- Distinguish between a z or t test
- Calculate a z and t test
- Use the steps of hypothesis testing
- Make decisions about the null and alternative hypotheses
- Utilize real-world data connecting stats and geography
Part 1: Significance Testing with Z and T tests
The chart above was completed using the data from columns B-D to calculate the values in columns E-G.
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:
- Test the hypothesis for each of these products. Assume that each are 2 tailed with a Confidence Level of 95% *Use the appropriate test
- Be sure to present the null and alternative hypotheses for each as well as conclusions
- What are the probabilities values for each crop?
- What are the similarities and differences in the results
Alternative Hypothesis-Ground Nuts: There is a difference between the yield of ground nuts in the certain district in Kenya and the overall averages of crop yields in Kenya.
Conclusion-Ground Nuts: Reject the null hypothesis, so there is no difference between the yield of ground nuts in the certain district in Kenya and the overall averages of crop yields in Kenya.
Null Hypothesis-Cassava: There is no difference between the yield of cassava in the certain district in Kenya and the overall averages of crop yields in Kenya.
Alternative Hypothesis-Cassava: There is a difference between the yield of cassava in the certain district in Kenya and the overall averages of crop yields in Kenya.
Conclusion-Cassava: Reject the null hypothesis, there is no difference between the yield of cassava in the certain district in Kenya and the overall averages of crop yields in Kenya.
Null Hypothesis-Beans: There is no difference between the yield of beans in the certain district in Kenya and the overall averages of crop yields in Kenya.
Alternative Hypothesis-Beans: There is a difference between the yield of beans in the certain district in Kenya and the overall averages of crop yields in Kenya.
Conclusion-Beans: Reject the null hypothesis, there is no difference between the yield of beans in the certain district in Kenya and the overall averages of crop yields in Kenya.
Probabilities:
Ground Nuts: 0.78344
Cassava: 0.99144
Beans: 0.96403
Similarities and Differences:
-The test statistic that was calculated for ground nuts as well as the one calculated for cassava both fell on the left side of the negative critical value of -0.0192.
-Ground nuts, cassava, and beans are all similar because they all had the same conclusion, which was to reject the null hypothesis, stating that there was no difference between each of these specific crop yields and the overall average yields for Kenya.
-All three crops are also similar because they all used the same two critical values to compare the test statistics to.
-One difference is that the probability for ground nuts was 0.78344, which was significantly lower than the probabilities found for cassava and beans.
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.
Null Hypothesis: There is no difference between the allowable limit of stream pollutants in the particular stream and the mean pollutant level.
Alternative Hypothesis: There is a difference between the allowable limit of stream pollutants in the particular stream and the mean pollutant level.
Probability: 0.97403
Conclusion: Fail to reject the null hypothesis. There is a difference between the allowable limit of stream pollutants in the particular stream and the mean pollutant level.
Part 2
Using block group data (2 shapefiles) for the City of Eau Claire and Eau Claire County, the hypothesis testing steps were followed in order to answer the study 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? After noting the study question for this part of the assignment, the null and alternative hypotheses were created. The null hypothesis is; there is no significant difference between the average value of homes for the City of Eau Claire and the block groups for Eau Claire County. The alternative hypothesis is; there is a significant difference between the average value of homes for the City of Eau Claire and the block groups for Eau Claire County. The statistical test chosen for this problem was a z test, and it was a one tailed test with a significance level of 95%. The test statistic was then calculated using the z test equation and came out to be -2.572. Using the significance level of 95% and the standard statistical tables (areas under a normal distribution) chart, the critical value was calculated to be 1.64. This value was placed on a graph and compared to the test statistic. The conclusion of this problem was to fail to reject the null hypothesis, stating that there is no significant difference between the average value of homes for the City of Eau Claire and the block groups for Eau Claire County. The map below was created to show the patterns of average home value for the City of Eau Claire and Eau Claire County. The concentration of the highest average home values appears to be centralized around the city area of Eau Claire, in the northwest corner on the County. As you move farther away from the city, the average home values appear to be much lower. The general area around the main city in Eau Claire appears to be the most populated area in the county, which coincides with the results of the z test that concluded there is no difference between the average home value for the City of Eau Claire and that of homes in Eau Claire County. The similar patterns of average home value for the city and the county, which were centralized in the city/northwest corner of the map, help to further explain the results of this problem, which indicated no significant difference in value.
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