|
Interval Type
|
Confidence Level
|
n
|
α
|
z or t?
|
z or t value
|
A
|
Two tailed
|
90
|
45
|
5%
|
Z
|
+- 1.96
|
B
|
Two Tailed
|
95
|
12
|
2.5%
|
T
|
+- 2.17
|
C
|
One Tailed
|
95
|
36
|
5%
|
Z
|
1.64
|
D
|
Two Tailed
|
99
|
180
|
.5%
|
Z
|
+- 1.96
|
E
|
One Tailed
|
80
|
60
|
20%
|
Z
|
-.84
|
F
|
One Tailed
|
99
|
23
|
1%
|
T
|
2.51
|
G
|
Two Tailed
|
99
|
15
|
.5%
|
T
|
+- 3.33
|
Scenario 1
The department of agriculture and Live Stock development in Kenya is curious to see how a survey of 23 farmer's crop yields per hectare compared to the rest of the countries national averages. Three crops were looked at. These crops were ground nuts, cassava, and beans. A two tailed t test was used to see compare their yields to the national averages.
- Null Hypothesis: There is no real statistical difference between the sample averages for the sampled farmers versus the country's national averages.
- Alternative Hypothesis: There is a real statistical difference between the sample averages for the sampled farmers versus the countries national averages.
- Statistical Test: A two tailed t-test
- Significance Level of .025
Crop
|
μ (sample)
|
x(population)
|
σ (Standard
Deviation)
|
α (Significance level)
|
T Value
|
CV
|
P-value
|
Ground Nuts
|
.51
|
.55
|
.3
|
.025
|
+- 0.639
|
2.074
|
.53
|
Cassava
|
3.4
|
3.8
|
.74
|
.025
|
+- 0.259
|
2.074
|
.79
|
Beans
|
.33
|
.28
|
.13
|
.025
|
+- 1.844
|
2.074
|
.08
|
(Figure. 1)
When looking at the results (Figure. 1) above we would fail to reject the null hypothesis. This means that there was no significant statistical difference between the sample of farmers and the national average. This is because the t-values are smaller than the critical value and therefore the null hypothesis is not rejected. When looking at the p-values it is also clear that the results are consistent as the p-values are larger than the significance level of .025.
Scenario 2
A research has a suspicion that a stream has higher levels of pollution that is permitted (4.4 mg/l. To test this, sample of 17 was taken. This sample had a mean of 6.8 mg/l and standard deviation of 4.2. A one tailed t-test was conducted with a significance level of .05.
- Null Hypothesis: There is no significant difference between the sampled streams pollution levels and the streams allowable limit.
- Alternative Hypothesis: There is a significant difference between the sampled streams pollution levels and the streams allowable limit.
- Statistical Test: a one tailed t-test
- Significance Level: level of .05
After conducting the t-test, a t-value of 2.356 was determined and this t-value was larger than the critical value of 1.746. These means that the null hypothesis is to be rejected, meaning that there is a significant difference between the polluted streams and the allowable limit. With a p-value of .0157 we are 98.42 percent sure that our result is correct.
Scenario 3
The question was of whether average value of homes for the city of Eau Claire is statistically different than surrounding county. This was done by looking at the statistics block groups for the city and county. The cities average property value was 151,876.51 and a standard deviation of 49,706.34. The counties average property value was 169,438.13.
- Null Hypothesis: There is not a significant statistical difference between the average property value in the city of Eau Claire compared to the surrounding County property value averages.
- Alternative Hypothesis: There is a significant statistical difference between the average property value in the city of Eau Claire compared to the surrounding County property value averages.
- Statistical Test: one tailed z-test (left)
- Significance Level: .05
After conducting the z-test it was deemed that there is a significant difference between the average property values in the Eau Claire and the surrounding county. It was determined that the city of Eau Claire's z-score was -2.57 which is smaller than the critical value of -1.64. The p-value was .005, meaning that that we are 99.5 percent confident in our observation. The spatial distribution of the block averages can be seen in the map below (Figure 2) .
Figure. 2 The map above shows the spatial distribution of the average property values for the City of Eau Claire and the surrounding county |
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