Assignment 3

Significance Testing

Part 1:

2. A Department of Agriculture and Live Stock Development organization in Kenya estimate that yields in a certain district should approach the following amounts in metric tons (averages based on data from the whole country) per hectare: groundnuts. 0.5; cassava, 3.70; and beans, 0.30.  A survey of 100 farmers had the following results: 

       μ             σ

               Ground Nuts     0.40        1.07

               Cassava             3.4          1.42

                   Beans                    0.33          0.14
Ground Nuts
     - Null Hypothesis: there is no difference between the yields of ground nuts and the estimated yield. 
     - Alternative Hypothesis: there is a difference between the yields of ground nuts and the estimated yield. 
     - A Z test was conducted to determine whether we would reject or fail to reject the null hypothesis.
     - There was a significance level of 95%, but since it was a two tailed test, we would use + or - 1.96. 
     - The Z value we got was -0.934. This means that we would fail to reject the null hypothesis or that there is not a difference. 
Cassava
     -Null Hypothesis: there is no difference between the yields.
     -Alternative Hypothesis: there is a difference between the yields.
     - A Z test was conducted to determine whether we would reject or fail to reject.
     - There once again was a significance level of 95%, leaving us with + or - 1.96.
     - The Z value we got was -2.11. Since this was less than -1.96 it would not fall within the 95% leaving us to reject the null hypothesis while stating there is a difference in the yields. 
Beans
     -Null Hypothesis: there is no difference between the yields.
     -Alternative Hypothesis: there is a difference between the yields.
     - Once again a Z test was conducted to determine whether we would reject or fail to reject.
     - The Z value we came up with was 21.42. This lead us to reject the null hypothesis since it fell outside the +1.96 range. 

3. An exhaustive survey of all users of a wilderness park taken in 1960 revealed that the average number of persons per party was 2.8.  In a random sample of 25 parties in 1985, the average was 3.7 persons with a standard deviation of 1.45 (one tailed test, 95% Con. Level) 

     -Null Hypothesis: there is no difference between the two parties.
     -Alternative Hypothesis: there is a difference between the two parties.
     -A T-test was conducted to determine if there was or wasn't a difference.
     -There is a significance level of 95% as a one tailed test.
     -The T value we came up with was 1.711 leaving us to reject the null hypothesis and that there is a difference between the two parties. 

Part 2:
    In the second part of our assignment we were given the task to look at data regarding the northern and southern halves of Wisconsin. Now when thinking of the term "up north", many people have different ideas. It is hard to think what a person from Florida would consider the term "up north" to be. Personally, when thinking of this term, I think of big woods, wolves, and lots of snow and cold for northern Wisconsin. Many people from the state would have different ideas as well that would associate a difference, because there is one, between northern and southern Wisconsin. 
    When dividing the state into two halves, a common parameter is highway 29, that runs east to west across the state at relatively the halfway point. This is the dividing line that I used in this assignment.  This left me with 27 northern counties and 45 southern counties. 

    Upon dividing the state into two halves, we were asked to look at SCORP data collected from the Wisconsin DNR. This data provided a number of characteristics that were unique to the state of Wisconsin. Some of this data reflected the term of "up north" while others pertained to the entire state as a whole. We were asked to choose three sets of this data and map it, which would show the areas throughout the state there were higher in these sets. The three sets of data I chose were the number of beaches, number of picnic areas, and the number of cottages. 
    The first map I made was the number of inland beaches. The minimum number of beaches in a county was one with the max coming in at 26 beaches. At first I figured that the higher number of beaches would all be in the northern half of the state considering that there is more lakes up north. Upon making my map I came to the conclusion that this was half right. Up north did have a lot more lakes and beaches but not the most beaches for a county. 

    The next map I chose to make was the number of picnic areas per county. The minimum number of picnic areas was 1 with the max coming in at 301. Instantly I thought about the University of Wisconsin Madison when talking about picnic areas. I figured this area would have the most considering all of the college aged students living there. I also knew that much of the northern half would be very low in picnic areas due to the fact of early winters. If it was campgrounds, then yes the northern half would have much more in my opinion. After creating the map, as I predicted, Dane county which features the University of Madison was one of the highest counties with picnic areas. 

    The final map I created with the term "up north" in mind, was the number of cottages. Cottages goes hand an hand with this term in my mind. Whenever growing up and my parents would talk about going to the cottage I instantly thought about lakes and going up north to grandmas. I figured before making my map of the number of cottages was that most of these cottages would be located in the northern half of the state. After making the map I stood correct. Looking at the data I found it very intriguing that  some of the counties had upwards of 12,500 cottages in their county. This seemed like a lot for a county, but shows that people are still willing to travel to northern Wisconsin during the summer and fall times to keep these cities and towns alive with tourism. 

Part 4:

    The final part of this assignment dealt with computing Chi-Square in SPSS. SPSS was new to many of us including myself. Chi-Square testing gives a numeric value for each variable comparing the observed distribution of each variable with the expected distribution. It also provides a statistical measure of how the observed variables are distributed throughout the state in respect to the expected distribution, with a significance level of 95 percent. 

    After computing the Chi square for inland beaches, with the number falling outside the 95 percent significance level, it is clear to state the number of beaches correlates with the northern and southern halves. Since there is more lakes in the northern half, it is safe to say that there would be more beaches as well in the northern half. 

    Upon completing chi square for picnic areas, with the number as well falling outside the 95 percent significance level as well as looking at the map, it is easy to see that picnic areas correlates with the southern half of the state. As I predicted earlier with Dane county being one of two counties with the most picnic areas. 

    The last chi square I conducted dealt with the number of cottages. This number as well fell outside the 95 percent significance level, which told me that it had a direct influence from the northern half of the state. More lakes in the northern half, leads to more cottages on these lakes. The two counties with the most cottages as well were in the northern half of the state.