Geography 370 Assignment 4

Part I. T and Z Tests

1. Question number one involved filling out the rest of the chart given the information provided.

2    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. 0.57; cassava, 3.7; and beans, 0.29.  A survey of 23 farmers had the following results." The following results are assuming that each set of crops are two-tailed and have a confidence level of 95%. A t test will be used because the information provided is under thirty, as 23 farmers' crops were used in the surveying. 

            μ               σ

                  Ground Nuts       0.52         0.3

                  Cassava               3.3           .75

                  Beans                  0.34        0.12

Null hypothesis: There will be no differences between the expected average of the country  (.57, 3.7, .29) versus the actual averages (.52, 3.3, .34) of the per hectare crops. 

Alternative hypothesis: There are differences between the country's expected averages (.57, 3.7, .29) versus the actual averages (.52, 3.3, .34) of the region's per hectare crops. 

Critical Values: -2.074, 2.074

Ground Nuts

     T score: -.799

     Probability value: .??

     Ground Nuts crops fails to reject the null hypothesis. This is because -.799 is within the critical   

     values of ± 2.074

Cassava

     T score: -2.558

     Probability value: ??

     Cassava crops rejects the null hypothesis, as -2.558 is within the critical values of  ± 2.074

Beans

     T score: 1.998

     Probability value: ??

     The beans also fails to reject the null hypothesis, as 1.998 is within the critical values of  ± 2.074.

Conclusion: The Cassava was the only crop that rejected the null hypothesis. This means that it ended up outside of the critical values, meaning there was a significant difference between the country's averages and the district's averages. Ground nuts and Beans fell within the critical values, even though the ground nuts had a negative t value. 

3. " A researcher suspects that the level of a particular stream’s pollutant is higher than the allowable limit of 4.2 mg/l.  A sample of n= 17 reveals a mean pollutant level of 6.4 mg/l, with a standard deviation of 4.4.  What are your conclusions? Please follow the hypothesis testing steps.  What is the corresponding probability value of your calculated answer?"

n=17

μ= 6.4 mg/l

μh=4.1 mg/l

σ= 4.4

Hypothesis testing

     Null hypothesis:  There is not a difference between the suspected average (4.2 mg/l) and the                actual average sample of 6.4 mg/l. 

     Alternative hypothesis: There is a difference between the suspected average (4.2 mg/l) and the    

     actual average sample (6.4 mg/l). 

     

     For this survey, a T test will be used, as only 17 surveys were used. The significance level of this   

     survey is 95% while being a one tailed test. 

          Critical value: 1.746

           T score: 2.06                                                                                                                                              Probability Value: .97403


a   Conclusion: Since the t score of 2.06 falls outside of the critical value of 1.746, the null hypothesis is rejected. This means that there is a difference between the suspected average and the actual average sample of the data. The stream is witnessing higher pollution levels than the expected average.

PPart II. Eau Claire City vs. County
 
      For this part, shapefiles were provided of block groups for the city of Eau Claire and of all block groups for the Eau Claire county. The test, since there were over 30 samples, was a z test with a confidence level of 95% and on a one-tailed curve.
d  
      City block group mean = 151876.51
      County block group mean= 169438.13
      Standard deviation of city block groups = 49706.92
      Probability value: .9949
      Critical value= 1.64