fit line crosses the Y axis or when x=0
• b= the slope of the line or the Regression Coefficient: shows how responsive the dependent variable is to change in the independent variable.
• Y= dependent variable
• X= independent variable

Ordinary Least Squares (OLS): Fighting a straight line through a set of points in such a way that the sum of the squared vertical distances from the observed points to the fitted line is minimized.

Dependent: What is explains by the independent variable always found on the Y axis.

Independent: What explains the dependent variable always found on the X axis. 

Part I:

Introduction: There was a claim by Town X's local news that as the number of kids that receive free lunches increases so does crime. This claim was based on data from a study the town conducted on crime rates and poverty per 100,000 people. This claim seems to put a far stretch for the data given. Running a regression equation would be a good way to either falsify or approve this claim by the local news. SPSS is a good medium to run the regression equation of if 30% of a new area of town gets free lunch what would the corresponding crime rate be. As well as to find out how confident the results are from this question.

Methodology: To complete this task you would run a regression analysis in SPSS. First realize that crime rate is the dependent variable while kids with free lunches is the independent variable. Then you would find your regression equation from the analysis finding the constant and the slope, and then plug in the 30% for the X value to find the Y value of the corresponding crime rate.

Results: The results show a R squared of .173 which means that there very little strength showing that the independent variable (free lunches) explains the dependent variable (crime rate). Also from the results we see that the constant is 21.819 and the slope is a positive 1.685.

Conclusions: We can conclude by these results that the claim by the news station was incorrect. The coefficient of determination is .173 on a 0 to 1 scale of strength meaning that it is weak. Other factors to take into account is that the significance level is .005 meaning that the two are correlated, but you can not tell causation from this. In terms of the new are with 30% free lunch the corresponding crime rate would be 72.5%, but I am not very confident in the results because of how weak the coefficient of determination value was. 

Part II:

Introduction: Being provided with 911 calls for Portland, OR , a company is curious as to what factors provide explanations about response times to these calls. With this information they are looking to build a new ER and would like to know how large to build it along with where the best location for it might be. The size of the ER is a question that is unanswerable with the data given, but the question that will be answered through this section is the best location to build this new ER.
The other data given was:

• Number of 911 calls per census tract
• Jobs
• Renters
• Low education: Number of people with no High School Degree
• Alcohol sales
• Unemployed
• Foreign Born population
• Medium Income
• Number of College graduates

Choosing three of many data given analyze those three using a Regression Analysis to figure out the relationships.

Methods: I ran each independent variable through SPSS using the Regression Analysis where the dependent variable is the Calls data. Shown in Figure 1 is what is run in SPSS for each three with the independent variable being the one changed out.

Figure 1: Linear Regression Analysis Sample

After going through all the Regression Analyzes then I found the variable with the highest R-squared value. Then I made a Choropleth Map of just the 911 calls along with a Residual Map of the the highest R-squared value which was for the variable of Renters. To make the Residual Map you go into the Spatial Statistics Tools and make it in Ordinary Least Squares.

Results:

Figure 2: Low Education Regression Analysis

Figure 3: Renters Regression Analysis

-This was the Independent Variable with the highest Coefficient of Determination which was .616 as the r squared. The hypothesis for this would be:

Null Hypothesis= There is no linear relationship between 911 Calls and Renters

Alternative Hypothesis= There is a linear relationship between 911 Calls and Renters

-This would reject the null hypothesis because of the significance being .000  meaning there is a linear relationship between the two. This is also true for the other two independent variables I chose, so they all had linear relationships with 911 Calls, but Renters just had a higher Coefficient of Determination. 

Figure 4: Unemployment Regression Analysis 

Figure 5: Choropleth Map of 911 Calls

The Blue and Green areas are that of higher 911 calls. So they are above the trend-line.

Figure 6: Standard Deviation for Choropleth Map

Figure 7: Standard Residuals Map of Renters

The reddish to orange color is areas that above the trend-line meaning they are higher than the average for Renters. This correlates with Figure 5 where the blue colors are on that map of higher average 911 calls these two have a higher relationship in that area.

Conclusions: For the first question of what factors provide response times to 911 calls. My answer to that in terms of my independent variable would be that the number of response calls is highly correlated with the Renters in the area. The places of highest 911 calls is where the higher averages are for renters seen in Figure 7. For the second question about where the best location is for the ER I chose the spot represented by the circle on Figure 7. It was in between the two areas of the highest points from away from the trend-line like I discussed in the results section about the Standard Residuals Map along with it being centered among most of the reddish/peach areas. The ER should be fairly big in size compared to if it was located else where in Portland, but like stated in the Introduction there is no way to actually answer the size question. 

Assignment 5

Assignment 5

Correlation and Spatial Auto-correlation 

Part I: Correlation 

Definitions:

Correlation= Measures the association between pairs of variables, but does not imply causation

-There is two factors that go into correlation, one is strength and the other is direction

Strength= The strength between two variables ranges from a positive to a negative one where positive one is the strongest.

Direction= The relationship can either be positive, null (no difference), or negative. 

Co-variation= the measure of how two variables change together

Hypothesis Testing with Correlation= To determine if an association exists between two variables 

Spatial Auto-correlation= Correlation of a variable with its self through space. So if neighboring areas are more a like it is a positive and negative if unalike. 

Moran's I: A chart of 4 quadrants of comparison ranging between positive and negative 1 showing the strength of the auto correlation. High, High = (+,+), High, Low = (+,-), Low, High = (-,+), Low, Low = (-,-). 

LISA map: The purpose of these maps is show a spatial component of the spatial auto-correlation. 4 categories of color that correlate with the strength of the relationships described in Moran's I. Red related to high and blue related to low. 

1. Using the data in Image 1, create a scatter-plot and place a trend line on that scatter-plot in Excel. The scatter-plot created is seen in Image 2. To find the Pearson correlation use SPSS which is shown in Image 3. 

-Null Hypothesis= There is no significant difference between distance and sound level

-Alternative Hypothesis= This is a significant difference between distance and sound level

Image 1: Scatter-plot Data

Image 2: Scatter-plot

Image 3: Correlation

Results Summary: From Image 2 and Image 3 you will find that there is a significant (because of the two asterisks) strong relationship because of the Pearson Correlation being -.896 along with the direction being negative because of the -.896 as well. The Sig 2-tailed test is .000, so that means that you reject the null hypothesis which is there is no significant difference between distance and sound level. 

2. Census Tracts and Population in Detroit, Mi

Results:
From this correlation matrix of Detroit there are a lot of things to infer. There are only two categorizes for whites that were not significant and that was the manufacturing and Finance categories. Looking at there sig 2-tailed test they both failed to reject the null  hypothesis which make sense then. For the black population all the categories have significant relationships besides that of the number of Finance Employees. For the Asian population all the relationships are significant besides with the Hispanic population category. With the Hispanic population there were five categorizes that were not significant with it. The categories were Asians, number of Bachelor's Degrees, number of manufacturing employees, number of retail employees, and number of Finance employees. There are certain correlations that stand out in terms of strength and direction. There is the strongest positive relationship between Asians and number of Retail employees, and Blacks had the strongest negative relationship. So, the probability you will see Asians being employed in retail is far higher than Blacks in Detroit.

Part II: Spatial Auto-correlation

Introduction: Given data of the percent Democratic votes and voter turnout from the Texas Election Commission for the years 1980 and 2012 Presidential Elections has there been any clustering patterns of these two values over the last 32 years? Using SPSS and GeoDa to answer this question and report the findings for the Texas Election Commission. Also, by using the online Census Data compare this to Hispanic population of Texas counties in 2010. This will all help to determine if there is a spatial auto-correlations for each of the elections with voting and voting turnout.

Method: To start off with this task for the TEC first you download the shape-file of Texas and the data of Hispanic Population 2010 from the online U.S. Census. For the Hispanic Data only the population percent is necessary so deleting row A2 and the rest of the columns is an important step to take once you have the excel spreadsheet of the data opened. In ArcMaps, open the shape-file of Texas downloaded from the Census, and join the Hispanic data along with the Texas voting patterns data provided by TEC to that Texas shape-file. While doing this make sure to use Geo_ID for joining. After joining export that shape-file with the data as a new shape-file so it is ready to use in GeoDa. Open GeoDa and click on "New Project From" and open this new Texas shape-file. You then have to create something refereed to as a "Spatial Weight".To do this go to tools, then create weight manager and ADD ID VARIABLE. Then under contiguity weight select Rook contiguity, and then hit create. After this is done begin to create Moran's I charts along with LISA cluster maps(make sure they are Uni-variate for both Moran's and LISA) for percent votes in 1980 and 2012 along with voter turnout in 1980 and 2012. Do this as well for Hispanic Voters, along with make a weights matrix's for this as well.

Results: Below are each of the Moran's I charts and each of the LISA cluster maps for voting and voter turnouts. There is also provided a Moran's I chart and LISA cluster map for Hispanic population. The last thing included is a weights matrix's for all of the categories.

Image 4: LISA map of Percent Democratic vote for 1980

Image 5: Moran's I chart of percent Democratic Vote for 1980

Image 6: LISA map percent Democratic Vote for 2012

Image 7: Moran's I chart of percent Democratic Vote for 2012

Image 8: LISA map of Voter Turnout for 1980

Image 9: Moran's map of Voter Turnout for 1980

Image 10: LISA map of Voter Turnout in 2012

Image 11: Moran's I chart of Voter Turnout in 2012

Image 12: LISA map of Percent Hispanic Population

Image 13: Moran's I chart of percent Hispanic Population

Conclusion:
There was a lower democratic vote increase over the past 32 years in northern and mid Texas, and an increase in Democratic vote in Southern Texas along the boarder. On the Moran's I charts in 2012 percent Democratic vote was much more clustered negatively than in 1980 which was much more spread out. There was still a higher number of Low-Low counties in 2012 as there were in 1980 just an increase in Low-Low in 2012. Voter turnout isn't clustered very tightly, but in 2012 there seems to be more of a cluster in High-High than there was in 1980. Comparing Democratic Vote to Voter turnout, in the areas of Democratic Voting High-High there is a voter turnout of Low-Low. This connects to the chart and map of Hispanic population because that is the area of High-High. Not much changed in voter turnout from 1980 to 2012 besides an increase in High-High in the northern tip counties along with a transition the most east part of the state from Low-Low to white. This means more people showed up to the poles in northern Texas along with in West Texas. There was a change from High-Low to Low-Low voter turnout in southern Texas from 1980 to 2012, so less people showed up to vote correlating to less Hispanics showing up to vote. Spatially there was not much change in voter turnout over 32 years, but for percent Democratic vote the area as to which were high and low didn't change but became more dense and gained surrounding counties.

Assignment 4

Hypothesis Testing

Introduction:
The objective of this lab is to be able to correctly distinguish if you need to use a z-test or a t-test, and to be able to calculate the answer to a z or t test. Then be able to apply the steps of hypothesis testing to come to a conclusion with the null and alternative hypotheses. The last objective is to be able to apply the data found to the real world and geography.

Terms: 
Z-test: Tests whether two population means are different usually used fro large samples over 30

Figure 1: Z-Test Equation

T-test: Tests whether the sample or samples fit a normal distribution with a sample size of less than 30

Figure 2: T-Test Equation

One Tailed: Is used when direction is given in the given study question
Two Tailed: Is used when you are unsure of the direction of your study question
Type 1 Error: occurs when a true null hypothesis is rejected
Significance Level: Set to determine the probability level of a null hypothesis being rejected
Null Hypothesis: States there is no difference between your sample mean and the hypothesized mean or it is equal to zero
Alternative Hypothesis: States that there is a difference between the sample mean and hypothesized mean
Hypothesis Testing: The way that inferences about a population are made from a sample

1. State the null hypothesis
2. State the alternative hypothesis
3. Choose a statistical test
4. Choose α or the level of significance
5. Calculate test statistic
6. Make a decision about the null and alternative hypotheses 

PART I

1)

For Two Tailed z or t values the answer is either + or - because of there being two separate values

2)

Testing the null and alternative hypothesis for estimated yields per hectare in certain areas in Kenya from the Department of Agriculture and Live Stock Development organization.
-First I found the T-Value shown in Figure 3 which is needed to be able to reject or fail to reject the hull hypothesis. Then I Figured out the T-Test values for Groundnuts, Cassava, and Beans. I did this by using the equation for T-Test in Figure 2. I then used the hypothesis testing to determine if there was or was not a difference between the sample mean and mean population.

Figure 3: T-Value for Question 2

2-Tailed with a confidence level of 95%
n= 23

       μ             σ         μh 

                Ground Nuts    0.51        0.3         .55

                Cassava            3.4          .74         3.8

                Beans               0.33        0.13       .28

T- Value= + or -2.074

Groundnuts:

1. State the null hypothesis

     - There is no significant difference between the sample mean and the mean for Groundnuts

2. State the alternative hypothesis

     - There is a significant difference between the sample mean and the mean for Groundnuts

3. Choose a statistical test

     - Use a T-Test because the value of n is less than 30

4. Choose α or the level of significance 

     - α= + or -.025, 2-tailed test

5. Calculate test statistic 

      - T-test= -.64

6. Fail to reject the null hypothesis

Cassava:

1. State the null hypothesis
     - There is no significant difference between the sample mean and the mean for Cassava
2. State the alternative hypothesis
     - There is a significant difference between the sample mean and the mean for Cassava
3. Choose a statistical test
     - Use a T-Test because the value of n is less than 30
4. Choose α or the level of significance
      - α= + or - .025, 2- tailed test
5. Calculate test statistic
      - T-test = -2.597
6. Reject the null hypothesis

Beans:

1. State the null hypothesis
      -There is no significant difference between the sample mean and the mean for Beans
2. State the alternative hypothesis
      - There is a significant difference between the sample mean and the mean for Beans
3. Choose a statistical test
     - Use a T-Test because the value of n is less than 30
4. Choose α or the level of significance
     - α= + or - .025, 2-tailed test
5. Calculate test statistic
     - T-test= 1.85
6. Fail to reject the null hypotheses

Similarities and Differences: The Groundnuts and Beans both failed to reject the null hypotheses while the Cassava rejected it. The Cassava had a greater standard deviation which backs up that it rejected because there is most likely more outliers that are pulling it from the mean. The Groundnuts and Beans both on the contrary have smaller standard deviations.

3)
Testing a suspicion a researcher had about the level of a particular stream's pollution being higher than the allowable limit which is 4.4 mg/l.
To figure this out I first again like I did in question two figured out the T-Value which is shown in Figure 4. Then I found out the T-Test stat using Figure 2. From there by using the Hypothesis testing I decided whether the suspicion the researcher had about the particular streams pollution being higher was correct.

Figure 4: T-Value for Question 3

One-tailed/95% level
Standard Deviation=4.2
u=6.8
uh=4.4
n=17
T-Value= 1.746

1. State the null hypothesis
    -There is no significant difference between the sample mean and the mean of the stream pollution
2. State the alternative hypothesis
    - There is a significant difference between the sample mean and the mean for the stream pollution
3. Choose a statistical test
    - Use a T=Test because the value of n is less than 30
4. Choose α or the level of significance
     - α= + or -.05, 1-tailed test
5. Calculate test statistic 
     -T-Test= 2.35
6. Reject the Null Hypothesis 

Part II:

Question: Is the average value of homes for the City of Eau Claire block groups significantly different from the block groups for Eau Claire County?
To answer this question I used the two shape files given and got the statistics needed from both shown in Figures 5 and 6. I got the statistics needed from Eau Claire City which was the standard deviation, mean and sample size. Then I got the only thing needed which was the population mean. I then concluded because the sample size for the sample mean was over 30 to use the Z-Test. From there I found the Z-Value shown in Figure 7. By doing that I had to decide that the Confidence level was 95 and that it was a 2-Tail test because the direction was unsure. I then calculated the Z-Score by using the equation shown in Figure 1 and by the statistics I found from the county and city. Finally I used hypothesis testing to figure out if there was a significant difference from the block groups for Eau Claire County and the City of Eau Claire.

Figure 5: Statistics of Eau Claire County

Figure 6: Statistics of the City of Eau Claire

Figure 7: Z-Test for Part II

1. State the null hypothesis

     - There is no significant difference between the sample mean and the mean of the average cost of housing 

2. State the alternative hypothesis 

     - There is a significant difference between the sample mean and the mean for the average cost of housing

3. Choose a statistical test

     - Use a Z-Test because the value of n is greater than 30

4. Choose α or the level of significance 

     - α= + or - .05, 2-tailed test

5. Calculate the test statistic

    - Z-Test= -2.57

6. Reject the Null Hypothesis  Z-Value=1.64 

Conclusion:

By saying that I rejected the null hypothesis that means that there is a significant difference between the average housing cost for the county and city of Eau Claire. This is shown in the map because there is a distinct pattern in the City of Eau Claire being cheaper housing (shown by the green and yellow colors) than the surrounding county. There isn't even any dark or light green shown on the map outside of the city and very little yellow. This would make sense because the Z-test calculation that I got was -2.57 meaning it falls on the left tail side of the Z-Value of -1.64. This would make sense to me because from what I know about the city of Eau Claire the dark green is student housing area, and also the farther out of the city you get the more expensive housing is because of increase in acreage. 

Assignment 3

Foreclosures in Dane County, Wisconsin

Introduction:

In this research question I will be investigating the changes in foreclosures in Dane County, Wisconsin from the years 2011 to 2012. I can not determine the reason behind this change spatially as there are a lot of reasons behind foreclosures, but I will be looking at the change from a spatial point of view. This information could possibly be helpful to others for instance families looking to by a house, relaters, and people in political offices. I am researching this question specifically because it has been a topic of concern among County Officials as to why this number has gone up, and if it will continue to do so. I will with my question be finding the patterns from one year to the next to see if I will be able to provide a prediction for the next year (2013).

Methodology:

To answer my research questions I will have to use a couple of different methods on the way.
First to find the change in the years 2011 to 2012 I had to add a field to my attribute table. I did this by adding a blank field add calculating it using the fields Count2011 and Count2012 and subtracting the year 2012 from 2011 to get the changes in house foreclosures.

Another method I used was calculating the Z-Score specifically of a certain Census Tract in the County. Z-Score is figured by taking your X like in this instance the number of foreclosures in a certain census tract and subtract that from your mean of the year and divide that by your standard deviation of that year. This equation is shown in Figure 1 below.
A method that relates to this is how I found the mean and standard deviation for the certain years I did this simply by going into the symbology part of Arcmaps and going to classify. This is represented by Figure 2.

2011:
Mean=11.39
Standard Deviation=8.78
2012:
Mean=12.30
Standard Deviation= 9.91

Figure 1: Z-Score

Figure 2: Standard Deviation and Mean

Along with Z-Score another method used was finding an X-value from probability. To do this you first use Figure 3 to find you percent probability and that from there you find the Z-Score found in the first row and column. When you have your Z-Score you can then use the Figure 1 equation to get the X value. I used this specifically to find the number of foreclosures that would be exceeded a percentage of the time. Which will become more clear further in the blog.

The data that I am using for these methods comes from the foreclosure data by the Census for the years 2011 and 2012 by individual house.

Figure 3: Standard Statistical Table

Results:

First for results I have the Map (Figure 1) that covered the change in foreclosures from the years 2011 to 2012 that I discussed in my methodology section. This showed that there was a more substantial change in an increase in foreclosures in 2012.

Figure 1: Map of Change form 2011 to 2012

My next results were that of the Z-Scores from 2011 and 2012 for three specific Census Tract areas of the county. I found from my results that in two of the areas (108 and 25) the Z-scores go up a small amount, but in the other one (120.02) it goes up a substantial amount. This goes along with my previous findings that the number of foreclosures goes up over the year. It also correlates to Figures 1 and 2 by the standard deviation and how it is correct with the up and down shown by what I calculated the Z-Scores as for those three Census Tracts.

Z-Scores:

Figure 2: Standard Deviation of 2011

2011: 

Census Tract 108 

Z-Score= 2.01

Census Tract 120.01

Z-Score= -.614

Census Tract 25

Z-Score= 1.78

Figure 3: Standard Deviation of 2012

2012:

Census Tract 108

Z-Score= 1.48

Census Tract 120.01

Z-Score= 2.99

Census Tract 25

Z-Score= -.938

Probability:

I found that the probability that the patterns from 2012 will stay the same into the next year:
From the data the number of foreclosures that will be exceeded 80% of the time is 3.98, and that will be exceeded 10% of the time is 24.98. If found this like I explained in the methodology section by using the Figure 3 in that section to find the Z-Score and then algebraically finding the X value which in this case is the number of house foreclosures. This shows that by the pattern the foreclosure numbers should increase the same spatially in 2013. Spatially I think it is also most likely to happen in the up and coming suburbs of Madison like Sun Prairie (which is on the northeast middle side of the county. From the map it shows more outer Census Tracts to be the locations of foreclosures so I believe the patterns will continue there.

Conclusions:

First I would like to restate my research question that is to find spatially the change in house foreclosures from the years of 2011 to 2012 in Dane County. A long with from these patterns will there also be an increase in the year 2013.
A summary of my results showed that at least one of my z-scores showed an increase in change, and that from Figure 1 of my results section the overall number went up from 2011 to 2012. Although comparing the mean and standard deviations of the two years the are only slightly different in numbers shown in my methodology section under Z-scores. The numbers match accurately with my results, but it is not as great of a difference as it seems to show on the map. In my conclusion of the results spatially, I believe that the increase happened more on the outer edges of the county and the suburbs of the main city of Madison in Dane County. This pattern is shown on all of the figures in the results section of the blog. The probability section made more sense of the patterns of the change and to if they it were to make sense for the same change to occur in 2013. It shows that 80 percent of the time for 2013 there will be rounding up to an increase of 4 houses more that will be foreclosed on. Although not a large number it still backs up the hypothesis.

Assignment 2

Part I:

Range: Range just refers to the greatest number in a group of observations minus the smallest number in that same group. It shows the disperses of the set of observations.

Mean: Mean refers to all the numbers in a group of observations added up divided by the total number of observations made. This is showing what number the observed group is more tending to out of all the numbers.

Median: Median is the number that is in the exact middle of a group of numerically ordered observations. Usually works best with a group that has an even number of observations, if there is an odd number of numbers then you take the difference between the even two.

Mode: Mode is the number that appears the most out of all the numbers in the group of observations.

Kurtosis: Kurtosis is shown on the graphed observations as steepness or the lack there of. This steepness is referring to the distribution of numbers away from the mean. There are two types flat or peaked. Peaked would be a lesser number distributed to that side and flat would be a greater number.

Skewness: Skewness is defined as how much the shape of the observations graphed is shifted from the one side of the mean of the group. It can either be a negative or positive skewness depending on if more observations fall below or above the mean.

Standard Deviation: This can be defined as how close or far away an observation is to the mean or the most common observation. It shows how far an observation "deviates" from the normal.

In past studies the highest score for standardized tests have always been higher at Eau Claire Memorial than Eau Claire North. This has led to a questioning in the staff at North. I will be looking to see if the top score is an accurate betrayal of the teachings at North. The test score max is 200 for this test and the highest score at North is 194 and at Memorial 198. I will be looking at more than the highest score, and instead the range, mean, median, mode, kurtosis, skewness, and standard deviation to tell the differences between the success of the students at the differing schools.

Eau Claire North:
Range= 83
Mean=160.92
Median=164.5
Mode=170
Kurtosis= -.55723
Skewness= -.5791

Standard Deviation:24.48

Eau Claire Memorial:
Range= 91
Mean=158.54
Median=159.5
Mode=120
Kurtosis= -1.17435
Skewness= -.1848

Standard Deviation: 27.58

These are the results I was able to compute of the two high schools. It turns out that in almost all categories North comes out on top of Memorial. Norths median and mean scores are higher than Memorials, but Memorials skewness is more negative than north which is better for scores. I don't think North teachers should be concerned about the performance of their students because over all they are testing about equal and even a little higher than Memorial based on the mean score. Mean score is in my opinion the best way to represent this data because it is the most fair way of averaging the students scores.

Part II:

Geographic or Spatial Mean Center: In my own words spatial mean center is where the push and pull of the land area (or latitude and longitude) ends up. It is the exact center of the area of a given place.

Weighted Mean Center: Weighted Mean takes in account other factors. You can do the weighted mean of a group of observations and it will be in the middle of where the greatest number of observations are pulling it.

This map is of the geographic mean (so unweighted) of the state of Wisconsin. This is where spatial the mean or center of the state is. It also shows two weighted means. One is the mean center of the population of the state in 2000 and the other is of the mean center of the population in 2015. The weighted mean centers make sense because it is relatively low in the state because of the pull from Madison and Milwaukee, but even though there is more land north of the center it is far less populated. You can tell by the graph that there was a shift from 2000 to 2015 in population. It has shifted southwest of where it was in 2000. This shows that it is moving in the direction of Dane County which is were Madison is which makes a lot of sense. Madison is the second biggest city in Wisconsin, and it is a very hip place for young people to move especially with it being a big college town.

Assignment 1

Assignment 1

Anna Turner

Part I

Nominal: 

Nominal data is data you can’t put numbers to it, but can be defined primarily by names and in mostly one but sometimes two categories. The map I choose is of the different names of all the Native American Tribes in the state of Wisconsin. This is an example of nominal data because it is the names of the tribes and if these were represented in numbers you would have no idea what it was talking about because it is something that is defined by name.

Map Created By: Paula Giese

Ordinal:

Ordinal data is data sets and numbers that are ordered and categorized in an exact or certain way. For example the map I choose I found of Breast Cancer in White Females in the United States. This relates to Ordinal because the map is showing that there is less cases some places in the U.S. and more cases other places in the U.S. by a numeric value. 

Map Created By: Jules Patry

Interval:

Interval data doesn’t start from an exact point like 0 it starts from any number and goes up to any number. For example the map I choose shows the pH levels of soils throughout the continental United States and Canada. The pH scale doesn’t start from 0 and is a quantitative data.

Map Created By: Mosiac Company in 2010

Ratio Data:

Ratio Data shows the differences and relationships between data and can be compared unlike nominal data that cannot be compared. For example, the map that I choose was the population of polish ancestry to that of the rest of the population of the states. This is an example of ratio because it is comparing and separating the whole population from the number that contain polish ancestry.

Map Created By: Ancestory.com 1990 

Part II

From the Data collected I think we should increase the number of Organic Farms in Green County. We should also use the Natural Breaks Map (Image 1) to represent Green County. This map represents how Green County gas far fewer organic farms than that of all of its surrounding counties, and that should draw in good marketing because if the surrounding areas can support such farms than so can Green. Green County is also in a historically successful area for farming, and would suggest a higher chance of the county being able to withstand more farms.

Image 1: Natural Breaks

Image 2: Quantile Breaks

Image 3: Equal Intervals