Question 1: Predict fuel consumption (mpg) based on variables in the dataset (use till number 6, acceleration). Investigate all the regression assumptions and follow all the procedure needed. Analyze your spss output (show interpretation of each table) and write down your conclusion after that stating any limitation.

Solution)

Variables Entered/Removeda
Model	Variables Entered	Variables Removed	Method
1	Acceleration, Weight, Cylinders, Horsepower, Displacementb	.	Enter
a. Dependent Variable: MPG
b. All requested variables entered.

In the above table the detail regarding the entered and removed variable is presented.

 

Model Summaryb
Model	R	R Square	Adjusted R Square	Std. Error of the Estimate	Change Statistics
					R Square Change	F Change	df1	df2	Sig. F Change
1	.841a	.708	.704	4.2471	.708	186.906	5	386	.000
a. Predictors: (Constant), Acceleration, Weight, Cylinders, Horsepower, Displacement
b. Dependent Variable: MPG

The value of R square in the above table is 0.708 The F value is 0.00 showing the significance of the model. The R square value is indicating that Acceleration, Horsepower and Weight does have impact on the dependent variable MPG.

 

ANOVAa
Model		Sum of Squares	df	Mean Square	F	Sig.
1	Regression	16856.526	5	3371.305	186.906	.000b
	Residual	6962.467	386	18.037
	Total	23818.993	391
a. Dependent Variable: MPG
b. Predictors: (Constant), Acceleration, Weight, Cylinders, Horsepower, Displacement

 

 

 

 

Coefficientsa
Model		Unstandardized Coefficients		Standardized Coefficients	t	Sig.
		B	Std. Error	Beta
1	(Constant)	46.264	2.669		17.331	.000
	Cylinders	-.398	.411	-.087	-.969	.333
	Displacement	-8.313E-5	.009	-.001	-.009	.993
	Horsepower	-.045	.017	-.223	-2.716	.007
	Weight	-.005	.001	-.564	-6.351	.000
	Acceleration	-.029	.126	-.010	-.231	.817
a. Dependent Variable: MPG

 

Through the above table the Std. error for Cylinders is 0.411. The value of beta has remain negative throughout the model. The lower t values showing higher amount of significance of the model.

 

Residuals Statisticsa
	Minimum	Maximum	Mean	Std. Deviation	N
Predicted Value	6.860	33.279	23.446	6.5659	392
Std. Predicted Value	-2.526	1.498	.000	1.000	392
Standard Error of Predicted Value	.231	1.840	.497	.170	392
Adjusted Predicted Value	6.584	33.311	23.443	6.5752	392
Residual	-11.5816	16.3416	.0000	4.2198	392
Std. Residual	-2.727	3.848	.000	.994	392
Stud. Residual	-2.752	3.860	.000	1.002	392
Deleted Residual	-11.7934	16.4443	.0030	4.2911	392
Stud. Deleted Residual	-2.776	3.931	.001	1.006	392
Mahal. Distance	.163	72.410	4.987	5.396	392
Cook's Distance	.000	.079	.003	.007	392
Centered Leverage Value	.000	.185	.013	.014	392
a. Dependent Variable: MPG

The value of Standard deviation shows deviation in the data. The higher value of standard deviation indicating dispersion in the data.

Conclusion on Statistics 

It is concluded that the value of R square in the above table is 0.708 The F value is 0.00 showing the significance of the model. The R square value is indicating that Acceleration, Horsepower and Weight does have impact on the dependent variable MPG. The higher values of standard deviation indicating dispersion in the data.

Question 2: Run an ANOVA analysis with the data. The dependent variable is the response to the advertisement. Factors are days of the week (Only use Monday and Friday) and type of news (Business, news and sports). Investigate the assumptions as well as the interactions. Write down your conclusions along with SPSS output interpretation.

Solution)

ANOVA
		Sum of Squares	df	Mean Square	F	Sig.
News	Between Groups	147.800	4	36.950	18.475	.000
	Within Groups	30.000	15	2.000
	Total	177.800	19
Business	Between Groups	32.300	4	8.075	4.750	.011
	Within Groups	25.500	15	1.700
	Total	57.800	19
Sports	Between Groups	102.500	4	25.625	16.356	.000
	Within Groups	23.500	15	1.567
	Total	126.000	19

 

The one-way ANOVA suggests that the relationship is significant it is significant for all three segments of news, the news itself, the business segment and the sports segment. Therefore, there is difference in days in terms of all of these segments.

Conclusion

It is concluded that suggests that the relationship is significant it is significant for all three segments of news, the news itself, the business segment and the sports segment. Difference in terms of day’s is evident in the above model.

Question 3: Run a logistic regression on the Bank Marketing data.  Use attributes (Variables) 1, 5, 6, 7, 8 and 12 only. Also use approximately 200 random data points from this dataset for analysis. Write down your conclusions.

Solution)

Logistic Regression

Categorical Variables Codings
		Frequency	Parameter coding
			(1)	(2)
contact	cellular	2896	1.000	.000
	telephone	301	.000	1.000
	unknown	1324	.000	.000
loan	no	3830	1.000
	yes	691	.000
housing	no	1962	1.000
	yes	2559	.000

The above table is showing the frequency of the variables. Cellular Contact, No loan and housing have the highest frequency.

Variables in the Equation
		B	S.E.	Wald	df	Sig.	Exp(B)
Step 0	Constant	-2.038-	.047	1915.134	1	.000	.130

The above table describes the Variables in the equation. It can be seen that there is no relationship between the variables in the marketing data. It can be said that the relationship is negative between the variables which include loan, housing, & contact. The value significance value of 0.00 shows significance of the model applied.

 

 

 

Variables not in the Equationa
			Score	df	Sig.
Step 0	Variables	age	9.192	1	.002
		balance	1.449	1	.229
		housing(1)	49.544	1	.000
		loan(1)	22.481	1	.000
		contact	87.870	2	.000
		contact(1)	63.765	1	.000
		contact(2)	3.027	1	.082
		campaign	16.904	1	.000
a. Residual Chi-Squares are not computed because of redundancies.

In the above table the relationship between the age and contact can be seen. The amount of balance and campaign are also showing significant amount of relationship.

Block 1: Method = Enter

Omnibus Tests of Model Coefficients
		Chi-square	df	Sig.
Step 1	Step	181.250	7	.000
	Block	181.250	7	.000
	Model	181.250	7	.000

The above table the value of Chi-square can be seen. The higher Chi-square value indicate low significance of the model.

Variables in the Equation
		B	S.E.	Wald	df	Sig.	Exp(B)
Step 1a	age	.007	.004	2.963	1	.085	1.007
	balance	.000	.000	.045	1	.832	1.000
	housing(1)	.469	.098	22.751	1	.000	1.599
	loan(1)	.760	.167	20.750	1	.000	2.138
	contact			65.020	2	.000
	contact(1)	1.155	.144	64.731	1	.000	3.173
	contact(2)	1.090	.216	25.486	1	.000	2.973
	campaign	-.104-	.024	18.389	1	.000	.901
	Constant	-3.912-	.274	204.273	1	.000	.020
a. Variable(s) entered on step 1: age, balance, housing, loan, contact, campaign.

It can be seen that the balance is insignificant and has no relationship, so the approval of loan has positive relationships with the variables Housing, loan and has negative relationship with the campaign, so it is advised that the company should not work its campaigns, instead it should focus on loans and housing of the client.

Conclusion on statistics

It is concluded that that the balance is insignificant and has no relationship, so the approval of loan has positive relationships with the variables Housing, loan and has negative relationship with the campaign, so it is advised that the company.