Assignment 4

Assignment 4
Jeff Hessburg
GEOG 370
2 April 2017

Part one of this blog requires completing the table below:

What is needed is

α = which is the significant level for a given test. To figure out this value; first look if the interval type is one tailed or two tailed. 

- If the interval type is One Tailed, than take 100 and subtract it from the confidence level, then times by (1/100) to get α

- If the interval type is two tailed, there is an extra step. Take 100 and subtract it from the confidence level, the times by (1/100), then finally divide by two, to get α.

Next it must be decided if the test is a z test or a t test. The test is will be a z test if n (the sample size) is greater than 30. If n is less than 30, a t test is required. 

Lastly, the Critical Value for the given Significance Level needs to be determined. This can be determined by looking at a table that gives the values for the given test type. If the test type is z, then the following table must be used. To read the table, take 1-α, then find the value on the graph, this will give the z value

http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf

If the test is a t test, the following table is used. to read it, first the degrees of freedom is needed to be calculated. This can be calculated by using the following equation: n-1. Use degrees of freedom and the α in the top row. This will determine the t value

http://d2r5da613aq50s.cloudfront.net/wp-content/uploads/451675.image0.jpg

For z and t test, If the interval type is two tailed, the negative and positive z or t values must be included.

The completed table is below:

For the second question in this assignment, an estimate from the Department of Agriculture and Live Stock Development is compared to the survey of 23 farmers. The estimate includes; ground nuts, cassava, and beans. Shown below are the calculations for t-value,  a visualization of where the t value places on a standard distribution with 95% confidence, the probability of the actual value, and if the hypothesized mean is rejected or failed to be rejected. 

There are not many similarities or differences. None of the hypothesis fell very close to the mean, but only one got rejected. One hypothesis fell above the mean and two below. 

For Part three the objective is to calculate if a researches suspicion of if a stream is polluted above the allowable limit. The calculations and results are shown below. 

PART II

Null hypothesis- There is no difference between the value of City of Eau Claire homes and the homes in all of Eau Claire County. 

Alternative hypothesis- There is a difference between the value of City of Eau Claire homes and the homes in all of Eau Claire County. 

Statistical Test- Two tailed Z test. Z test because the sample size is larger than 30 and two tailed because the null hypothesis could be rejected if the test gets results above or below the mean. 

An α of .05 was chosen because, after literature review, it was discovered that .05 is chosen for most average tests like this. There are worries of type I and type II error but not much of one over the other. 95% confidence level is perfect when trying to avoid both errors. 

Below are the calculations that determine if the null hypothesis can be rejected. 

It can be concluded that there is a difference between the average value of homes in the city of Eau Claire, and the County of Eau Claire as a whole.

Below is a visual representation of the average value of home in its given block group.