Assignment 4

Part 1

Question 1:

	Interval Type	Confidence Level	N	α	Z or t	Z or t value
A	Two-tailed	90	45	.05	Z	1.64, -1.64
B	Two-tailed	95	12	.025	t	2.201, -2.201
C	One-tailed	95	36	.05	Z	1.64
D	Two-tailed	99	180	.005	Z	2.57, -2.57
E	One-tailed	80	60	.2	Z	.84
F	One-tailed	99	23	.01	t	2.508
G	Two-tailed	99	15	.005	t	2.977, -2.977

Question 2:
-2 tailed t test, a (significance level)=.025

Ground Nuts:
        a) μ=.52   σ=.3     n=23   μh=.57
               t test statistic=.011
          b) Null Hypothesis:  There is no significant difference between the sample mean of ground nuts and the hypothesized mean of yields for the whole country.
              Alternative Hypothesis: There is a difference between the sample mean of ground nuts and the hypothesized mean of yields for the whole country.
              In this case because the t test statistic falls (.011) falls between the critical values (-2.074 and 2.074), we fail to reject the null hypothesis, stating that the result falls within a normal distribution or that there is little or no difference between the sample and hypothesis means.
         c) Probability Value: .54 or 54%.  This value is still below the confidence level of 97.5% so we can verify that it is true that we fail to reject the null.

Cassava:
          a)  μ=3.3   σ=.75   n=23   μh=3.7
                t test statistic=.099
         b)Null Hypothesis:  There is no significant difference between the sample mean of Cassava and the hypothesized mean of yields for the whole country.
            Alternative Hypothesis: There is a difference between the sample mean of Cassava and the hypothesized mean of yields for the whole country.
            The test statistic (.099) falls between the critical values (-2.074 and 2.074), we fail to reject the null hypothesis, meaning there is little or no difference between the sample and hypothesis means.
         c)Probability Value: .54 or 54%.  This value is still below the confidence level of 97.5% so we can verify that it is true that we fail to reject the null.

Beans:
         a)  μ=.34   σ=.12   n=23   μh=.29
              t test statistic=.049
         b) Null Hypothesis: There is no significant difference between the sample mean of Beans and the hypothesized mean of yields for the whole country.
             Alternative Hypothesis: There is a difference between the sample mean of Beans and the hypothesized mean of yields for the whole country.
            The test statistic (.049) falls between the critical values (-2.074 and 2.074), we fail to reject the null hypothesis, meaning there is little or no difference between the sample and hypothesis means.
         c)Probability Value: .54 or 54%.  This value is still below the confidence level of 97.5% so we can verify that it is true that we fail to reject the null.

d) The similarities for each crop as the results show are that all of the sample means for each fall very close to the actual mean for each.  We can expect this also because the standard deviations for each are quite small.  In each case there is little or no difference between means so we fail to reject the Null.

Question 3:

μ=6.4   σ=4.4   n=17   μh=.4.2
1)Null Hypothesis: There is no difference between the sample mean pollutant level of a stream and the hypothesized mean pollutant level.
2)Alternative Hypothesis: There is a difference between the sample mean pollutant level of a stream and the hypothesized mean pollutant level.
3)t test
4)Significance Level of .05
5)t test statistic=2.075
6) We reject the Null Hypothesis, stating that there is a difference between the sample and hypothesized stream pollutant levels because the test statistic falls outside the critical value of 1.746.
Probability Value: .97403 or about 97.4%.  This is over the 95% Significance Level so this is another indicator that we reject the null, or that there is a difference between the means.

Part 2

Looking at two different shapefiles, one of block groups for the City of Eau Claire and the other for block groups in Eau Claire County, the question we want to answer is if the average value of homes in the city blocks groups is significantly different from average home value in the block groups for the whole county.  To do this we can use the Steps for Hypothesis Testing:

1)Null Hypothesis:  There is no difference between the average home value of the city of Eau Claire block groups and the average home value of the Eau Claire County block groups.

2)Alternative Hypothesis:  There is a difference between the average home value in the City of Eau Claire and the average home value in Eau Claire County. 

3) Z Test

4)Level of Significance: .05

5)Test Statistic=  -2.57

      μ=151876.5   σ=49706.92  n=53  μh=169438.1

6)In this case we would fail to reject the null hypothesis, which means that there is little or no difference.  The test statistic (-2.57) fell below the critical value of 1.64.  We can then use this to answer the original question:  The average value of homes for the City of Eau Claire block groups is not significantly different from the block groups for Eau Claire County.  The map below differentiates which block groups out of the county belong to the City of Eau Claire and also shows the how each block groups average home value deviates from the average for the County.  Although it appears that some of the block groups in the City of Eau Claire are below average, it is not enough to make the difference statistically significant or fail to reject the null.