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Project Part 4
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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 T-test because my data is a sample.

One sample T hypothesis test:
μ : Mean of variable
H0 : μ = 700
HA : μ ≠ 700

Hypothesis test results:

VariableSample MeanStd. Err.DFT-StatP-value
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, t-stat is 2.5511836, Std err. 27.499491

Since the p-value 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

Chi-Square test:

StatisticDFValueP-value
Chi-square 20 14.425926 0.8083

Warning: over 20% of cells have an expected count less than 5.
Chi-Square 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 chi-square test is 14.425926 and the p-value is .8083.

Since my p-value 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

ColumnnMeanStd. 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

SourceDFSSMSF-StatP-value
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 F-stat is 457.92161 and the P-value is .0001.

Since our p-value 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
H0 : μ = 700
HA : μ ≠ 700
Standard deviation not specified.

Hypothesis test results:

VariablenSample MeanStd. Err.Z-StatP-value
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, z-stat is 2.5511836, p-value is .0107.

The only thing different from my t-test is that the p-values 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

Chi-Square test:

StatisticDFValueP-value
Chi-square 27 15.677908 0.9588

Warning: over 20% of cells have an expected count less than 5.
Chi-Square suspect.

My variables used now are state and distance

The value is 15.677908 and the p-value is .9588

The null hypothesis is that there is no relationship between variables.

Alternative Hypothesis is that there is a relationship.

Since the p-value 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

StatenMeanStd. 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

SourceDFSSMSF-StatP-value
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 f-stat is .62403364 and the p-value is .7643

The p-value 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.

Result 1: One sample T hypothesis test   [Info]
One sample T hypothesis test:
μ : Mean of variable
H0 : μ = 700
HA : μ ≠ 700

Hypothesis test results:
VariableSample MeanStd. Err.DFT-StatP-value
Walmart Incidents770.1562527.499491312.55118360.0159

Result 2: Contingency table (with data)   [Info]
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

Chi-Square test:
StatisticDFValueP-value
Chi-square2014.4259260.8083
Warning: over 20% of cells have an expected count less than 5.
Chi-Square suspect.

Result 3: One Way ANOVA   [Info]
Analysis of Variance results:
Data stored in separate columns.

Column statistics
ColumnnMeanStd. Dev.Std. Error
Target Incidents32174.687599.12404717.522821
Walmart Incidents32770.15625155.5606127.499491
Distance322.3031251.94297480.34347267

ANOVA table
SourceDFSSMSF-StatP-value
Columns2103882435194121.7457.92161<0.0001
Error931054882.111342.819
Total9511443126

Result 4: One sample Z hypothesis test Ec 1   [Info]
One sample Z hypothesis test:
μ : Mean of variable
H0 : μ = 700
HA : μ ≠ 700
Standard deviation not specified.

Hypothesis test results:
VariablenSample MeanStd. Err.Z-StatP-value
Walmart Incidents32770.1562527.4994912.55118360.0107

Result 5: Contingency table (with data) ec2   [Info]
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

Chi-Square test:
StatisticDFValueP-value
Chi-square2715.6779080.9588
Warning: over 20% of cells have an expected count less than 5.
Chi-Square suspect.

Result 6: One Way ANOVA ec3   [Info]
Analysis of Variance results:
Responses: Walmart Incidents
Factors: State

Response statistics by factor
StatenMeanStd. Dev.Std. Error
AZ4696.25101.3981450.699071
CA4774.75118.8847559.442374
CT1685NaNNaN
FL5769.8166.1270674.29428
IL1844NaNNaN
KS1607NaNNaN
MO1595NaNNaN
NC2724.585.55992160.5
NM3760.33333193.87195111.93202
TX10845.1191.9140160.68854

ANOVA table
SourceDFSSMSF-StatP-value
State9152561.8516951.3170.624033640.7643
Error22597610.3727164.108
Total31750172.22

Data set 1. Crime: Walmart vs. Target   [Info]
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HTML link:
<A href="https://www.statcrunch.com/5.0/viewreport.php?reportid=74552">Project Part 4</A>

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 By xg15Dec 12, 2017 need to specify the mean as population mean By michael.roberts97Dec 11, 2017 No it is not my full report. By xg15Dec 11, 2017 Is this your full report?

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