MODULE 6_7 Application Assignments

Example Data

Example Data

A Survey of 50 Companies

In January 08, fifty customers of a lumber manufacturer were surveyed regarding their satisfaction with products and service. These customers buy from the supplier and sell to retail chains like Home Depot and Lowes. Shortly after, the manufacturing company was sold. In June 08, the customers were telephoned and interviewed and were asked to rate overall satisfaction again.

VariablePositionLabelMeasurement Level

id1IDScaleParticipant ID number

delivery2Delivery ReliabilityScaleOn a scale of 1 to 10, how would you rate the reliability of delivery of your orders?

Prodsat3Product SatisfactionScaleOn a scale of 1 to 10, how would you rate your satisfaction with the quality of your most recently purchased products?

Techsat4Technical SupportScaleOn a scale of 1 to 10, how would you rate your satisfaction with the technical support?

Salesat5Salesforce ScaleOn a scale of 1 to 10, how would you rate your satisfaction with the sales support?

Size6Firm SizeOrdinal0 = small (less than 100 emp.) 1 = large (100 or more)

Usage7Usage LevelScaleWhat percent of your purchases are from our company?

Satjan8Overall Satisfaction in JanuaryScaleOn a scale of 1 to 7, rate your overall satisfaction with your most recent purchasing experience.

Satjun8Overall Satisfaction in JuneScaleOn a scale of 1 to 7, rate your overall satisfaction with your most recent purchasing experience.

Structure10Structure of ProcurementNominalHow your purchasing is structured?0 = Decentralized; 1 = Centralized

OwnType11Type of OwnershipNominal0 = Publicly Traded; 1 Privately owned

PurType12Type of PurchasingNominal1 = Private Label; 2 = Company Brand; 3 = Both

Variables in the working file

For each research question, describe in your Microsoft Word document the application of the seven steps of the hypothesis testing model.

Step 1: State the hypothesis (null and alternate).Step 2: State your alpha (unless requested otherwise, this is always set to alpha = .05).Step 3: Collect the data (use one of the data sets).Step 4: Calculate your statistic and p-value. (This is where you run spss and examine your output files.) Step 5: Accept or reject the null hypothesis. (This is where you report the results of your analyses t (df) = t-value, p = sig. level.) Step 6: Assess the Risk of Type I and Type II Error. (Did the data meet the assumptions of the statistic, effect size, and sample size?)Step 7: State your results in APA style and format.

Example 7

Question 1: Is there a relationship among the variables measuring different aspects of customer satisfaction?1. Run a Pearson correlation matrix using delivery reliability, product satisfaction, technical support, sales satisfaction, overall satisfaction in January and overall satisfaction in June.

Correlations

Delivery ReliabilityProduct SatisfactionTechnical SupportSalesforceOverall Satisfaction in JanuaryOverall Satisfaction in June

Delivery ReliabilityPearson Correlation1.193.436**.112.206.628**

Sig. (2-tailed).180.002.439.152.000

N505050505050

Product SatisfactionPearson Correlation.1931.317*.726**.195.484**

Sig. (2-tailed).180.025.000.174.000

N505050505050

Technical SupporPearson Correlation.436**.317*1.133.340*.555**

Sig. (2-tailed).002.025.356.016.000

N505050505050

SalesforcePearson Correlation.112.726**.1331.173.326*

Sig. (2-tailed).439.000.356.229.021

N505050505050

Overall Satisfaction in JanuaryPearson Correlation.206.195.340*.1731.479**

Sig. (2-tailed).152.174.016.229.000

N505050505050

Overall Satisfaction in JunePearson Correlation.628**.484**.555**.326*.479**1

Sig. (2-tailed).000.000.000.021.000

N505050505050

**. Correlation is significant at the 0.01 level (2-tailed).

*. Correlation is significant at the 0.05 level (2-tailed).

2. Create a scatter plot for the following pairs: (1) delivery reliabilityoverall satisfaction in June; (2) product satisfactionoverall satisfaction in June; and delivery reliabilityproduct satisfaction.

The scatter diagram suggest that there is a weak positive correlation between the two variables.



3. Report the descriptive statistics, assumptions tests, as well as tests of statistical significance identify of positive and negative relationships.

Descriptive Statistics

MeanStd. DeviationN

Delivery Reliability4.341.67350

Product Satisfaction5.341.15450

Technical Support2.88.89550

Sales force2.66.84850

Overall Satisfaction in January3.581.37250

Overall Satisfaction in June4.70.95350

Students t test is adopted to check whether there is any significant positive correlation between the variables.

H0: Correlation coefficient =0H1: Correlation coefficient >0 (One sided hypothesis)

Test Statistic used is t test Significance level =0.05 Decision rule : Reject the null hypothesis if the p value is less than the significance level.

The Correlation coefficient with p value of the one tailed test is given below. Correlations

Delivery ReliabilityProduct SatisfactionTechnical SupportSalesforceOverall Satisfaction in JanuaryOverall Satisfaction in June

Delivery ReliabilityPearson Correlation1.193.436**.112.206.628**

Sig. (1-tailed)0.090.0010.2190.075.000

Product SatisfactionPearson Correlation.1931.317*.726**.195.484**

Sig. (1-tailed)0.0900.0125.0000.087.000

Technical SupportPearson Correlation.436**.317*1.133.340*.555**

Sig. (1-tailed).0010.01250.178.016.000

Sales forcePearson Correlation.112.726**.1331.173.326*

Sig. (1-tailed)0.219.0000.1780.11450.015

Overall Satisfaction in JanuaryPearson Correlation.206.195.340*.1731.479**

Sig. (1-tailed)0.0750.087.0160.1145.000

Overall Satisfaction in JunePearson Correlation.628**.484**.555**.326*.479**1

Sig. (1-tailed).000.000.000.0.015.000

**. Correlation is significant at the 0.01 level (2-tailed).

*. Correlation is significant at the 0.05 level (2-tailed).

Conclusion The t test for the significant correlation indicates that the correlation between Product satisfaction- Delivery reliability, Sales force- Delivery reliability, Overall satisfaction- Delivery reliability, Over all satisfaction Product satisfaction ,Sales force Technical support, Overall satisfaction in January Technical support are insignificant.

Question 2: Does delivery reliability impact overall satisfaction in June?1. Run a simple regression using delivery reliability as the independent variable and overall satisfaction in June as the dependent variable.

Coefficientsa

ModelUnstandardized CoefficientsStandardized CoefficientstSig.

BStd. ErrorBeta

1(Constant)3.147.29710.595.000

Delivery Reliability.358.064.6285.596.000

a. Dependent Variable: Overall Satisfaction in June

The estimated regression model is Overall Satisfaction in June = 3.147 +0.358 * Delivery Reliability

Model Summaryb

ModelRR SquareAdjusted R SquareStd. Error of the Estimate

1.628a.395.382.749

a. Predictors: (Constant), Delivery Reliability

b. Dependent Variable: Overall Satisfaction in June

The model adequacy measure R2 suggests that 39.5% variability in Overall Satisfaction in June can be explained by the regression model.

2. Report the descriptive statistics, assumptions tests (scatter plots), as well as tests of statistical significance.


MeanStd. DeviationN



Correlations

Overall Satisfaction in JuneDelivery Reliability

Pearson CorrelationOverall Satisfaction in June1.000.628

Delivery Reliability.6281.000

Sig. (1-tailed)Overall Satisfaction in June..000

Delivery Reliability.000.

NOverall Satisfaction in June5050

Delivery Reliability5050

The Correlation coefficient between Overall Satisfaction in June and delivery reliability is positive with 0.628. The regression coefficient of Delivery Reliability on Overall Satisfaction in June can be interpreted as For a unit increase in Delivery Reliability, the Overall Satisfaction in June increase by 0.358 unitsThe significance of this regression coefficient is tested using the t testH0: Regression coefficient =0H1: Regression coefficient > 0

Significance level =0.05Decision rule: Reject the null hypothesis if the p value is less than the significance level.Details T statistic =5.596P value =0.000Conclusion: Reject the null hypothesis. The sample provides enough evidence to support the claim that Delivery Reliability has a significant effect on Overall Satisfaction in June.

The assumption for the validity of regression analysis is checked using the residual analysis. The histogram and normal probability plots suggest that the residuals are normally distributed. The homogeneity of variance assumption is valid as the plots of residuals against the predicted values are random.

Question 3: Does delivery reliability and product satisfaction impact overall satisfaction in June?1. Run a multiple regression using delivery reliability as the independent variable and overall satisfaction in June as the dependent variable.

Coefficientsa

ModelUnstandardized CoefficientsStandardized CoefficientstSig.

BStd. ErrorBeta

1(Constant)1.662.4793.467.001

Delivery Reliability.316.058.5565.464.000

Product Satisfaction.312.084.3773.712.001

a. Dependent Variable: Overall Satisfaction in June

The estimated regression model is

Overall Satisfaction in June =1.662+ 0.316 * Delivery Reliability+0.312* Product Satisfaction

Model Summaryb

ModelRR SquareAdjusted R SquareStd. Error of the Estimate

1.729a.532.512.666

a. Predictors: (Constant), Product Satisfaction, Delivery Reliability

b. Dependent Variable: Overall Satisfaction in June

The model adequacy measure R2 suggests that 53.2 % variability in Overall Satisfaction in June can be explained by the regression model.

2. Report the descriptive statistics, assumptions tests (scatter plots), as well as tests of statistical significance.


MeanStd. DeviationN



Product Satisfaction5.341.15450

Correlations

Overall Satisfaction in JuneDelivery ReliabilityProduct Satisfaction

Pearson CorrelationOverall Satisfaction in June1.000.628.484

Delivery Reliability.6281.000.193

Product Satisfaction.484.1931.000

Sig. (1-tailed)Overall Satisfaction in June..000.000

Delivery Reliability.000..090

Product Satisfaction.000.090.

NOverall Satisfaction in June505050

Delivery Reliability505050

Product Satisfaction505050

The Correlation coefficient between Overall Satisfaction in June and delivery reliability is positive with 0.628 and Overall Satisfaction in June and Product satisfaction is 0.484 . The regression coefficient of Delivery Reliability on Overall Satisfaction in June can be interpreted as For a unit increase in Delivery Reliability, the Overall Satisfaction in June increase by 0.358 units,. For a unit increase in product satisfaction, the Overall Satisfaction in June increase by 0.312 units,.

The significance of this regression coefficient is tested using the t testH0: Regression coefficient =0H1: Regression coefficient > 0

Significance level =0.05Decision rule: Reject the null hypothesis if the p value is less than the significance level.Details Delivery ReliabilityProduct SatisfactionT statistic5.4643.712P value 0.0000.001

Conclusion: Reject the null hypothesis. The sample provides enough evidence to support the claim that Delivery Reliability and product satisfaction has a significant effect on Overall Satisfaction in June.

The assumption for the validity of regression analysis is checked using the residual analysis. The histogram and normal probability plots suggest that the residuals are normally distributed. The homogeneity of variance assumption is valid as the plots of residuals against the predicted values are random.

Write a brief conclusion statement summarizing your results. What can you tell this manufacturing company about the relationship among satisfaction variables? Are there any areas they need to improve? Does adding a second variable to the regression equation increase prediction of customer satisfaction?

The regression analysis indicates that both Delivery Reliability and Product SatisfactionSatisfaction variables have a significant effect on overall satisfaction in June. The multiple regression models is able to explain 52.3% variability in overall satisfaction in June. We may add more explanatory variables to improve the model adequacy to a higher level.It can be noted that the model adequacy increase from 39.5% to 52.3% due to the addition of Product Satisfaction as the second explanatory variable .

Example 8

Question 1: Before the change of ownership, the company was encouraging its customers to reduce private labeling as a way to reduce cost of goods sold. Explore the distribution of customers by purchase type. Does the distribution of customers (private label, brand label, or both) differ from what one would expect by chance? Does if differ they expect more brand labeling?

Type of Purchasing

Observed NExpected NResidual

Private label1816.71.3

Company Brand1816.71.3

Both1416.7-2.7

Total50

H0: There is no significant difference in the number of customers in the three categories.H1: There is significant difference in the number of customers in the three categories.Test Statistic used is Chi square test for goodness of fit.Significance level =0.05Decision rule: Reject the null hypothesis if the p value is less than the significance level.Details

Test Statistics

Type of Purchasing

Chi-Square.640a

df2

Asymp. Sig..726

a. 0 cells (.0%) have expected frequencies less than 5. The minimum expected cell frequency is 16.7.

Conclusion: Fails to reject the null hypothesis. The sample does not provides enough evidence to support the claim that there is significant difference in the number of customers in the three categories.

Question 2: Run a chi square goodness of fit using purchase type as the variable with all categories equal for the expected value.1. Run a chi square goodness of fit using purchase type as the variable with all categories unequal with 12, 26, and 12 as the expected values.

Type of Purchasing

Observed NExpected NResidual

Private label1812.06.0

Company Brand1826.0-8.0

Both1412.02.0

Total50

2. Report the observed and expected values and the tests of statistical significance.

H0: The number of customers in the three categories are (12,26,12).H0: The number of customers in the three categories are different from (12,26,12).

Significance level =0.05Decision rule: Reject the null hypothesis if the p value is less than the significance level.Details Test Statistics

Type of Purchasing

Chi-Square5.795a

df2

Asymp. Sig..055

a. 0 cells (.0%) have expected frequencies less than 5. The minimum expected cell frequency is 12.0.

Conclusion: Fails to reject the null hypothesis. The sample does not provides enough evidence to support the claim that there is significant difference in the number of customers in the three categories are different from (12,26,12).

Question 3: Is there a relationship between the company size and type of procurement?Run chi-square independence test (crosstabs) using company size and type of procurement. Use the chi square and the phi coefficient to evaluate the relationship and statistical significance. Report the observed and expected values and the tests of statistical significance.

H0: There is no association between company size and type of procurement.H1: There is association between company size and type of procurement.

Test Statistic used is Chi square test for independence Significance level =0.05Decision rule: Reject the null hypothesis if the p value is less than the significance level.Details

Firm Size * Structure of Procurement Crosstabulation

Count

Structure of ProcurementTotal

DecentralizedCentralized

Firm SizeSmall141327

Large121123

Total262450

Firm Size * Structure of Procurement Crosstabulation

Expected Count

Structure of ProcurementTotal

DecentralizedCentralized

Firm SizeSmall14.013.027.0

Large12.011.023.0

Total26.024.050.0

ValuedfAsymp. Sig. (2-sided)

Pearson Chi-Square.001a1.982

Continuity Correctionb.00011.000

Likelihood Ratio.0011.982

Fisher's Exact Test

Linear-by-Linear Association.0011.982

N of Valid Cases50

Conclusion: Fails to reject the null hypothesis. The sample does not provide enough evidence to support the claim that there is association between company size and type of procurement. The phi and chi square coefficients indicate jointly the strength and the significance of a relationship. The value of Phi is very small indicating that there is no relationship between company size and type of procurement.

Symmetric Measures

ValueApprox. Sig.

Nominal by NominalPhi-.003.982

Cramer's V.003.982

N of Valid Cases50

Documents

MODULE 6_7 Application Assignments