AJ Davis Department Store Part B
AJ Davis 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 is larger than the significance level of 0.05. Therefore, we will not reject the null hypothesis. 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%. Based on the results provided by MINITAB below, the lower 95% confidence limit is 0.28. Since 0.42 is greater than the 95% lower confidence limit, hence, we cannot support the claim that the true population proportion of customers who live in an urban area is greater than 45%. Confidence Interval:

...
Course Project Part A
September 15, 2013
Applied Managerial Statistics
Professor Mayers
Brief Introduction
The following report presents a detailed statistical analysis of AJDavisDepartmentStore 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. AJDavisDepartmentStore 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...

...AJDavisDepartmentStore
Project PartB: 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...

...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...

...AJDavisDepartmentStores - Project Part A, B, and C
Stacie Borowicz
June 14, 2013
Math 533
Project Part A – Exploratory Data Analysis
Credit Balance ($)
Based on a sample of 50 customers, the credit balance for customers of DavisDepartmentstores is on average $3970.00. Based on the graph, 18 of the 50 sampled fall below and 17 fell above the average. The standard deviation for credit balance is 931.9.
Income
Annual Income of DavisDepartmentStores customers range anywhere from $22,000 to $67,000. Majority of their customers have an annual income of about $43,740. The standard deviation for income is $14,640.
Size
Average household size is about 3.4. Based on the graph above, 15 of the 50 sampled have a household size of 2 (the most common household size). The standard deviation is 1.739.
Credit Balance and Location
Comparing income to credit balance shows the higher the income the higher the credit balance. Although a few of the sampled customers have a low income and a high balance this is not a normal scenario based on the graph above.
Years and Credit Balance
Location and Size
Based on the graph, it is clear that the household size is normally 2. Urban and Rural areas seem to have the most households with 2 whereas the suburban location has...

...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....

...The following report presents a detailed statistical analysis of AJDAVISdepartmentstore 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...

...The following report presents the detailed statistical analysis of the data collected from a sample of credit customers in the department chain storeAJDAVIS.
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: |
Location | Frequency |
Urban | 21 |
Suburban | 15 |
Rural | 14 |
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: |
Size |
Mean | 3.42 |
Standard Error | 0.24593014 |
Median | 3 |
Mode | 2 |
Standard Deviation | 1.73898868 |
Sample Variance | 3.02408163 |
Kurtosis | -0.7228086 |
Skewness | 0.52789598 |
Range | 6 |
Minimum | 1 |
Maximum | 7 |
Sum | 171 |
Count | 50 |
Frequency Distribution: |
Size | Frequency |...

...Project PartB: Hypothesis Testing and Confidence Intervals
a. The average (mean) annual income was less than $50,000
Null Hypothesis: The average annual income was greater than or equal to $50,000
H₀: µ > 50000
Alternate Hypothesis: The average annual income was less than $50,000.
Ha: µ > 50000
Analysis Plan: Significance Level, α=0.05.
Since the sample size, n > 30 I will use z-test for mean to test the given hypothesis.
As the alternative hypothesis is Ha: µ > 50000, the given test is a one-tailed z-test.
Critical Value and Decision Rule:
The critical value for significance level, α=0.05 for a lower-tailed z-test is given as-1.645.
Decision Rule: Reject H₀, if z – statistic, -1.645
Test Statistic - minitab
One-Sample Z: Income ($1000)
Test of mu = 50 vs < 50
The assumed standard deviation = 14.55
95% Upper
Variable N Mean StDev SE Mean Bound Z P
Income ($1000) 50 43.48 14.55 2.06 46.86 -3.17 0.001
Interpretation of Results and Conclusion:
Since the P-value (0.0001) is smaller than the significance level (0.05), we reject the null hypothesis. The p-value implies the probability of rejecting a true null hypothesis.
The significance level of 0.05, there is enough evidence to support the claim that the average annual income was less than $50,000.
Confidence Interval - minitab
One-Sample Z
The assumed standard deviation = 14.55
N Mean SE Mean 95% CI
50 43.48 2.06 (39.45, 47.51)
The 95% upper confidence limit is 47.51....

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