My data is a sample because it just takes some of the incidents from some states. It is not an entire population.
So I conducted a Ttest because my data is a sample.
One sample T hypothesis test:
μ : Mean of variable
H_{0} : μ = 700
H_{A} : μ ≠ 700
Hypothesis test results:
Variable  Sample Mean  Std. Err.  DF  TStat  Pvalue 

Walmart Incidents  770.15625  27.499491  31  2.5511836  0.0159 
Variable is Walmart incidents
Null is that the sample mean is 700
Alternative is that the sample mean is less than 700 or does not equal 700
Statistics p value is .0159, tstat is 2.5511836, Std err. 27.499491
Since the pvalue is less than .05 you would reject the null hypothesis. This makes sense since we found that the sample mean is 770.15625.
Contingency table results:
Rows: Walmart Incidents
Columns: Target Incidents
0 to 100  100 to 200  200 to 300  400 to 500  500 to 600  Total  
500 to 600  0  1  1  0  0  2 
600 to 700  3  4  2  1  0  10 
700 to 800  1  3  4  0  1  9 
800 to 900  0  3  1  0  0  4 
900 to 1000  2  3  0  0  0  5 
1100 to 1200  0  2  0  0  0  2 
Total  6  16  8  1  1  32 
ChiSquare test:
Statistic  DF  Value  Pvalue 

Chisquare  20  14.425926  0.8083 
Warning: over 20% of cells have an expected count less than 5.
ChiSquare suspect.
My variables used were Walmart and Target Incidents.
Null hypothesis is that there is no relationship between variables.
Alternative ehypothesis is that there is a relationship.
The value of the chisquare test is 14.425926 and the pvalue is .8083.
Since my pvalue is higher than my signifigance level of .05 we would not reject the null hypothesis. This suggest that there is no relationship, which makes sense.
Analysis of Variance results:
Data stored in separate columns.
Column statistics
Column  n  Mean  Std. Dev.  Std. Error 

Target Incidents  32  174.6875  99.124047  17.522821 
Walmart Incidents  32  770.15625  155.56061  27.499491 
Distance  32  2.303125  1.9429748  0.34347267 
ANOVA table
Source  DF  SS  MS  FStat  Pvalue 

Columns  2  10388243  5194121.7  457.92161  <0.0001 
Error  93  1054882.1  11342.819  
Total  95  11443126 
The variables used are Target incidents, Walmart incidents, and Distance.
The null hypothesis is that the Walmart and Target incidents population mean equal each other.
The alternative is that they do not equal each other.
From the ANOVA above you can see that the Fstat is 457.92161 and the Pvalue is .0001.
Since our pvalue is smaller than my signifigance level of .05 you should reject the null hypothesis, meaning that the means are not equal.
Ec 1
Here I will be using a z test since I didn't ealrier beacuse my data set was a sample and not a population.
One sample Z hypothesis test:
μ : Mean of variable
H_{0} : μ = 700
H_{A} : μ ≠ 700
Standard deviation not specified.
Hypothesis test results:
Variable  n  Sample Mean  Std. Err.  ZStat  Pvalue 

Walmart Incidents  32  770.15625  27.499491  2.5511836  0.0107 
Variable used is Walmart Incidents
Null is that the sample mean is equal to 700
Alternative is that it does not equal 700
Staistics here are the sample mean is 770.15625, std. err. is 27.499491, zstat is 2.5511836, pvalue is .0107.
The only thing different from my ttest is that the pvalues are different but they are still lower than my signifigance level of .05 which means we reject the null hypothesis.
Ec 2
Contingency table results:
Rows: State
Columns: Distance
0 to 2  2 to 4  4 to 6  6 to 8  Total  
AZ  3  1  0  0  4 
CA  3  1  0  0  4 
CT  0  1  0  0  1 
FL  2  2  1  0  5 
IL  1  0  0  0  1 
KS  0  1  0  0  1 
MO  1  0  0  0  1 
NC  1  1  0  0  2 
NM  1  1  1  0  3 
TX  5  2  1  2  10 
Total  17  10  3  2  32 
ChiSquare test:
Statistic  DF  Value  Pvalue 

Chisquare  27  15.677908  0.9588 
Warning: over 20% of cells have an expected count less than 5.
ChiSquare suspect.
My variables used now are state and distance
The value is 15.677908 and the pvalue is .9588
The null hypothesis is that there is no relationship between variables.
Alternative Hypothesis is that there is a relationship.
Since the pvalue is greater than my signifigance level of .05 we can not reject the null hypothesis.
Ec 3
Analysis of Variance results:
Responses: Walmart Incidents
Factors: State
Response statistics by factor
State  n  Mean  Std. Dev.  Std. Error 

AZ  4  696.25  101.39814  50.699071 
CA  4  774.75  118.88475  59.442374 
CT  1  685  NaN  NaN 
FL  5  769.8  166.12706  74.29428 
IL  1  844  NaN  NaN 
KS  1  607  NaN  NaN 
MO  1  595  NaN  NaN 
NC  2  724.5  85.559921  60.5 
NM  3  760.33333  193.87195  111.93202 
TX  10  845.1  191.91401  60.68854 
ANOVA table
Source  DF  SS  MS  FStat  Pvalue 

State  9  152561.85  16951.317  0.62403364  0.7643 
Error  22  597610.37  27164.108  
Total  31  750172.22 
Here my variables are Walmart incidents and state.
The fstat is .62403364 and the pvalue is .7643
The pvalue is higher than my signifigance level of .05 so we shall not reject the null hypthesis.
Ec 4
A better way to do this survey to me would be to designate a certain area instead of just radomly choosing walmart and targets in different states. I say you pick one area of the U.S. at a time and then collect all the incident from each state then determine.
One sample T hypothesis test:
μ : Mean of variable H_{0} : μ = 700 H_{A} : μ ≠ 700 Hypothesis test results:

Contingency table results:
Rows: Walmart Incidents Columns: Target Incidents
ChiSquare test:
ChiSquare suspect. 
Analysis of Variance results:
Data stored in separate columns. Column statistics
ANOVA table

One sample Z hypothesis test:
μ : Mean of variable H_{0} : μ = 700 H_{A} : μ ≠ 700 Standard deviation not specified. Hypothesis test results:

Contingency table results:
Rows: State Columns: Distance
ChiSquare test:
ChiSquare suspect. 
Analysis of Variance results:
Responses: Walmart Incidents Factors: State Response statistics by factor
ANOVA table

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Dec 12, 2017
need to specify the mean as population mean
Dec 11, 2017
No it is not my full report.
Dec 11, 2017
Is this your full report?