The following report presents the detailed statistical analysis of the data collected from a sample of credit customers in the department chain store AJ DAVIS. The 1st individual variable considered is Location. It is a qualitative variable. The three subcategories are Urban, Suburban and Rural. Since this is a qualitative variable, the measures of central tendency and descriptive statistics has not been computed for this variable. The frequency distribution and pie chart are given as follows: Frequency Distribution:|

From the frequency distribution and pie chart, it is clearly stated that the maximum number of customers belongs to the rural category (42%),consequently by those in the suburban category (30%). Only 28% of the customers belong to the urban category.

The 2nd individual variable considered is Size. It is a quantitative variable. The measures of central tendency, variation and other descriptive statistics have been intended for this variable and are given as follows: Descriptive Statistics:|

The mean household size of the customers averages 3.42. The median of the data is 3 and the mode is 2. The standard deviation is given approximately as 1.74. Maximum number of customers has a household size of 2 as is states from the frequency distribution and the bar graph.

The 3rd individual variable considered is Credit Balance. It is a quantitative variable. The methods of central tendency, variation and other descriptive statistics have been analyzed for this variable and...

...
Course Project Part A
September 15, 2013
Applied Managerial Statistics
Professor Mayers
Brief Introduction
The following report presents a detailed statistical analysis of AJDavis Department Store credit customers. Data was collected from a sample of 50 AJDavis credit customers on five variables which are Location, Size, Income, Years, and Credit Balance. Out of the five variables, Location,Size, and Income is emphasize more in this analysis. AJDavis Department Store is very determined to find out more information about their credit customers. So by doing a in-depth analysis of the variables and their relationships through graphical, numerical summary and interpretation should give a detailed summary of their customers.
Individual Variables
Location is the 1st individual variable considered. The location of AJDavis’ customers is distributed between three classes of Urban, Suburban and Rural Areas. It is a qualitaive variable that can not be measured numerical so central tendency and other descriptiive statistics can not be computed . The frequency distribution table and pie chart are given as followed:
Frequency Distribution Table
Location
Frequency
Urban
22
Suburban
15
Rural
13
Total
50
Following calculating the frequency, a pie chart was developed to determine which particular category had the...

...AJDavis Department Store
Project Part B: Hypothesis Testing and Confidence Intervals
A. The average (mean) annual income was less than $50,000
• Null Hypothesis is the average annual income is ≥ to $50,000.
o Ho: µ ≥ 50,000
• Alternate Hypothesis is the average annual income is < than $50,000.
o Ha: µ < 50,000
• Analysis Plan significance level is: a = 0.05
• n > 30 the z test was used to test the hypothesis
• Alternative Hypothesis Ha: µ ≥ 50,000, this means the test is a one tailed z test.
Critical Value and Decision Rule
• C.V: a = 0.05 the lower tailed z test is 1.645
• D.R: reject Ho if z – statistic is -1.645
Sample Z using Minitab:
• Income ($1000)
• Variable I
• Test of µ = 50 verse < 50
• Standard Deviation = 14.55 using
• Confidence Level = 95%
• Alternative = not equal
Results from Minitab:
• N = 50
• Mean = 43.48
• Standard Deviation = 14.55
• SE Mean = 2.06
• 95% CI (39.45, 47.51)
Interpretation of Results:
We reject the Null Hypothesis since the P-value .0001 is smaller than the significance level 0.05. The p-value indicates the probability of rejecting a true Null Hypothesis. There is significant support to claim that the average annual income was less than $50,000 since there is a significance level of 0.05. The 95% upper confidence limit is 47.51. Because 50 lies beyond...

...The following report presents a detailed statistical analysis of AJDAVIS department store customers. Data was collected from a sample of 50 AJDAVIS credit customers for the purpose of learning more about the customers of AJDAVIS.
The first variable considered is Location, a categorical variable. The three subcategories are Urban, Suburban and Rural. The frequency distribution and pie chart are included. Measures of central tendency and descriptive statistics are not calculated due to the categorical nature of the variable.
Frequency Distribution:
LOCATION | FREQUENCY |
Urban | 22 |
Suburban | 15 |
Rural | 13 |
The largest number of customers belong to the Urban Location category (44%), followed by those in the Suburban Location category (30%). The least number of customers belong in the Rural Location category (26%).
The next individual variable considered is Household Size, meaning the number of people living in the household. Size is a quantitative variable. The measures of central tendency and variation along with other descriptive statistics have been calculated for this variable.
Descriptive Statistics: AJDAVIS Customer Data - Household Size
Total
Variable Count N N* CumN Percent CumPct Mean SE Mean TrMean StDev
C1 50 50 0 50 100 100 3.420 0.246 3.341 1.739...

...INTRODUCTION
Base on the data collected in the previous samples, the manager has made an alternative hypothesis on the following:
A) The average (mean) annual income was less than $50,000
B) The true population proportion of customers who live in an urban area exceed 40%
C) The avarage (mean) number of years lived in the current home is less than 13 years D) The avarage (mean) credit balance for suburban costumers is more than $4,300
Using the sample data, we will perform hypothesis tests on the aforementioned situations above in order to determine if we can support the manager's belief in each case. The hypothesis tests will be computed with 95% confidence intervals to ensure accuracy in our alternative hypothesis opinion.
The avarage (mean) annual income was less than $50,000
The manager is correct. The avarage (mean) annual income is less than $50,000. After running the data located in Appendix A, notice that the avarage (mean) annual income came out to 43.74. This means that the avarage income is $43,740, which is less than $50,000.
Furthermore, after running a 95% confidence interval on this data, we can say that we are 95% confidence interval on this data and we are 95% certain that the avarage (mean) annual income falls at 47,210 on the higher end of the tail, which is less than 50,000. Since the p-value of .002 is less than the alpha value of .05, we can say beyond a...

...Problem N°1
1.Formulate the null and alternative hypotheses.
Null Hypothesis: The average (mean) annual income was greater than or equal to $50,000
H_0: μ≥50000
Alternate Hypothesis: The average (mean) annual income was less than $50,000.
H_a: μ 30 we will use the z-test.
As Ha:μ0.40 the, test is a right tailed z-test.
The critical value for significance level, α=0.05 for a right tailed z-test is given in the table as: 1.645.
Decision Rule: Reject H_0,if z>1.645
3. Calculate the test statistic.
In order to do the calculations by hand, we have p hat = (0.4*50)/50=0.4 and q=1-0.4=0.6 and n=50. P>0.4 mean that 21/50=0.42 then p > 0.42
Z- test statistic: z= (phat- P0 )/√((po*q)/n) = (0.4-0.42)/(/√((0.4*0.6)/50)) = -0.2887
4. Compare the test statistic to the rejection region and make a judgment about the null hypothesis.
Reject H_0,if z>1.645
-0.2887alpha(0.05) which mean that we fail to reject H0.
7. Based on the p-value, what decision would you make concerning the null hypothesis? Why?
Since the P-value (0.386) is greater than the significance level (0.05), we fail to reject the null hypothesis. The p-value implies the probability of rejecting a true null hypothesis.
At a significance level of 0.05, there is no sufficient evidence to support the claim that the true population proportion of customers who live in an urban area is greater than 40%.
Problem N°3: The average (mean) number of years lived in the current home is less than 13 years....

...AJDavis Department Store Part B
AJDavis Department Store
Introduction
The following information will show whether or not the manager’s speculations are correct. He wants to know the following information: Is the average mean greater than $45,000? Does the true population proportion of customers who live in an urban area exceed 45%? Is the average number of years lived in the current home less than 8 years? Is the credit balance for suburban customers more than $3200? Hypothesis testing and confidence intervals for situations A-D are calculated.
A. THE AVERAGE (MEAN) ANNUAL INCOME WAS LESS THAN $45,000.
Solution:
Step 1: Null Hypothesis: The average (mean) annual income was equal to $45,000.
H_0: μ=45,0000
Step2: Alternate Hypothesis: The average (mean) annual was less than $50,000.
H_a: μ 45 , a z-test for the mean will be used to test the given hypothesis.
As for the alternative hypothesis, which is Ha:μ0.45 and the given test is a one-tailed (upper-tailed) z-test.
Step 4: Critical Value and Rejection Region:
The critical value for significance level is ∝=0.05. The upper tail z-test is 1.645.
Rejection Region: Reject H_0,if z-statistic>1.645.
Step 5: Assumptions:
The sample size in this experiment is n 0.4
95% Lower
Sample X N Sample p Bound Z-Value P-Value
1 21 50 0.420000 0.305190 0.29 0.386
Step 7: Interpretation:
According to the calculations, the p-value is 0.386. This value...

...A. Brief Introduction
There are 50 credit customers who were selected for the data collection on five variables such as location, income, size, years, and credit balance. In order to understand more about their customer, AJDAVIS must use graphical, numerical summary to be able to interpret and better expand their business in the future.
B. Discuss your 1st individual variable, using graphical, numerical summary and interpretation
A histogram shows the distribution of data within the Income. In this Histogram graph of Income, it shows that the graph is not symmetrical. This histogram graph has a wider bell shape form. The graph shows that this graph is more like two graph because there is a clear difference between income generating from 20-40 and from 50-above. There are two separated cluster; therefore, the skewness of this graph is skewed right. Income has a lower value of kurtosis which indicates a lower, less distinct peak. The following table shows the numerical summary of Income:
Total
Variable Count N N* CumN Percent CumPct Mean SE Mean TrMean StDev
Income ($1,000) 50 50 0 50 100 100 46.02 1.96 45.70 13.88
Sum of
Variable Variance CoefVar Sum Squares Minimum Q1 Median Q3
Income ($1,000) 192.75 30.17 2301.00 115337.00 25.00 33.00 44.50 57.25...

...
Course Project: AJDavis Department Stores
Natasha Unaphum
MATH533: Applied Managerial Statistics
September 10th, 2014
Professor Rolston
Introduction:
AJDavis is a department store chain, they are trying to get to know more about their clientele and to further expand their business. A sample of 50 credit customers are selected for this research, information that includes, location (rural, urban or suburban), Income (in $1,000), size (household size), years (number of years lived in that location), and credit balance (customers current credit card balance on the store’s credit card).
Discuss your 1st variable, using graphical, numerical summary and interpretation
Numerical Summary of Credit Balance are as follows:
Mean: 3970.5 Minimum: 1864
Standard Deviation: 931.9 Q1: 3109.3
Variance: 868429.8 Median: 4090
Skew: -0.15043 Q3: 4747.5
N: 50 Max: 5678
The histogram above shows the Credit Balance variable of the 50 customers surveyed. The histogram is almost symmetrical with one outlier which is the credit balance of $2,000. While it being symmetrical you can almost fold the y-axis in half to have it look the same. While observing the histogram, its skewed to the left because of the outlier, and the skew is -.015043. Using the Anderson-Darling Normality Test, the P-value for Credit Balance is 0.400, and A^2 is...

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