spss

ONE SAMPLE T-TEST

OBJECTIVE

To conduct one sample T-test using SPSS.

PROBLEM

A major oil company developed a petrol additive that was supposed to increase engine efficiency. Twenty two cars were test driven both with and without the additive and the number of kilometer per liter was recorded. Whether the car was automatic or manual was also recorded and coded as 1 = manual and 2 = automatic.

During an earlier trial 22 cars were test driven using the additive. The mean number of kilometers per liter was 10.5.

NULL HYPOTHESIS

There is no significant difference in engine efficiency between the present trial and the earlier trial.

ALTERNATE HYPOTHESIS

There is a significant difference in engine efficiency between the present trial and the earlier trial.

PROCEDURE

1. Select the Analyze menu.

2. Click on compare means and then one-sample T Test…. To open the One-Sample T Test dialogue box.

3. Select ‘withadd’ and move the variable into the Test Variable(s): box

4. In the Test Value: box type the mean score (10.5).

5. Click Ok.

OUTPUT

ONE SAMPLE T- TEST

One-Sample Statistics

22 13.86 2.748 .586withaddN Mean Std. Deviation

Std. ErrorMean

One-Sample Test

5.741 21 .000 3.364 2.15 4.58withaddt df Sig. (2-tailed)

MeanDifference Lower Upper

95% ConfidenceInterval of the

Difference

Test Value = 10.5

INFERENCE

1. The difference between the sample mean and the hypothesized mean is determined by consulting the t-value, degree of freedom (df) and two-tail significance.

2. If the value for two-tail significance is less than .05 (p<.05), then the difference between the means is significant.

3. The cars in the present trial appear to have greater engine efficiency than that of those in the earlier trial – t (21) = 5.74, p<.05.

RESULT

The output indicates that there is a significant difference in engine efficiency between the present trial and the earlier trial.

INDEPENDENT SAMPLE T-TEST

OBJECTIVE:

To find out the difference in opinion among two sets of people by Independent sample t-test

using SPSS.

PROBLEM:

As marketers of brand jeans, we want to find out whether a set of customers in Delhi and set

of customers in Mumbai thought of our brand in the same way or not. A small survey was

conducted in both the cities and the ratings were obtained on an interval scale 1-7. We want

to find out whether the two sets of rating are significantly different.

S.NO. RATING CITY S.NO. RATING CITY1 2 1 16 3 22 3 1 17 4 23 3 1 18 5 24 4 1 19 6 25 5 1 20 5 26 4 1 21 5 27 4 1 22 5 28 5 1 23 4 29 3 1 24 3 210 4 1 25 3 211 5 1 26 5 212 4 1 27 6 213 3 1 28 6 214 3 1 29 6 215 4 1 30 5 2

NULL HYPOTHESIS

There is no significant difference between the ratings given by the customers in Mumbai and

Delhi at 95% confidence interval.


There is significant difference between the ratings given by the customers in Mumbai and

Delhi at 95% confidence interval.

PROCEDURE:

1. The variables are entered in the variable view of the SPSS data editor where city is a

categorical variable using nominal measure and respondent’s ratings in scale.

2. In the value cell for city, enter the label values as “1- Mumbai” and “2-Delhi”.

3. The given data is entered in the data view.

4. Choose Analyse from the main menu.

5. Then choose Compare means > Independent sample T-test.

6. In the Independent sample t-test dialogue box, ratings given by the respondents are

entered as a test variable and the city they belong to is entered as grouping variable.

7. Enter the specified values for the groups after clicking defining groups.

8. The output chart is generated and it is analyzed and inference is obtained.

OUTPUT:

Group statistics

Respondent's city N Mean

Std. Deviation

Std. Error Mean

Respondent's rating

Mumbai 15 3.73 .884 .228Delhi 15 4.73 1.100 .284

Independent Samples Test

Levene's Test for Equality

of Variances

t-test for Equality of Means

FSig

.T Df

Sig. (2-

tailed)

Mean Difference

Std. Error Difference

95% Confidence Interval of the

DifferenceUpper Lower

RatingsEqual

variances assumed

.727.401

-2.745 28 .010 -1.000 .364 -1.746 -.254

Equal variances

not assumed

-2.745 26.759 .011 -1.000 .364 -1.748 -.252

INFERENCE:

1. The Independent samples t-test procedure compares the two group means (both

Mumbai and Delhi).

2. The mean value for the two groups are displayed in the Group Statistics table

(3.73 – 4.73 = - 1.00)

3. One test assumes that the variances of the two groups are equal. Levene tests this

assumption.

4. The significance value for the Levene’s test is high (0.401 is typically greater than of

0.10), so the result is assumed that there is equal variance for both the groups and the

second test is ignored.

5. The significance value for the t-test 0.010 is less than 0.05 and the confidence

interval for the mean difference does not contain zero.

6. So, the Null hypothesis is rejected and the Alternate hypothesis accepted. This

indicates that there is a significant difference between the two group means.

RESULT:

There is a significant difference in the ratings on the brand, given by the respondents in the

cities of Mumbai and Delhi.

PAIRED SAMPLE T-TEST

OBJECTIVE

To conduct a Paired Sample T-Test using SPSS.

PROBLEM

A major oil company developed a petrol additive that was supposed to increase engine efficiency. Twenty two cars were test driven both with and without the additive and the number of kilometer per liter was recorded. Whether the car was automatic or manual was also recorded and coded as 1 = manual and 2 = automatic

Does engine efficiency improve when the additive is used? This is a repeated measure t-test design.

NULL HYPOTHESIS

There is no significant difference exists between engine efficiency with and without the additive.


There is a significant difference exists between engine efficiency with and without the additive.

PROCEDURE

1. Select the Analyze menu.

2. Click on Compare Means and then Paired-Samples T Test… to open the Paired-Sample T Test dialogue box.

3. Select the variables ‘without and withadd’ and move the variables into the Paired Variables: box

4. Click Ok.

OUTPUT

PARIED SAMPLE T-TEST

Paired Samples Statistics

8.50 22 3.335 .711

13.86 22 2.748 .586

without

withadd

Pair1

Mean N Std. DeviationStd. Error

Mean

Paired Samples Correlations

22 .559 .007without & withaddPair 1N Correlation Sig.

INFERENCE

1. It can be determined that whether the groups come from the same or different populations

2. The significance is determined by looking at the probability level (p) specified under the heading ‘two tail significance’.

3. If the probability value is less than the specified alpha value, then the observed t-value is significant

4. The 95 percent confidence interval indicates that 95 percent of the time the interval specified will contain the true difference between the population means

5. The additive significantly improves the number of kilometers to the liter, t(21) = 8.66, p<.05

RESULT

The output shows that there is a significant difference exists between engine efficiency with and without the additive.

Paired Samples Test

-5.364 2.904 .619 -6.651 -4.076 -8.663 21 .000without - withaddPair 1Mean Std. Deviation

Std. ErrorMean Lower Upper

95% ConfidenceInterval of the

Difference

Paired Differences

t df Sig. (2-tailed)

ONE WAY ANOVA

OBJECTIVE:

To test the preferred ad copy by the target population before the launch of its campaign.

PROBLEM:

There are three different versions of advertising copy created by an advertising agency for a

campaign. Let us call these versions of copy as adcopy 1, 2 and 3. A sample of 18

respondents is selected from the target population in the nearby areas of the city. At random,

these 18 respondents are assigned to the 3 versions of ad copy. Each version of ad copy is

thus shown to six of the respondents. The respondents are asked to rate their liking for the ad

copy shown to them on a scale of 1 to 10. (1 = Not liked at all, 10 = Liked a lot, and other

values in between these two).

S. No. Ad copy Rating1 1 6.002 1 7.003 1 5.004 1 8.005 1 8.006 1 8.007 2 4.008 2 4.009 2 5.0010 2 7.0011 2 7.0012 2 6.0013 3 5.0014 3 5.0015 3 4.0016 3 7.0017 3 8.0018 3 7.00

Null Hypothesis

There is no difference in the ratings between the three versions of the ad copy at 95%

confidence level.

Alternative Hypothesis

There is a significant difference between the three versions of the ad copy at 95%

confidence level.

PROCEDURE:

1. The given data is entered in the variable view and then in the data view.

2. Choose Analyse > Compare means > One-way ANOVA.

3. In the one-way ANOVA dialog box, select ‘ratings’ as the dependent list and ad copy

as its factor.

4. Select other variables as required.

5. The output chart is generated and analysed and inference obtained.

OUTPUT:

DescriptivesRatings

NMea

n

Std. Deviatio

n

Std. Erro

r

95% Confidence Interval for Mean Minimu

mMaximu

mLower Bound

Upper Bound

Ad copy1

67.0000

1.26491.51640

5.6726 8.3274 5.00 8.00

Ad copy2

65.5000

1.37840.56273

4.0535 6.9465 4.00 7.00

Ad copy3

66.0000

1.54919.63246

4.3742 7.6258 4.00 8.00

Total18

6.1667

1.46528.34537

5.4380 6.8953 4.00 8.00

Test of Homogeneity of Variances

Ratings Levene Statistic

df1 df2 Sig.

.536 2 15 .596

ANOVARatings

Sum of Square

sdf

Mean Square

F Sig.

Between Groups

7.000 2 3.5001.78

0.203

Within Groups

29.500 15 1.967

Total 36.500 17

INFERENCE:

1. The descriptive of the ratings are obtained in terms of mean and standard deviation.

2. The mean values of the three versions of ad copy are displayed.

3. The significance value for the Levene’s test of homogeneity of variables is high

(0.596 >0.05) and the ANOVA table, sig represents the significance level of F-test.

4. Therefore the null hypothesis is not rejected and alternate hypothesis is not accepted.

Hence the variances for the three versions are equal and the assumption is justified.

RESULT:

There is no significant difference in the preferences over the three versions of the ad

copy by a target population before the launch of its campaign.

CORRELATION

OBJECTIVE:

To find the interrelationship between the dependent and the independent variables.

PROBLEM:

A manufacturer and the marketer of electric motors would like to build a regression

model consisting of 5 or 6 independent variables to sales. Past data has been collected for 15

sales territories, on sales and 6 different independent variables. Build a regression model and

recommend whether or not it should be used by the company.

Dependent Variable:

Y Sales (in Rs. Lakhs) in the territory.

Independent Variables:

X1 Market potential of the territory.

X2 No. of dealers of the Company in the territory.

X3 No. of sales person in the territory.

X4 Index of the competitor activity in the territory on a 5 point scale.

X5 No. of service people in the territory.

X6 No. of existing customers in the territory.

S. No. Sales (Y) Potential (X1)

Dealers (X2)

People (X3)

Competition (X4)

Service (X5)

Customers (X6)

1 5 25 1 6 5 2 202 60 150 12 30 4 5 503 20 45 5 15 3 2 254 11 30 2 10 3 2 205 45 75 12 20 2 4 306 6 10 3 8 2 3 167 15 29 5 18 4 5 308 22 43 7 16 3 6 409 29 70 4 15 2 5 3910 3 40 1 6 5 2 511 16 40 4 11 4 2 1712 8 25 2 9 3 3 1013 18 32 7 14 3 4 3114 23 73 10 10 4 3 4315 81 150 15 35 4 7 70

Null Hypothesis

There is no significant relationship between the independent and the dependent

variables at 95% confidence interval.


There is significant relationship between the independent and dependent variables at

95% confidence interval.

PROCEDURE:

Let the estimating equation be Y= a1X1+a2X2+a3X3+a4X4+a5X5

1. The variables are defined in the variable view of the SPSS data editor.

2. Enter the data in the data view.

3. Choose Analyze > Correlate > Bivariate from the main menu.

4. In the bivariate correlations dialogue box select all the dependent and independent

variables.

5. Select Pearson’s correlation coefficient with test of significance being one tailed.

6. Also include the statistics for mean and standard deviation.

7. The output chart is generated, analyzed and inference obtained

OUTPUT:

Descriptive Statistics

CorrelationSales

in Rs.lakh

s in the

territory

Market potential

in the territory(in Rs. lakhs)

No. of dealers of

the company

in the territory

No. of sales

people in the

territory

Index of competit

or activity in the

territory

No. of service

people in the

territory

No. of existing customers in the territory

Sales in Rs.lakhs

in the territory

Pearson Correlation

1 .945(**) .908(**).953(*

*)-.046 .726(**) .878(**)

Sig. (1-tailed) .000 .000 .000 .436 .001 .000N 15 15 15 15 15 15 15

Market potential

in the territory (in Rs. Lakhs)

Pearson Correlation

.945(**)

1 .837(**).877(*

*).140 .613(**) .831(**)

Sig. (1-tailed) .000 .000 .000 .309 .008 .000

N 15 15 15 15 15 15 15

No. of dealers of

the company

in the territory

Pearson Correlation

.908(**)

.837(**) 1.855(*

*)-.082 .685(**) .860(**)

Sig. (1-tailed) .000 .000 .000 .385 .002 .000

N 15 15 15 15 15 15 15

No. of sales

people in the

Pearson Correlation

.953(**)

.877(**) .855(**) 1 -.036 .794(**) .854(**)

Sig. (1-tailed) .000 .000 .000 .449 .000 .000N 15 15 15 15 15 15 15

MeanStd.

Deviation N

Sales in Rs.lakhs in the territory 24.13 21.980 15Market potential in the territory (in Rs.

lakhs)55.80 42.543 15

No. of dealers of the company in the territory

6.00 4.408 15

No. of sales people in the territory 14.87 8.340 15Index of competitor activity in the

territory3.40 .986 15

No. of service people in the territory 3.67 1.633 15

No. of existing customers in the territory 29.73 16.829 15

territoryIndex of competit

or activity in the

territory

Pearson Correlation

-.046 .140 -.082 -.036 1 -.178 -.015

Sig. (1-tailed) .436 .309 .385 .449 .263 .479

N 15 15 15 15 15 15 15

No. of service

people in the

territory

Pearson Correlation

.726(**)

.613(**) .685(**).794(*

*)-.178 1 .818(**)

Sig. (1-tailed) .001 .008 .002 .000 .263 .000

N 15 15 15 15 15 15 15


Pearson Correlation

.878(**)

.831(**) .860(**).854(*

*)-.015 .818(**) 1

Sig. (1-tailed) .000 .000 .000 .000 .479 .000

N 15 15 15 15 15 15 15

INFERENCE:

The correlations table shows Pearson correlation coefficients, significance values, and

the number of cases with non missing values.

1. The Pearson correlation coefficient measures the linear association between two

variables if the value of the correlation coefficient ranges from -1 to 1.

2. The sign of the correlation coefficient indicates the direction of the relationship.

Hence from the inference there is a negative relation between sales and the index of

the competitor activity and the positive relationship with market potential, number of

dealers, no of salespersons, number of service people and the no of existing

customers.

3. The absolute value of the correlation coefficient indicates the strength, with larger

absolute values indicating stronger relationships.

4. The significance levels of market potential is 0.000, no of service people is 0.001 and

no of existing customers is 0.000 which is less than 0.05. So null hypothesis is

rejected and alternate hypothesis is accepted. Hence it indicates that the correlation is

significant and the variables are linearly related with sales.

5. The significance level of the index of the competitor 0.436 is greater than 0.05 then

the correlation is not significant and the variable is not linearly related.

6. This indicates that the manufacturer should not consider the index of the competitor

since it does not affect the sales.

RESULT:

Hence there is dependence between the sales (dependent variable) and the market

potential of the territory, number of dealers of the company in the territory, number of sales

person in the territory, number of service people in the territory, number of existing

customers in the territory (independent variables). Index of the competitor activity in the

territory and sales are negatively correlated.

FACTOR ANALYSIS

OBJECTIVE:

To find the factors which are fewer but linear combinations of original 10 variables.

PROBLEM:

A two wheeler manufacturer is interested in determining which variables potential

customers think about when they consider his product. Twenty two-wheeler owners were

surveyed by the manufacturer. They were asked to indicate on a 7 point scale,

1- Completely agree to 7- Completely disagree. Their agreement or disagreement with a set

of 10 statements relates to their perception and some attributes of the two-wheeler. Use

factorial analysis to find underlying factors which are fewer but are linear combinations of

original 10 variables.

TEN STATEMENTS:

1. I use a two-wheeler because it is affordable.

2. It gives me a sense of freedom to own a two-wheeler.

3. Low maintenance cost makes a two-wheeler very economical in a long run.

4. Two-wheeler is essentially a man’s vehicle.

5. I feel very powerful when I am on my two-wheeler.

6. Some of my friends who don’t have their own vehicle is jealous of me.

7. I feel good whenever I see the ad of my two-wheeler.

8. My vehicle gives me a comfortable ride.

9. I think two-wheelers are safe way to travel.

10. Three people should be legally allowed to travel on a two-wheeler.

S.No. Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q101 1 4 1 6 5 6 5 2 3 22 2 3 2 4 3 3 3 5 5 23 2 2 2 1 2 1 1 7 6 24 5 1 4 2 2 2 2 3 2 35 1 2 2 5 4 4 4 1 1 26 3 2 3 3 3 3 3 6 5 37 2 2 5 1 2 1 2 4 4 58 4 4 3 4 4 5 3 2 3 39 2 3 2 6 5 6 5 1 4 110 1 4 2 2 1 2 1 4 4 111 1 5 1 3 2 3 2 2 2 112 1 6 1 1 1 1 1 1 2 213 3 1 4 4 4 3 3 6 5 314 2 2 2 2 2 2 2 1 3 215 2 5 1 3 2 3 2 2 1 616 5 6 3 2 1 3 2 5 5 417 1 4 2 2 1 2 1 1 1 318 2 3 1 1 2 2 2 3 2 219 3 3 2 3 4 3 4 3 3 320 4 3 2 7 6 6 6 2 3 6

PROCEDURE:


2. Enter the given data in the data view.

3. Choose Analyze > Data reduction > Factor analysis from the main menu and enter the

variables.

4. In the factor analysis dialogue box select the analysis variables and check the options

as required.

5. The output chart is generated, analyzed and inference obtained.

OUTPUT:


MeanStd.

DeviationAnalysis

NIt is affordable 2.35 1.309 20Gives sense of freedom 3.25 1.482 20Economical 2.25 1.118 20Man's vehicle 3.10 1.804 20Feel powerful 2.80 1.508 20Friends would be jealous 3.05 1.605 20Feel good to see ad of my vehicle

2.70 1.455 20

Comfortable driving 3.05 1.905 20Safe way to travel 3.20 1.508 203 people should be legally allowed

2.80 1.473 20

Communalities

InitialExtractio

nIt is affordable 1.000 .722Gives sense of freedom 1.000 .452Economical 1.000 .731Man's vehicle 1.000 .945Feel powerful 1.000 .950Friends would be jealous 1.000 .914Feel good to see ad of my vehicle

1.000 .955

Comfortable driving 1.000 .799Safe way to travel 1.000 .7773 people should be legally allowed

1.000 .789

Extraction Method: Principal Component Analysis.

Component Matrix (a)

Component1 2 3

It is affordable .176 .670 .493Gives sense of freedom -.136 -.608 .254Economical -.107 .820 .218Man's vehicle .966 -.036 -.097Feel powerful .951 .166 -.136Friends would be jealous .952 -.084 -.025Feel good to see ad of my vehicle

.971 .096 -.046

Comfortable driving -.322 .775 -.308Safe way to travel -.069 .735 -.4823 people should be legally allowed

.161 .319 .814


1. 3 components extracted.

Component Score Coefficient Matrix

Component 1 2 3It is affordable .004 .023 .434Gives sense of freedom -.063 -.278 .043Economical -.041 .176 .283Man's vehicle .256 .003 -.042Feel powerful .257 .081 -.030Friends would be jealous .245 -.038 -.007Feel good to see ad of my vehicle

.253 .026 .014

Comfortable driving -.047 .360 -.057Safe way to travel .033 .406 -.1663 people should be legally allowed

-.032 -.203 .568


Rotation Method: Varimax with Kaiser Normalization.

Component Scores.

Component Score Covariance Matrix

Component

1 2 3

1 1.000 .000 .0002 .000 1.000 .0003 .000 .000 1.000


Rotation Method: Varimax with Kaiser Normalization.

Component Scores.

INFERENCE:

Factor analysis is primarily used for data reduction or structure detection.

1. Communalities indicate the amount of variance in each variable that is accounted for.

2. Communalities table reports the factor loadings for each variable on the unrotated

components or factors.

3. Rotated component matrix table (called the Pattern Matrix for oblique rotations)

reports the factor loadings for each variable on the components or factors after

rotation.

4. Group the factors which have high values.

5. Here man’s vehicle, feel powerful, friend would be jealous and feel good to see ad of

my vehicle have high value (greater than 0.5). So we can group them into component

1.

6. Similarly economical, comfortable driving and safe way to travel have high value and

hence are grouped in component 2.

7. Finally, it is affordable and three people should be legally allowed are grouped into

component 3.

8. Since value of sense of freedom is negative in all the three components this factor is

eliminated.

RESULT:

The ten factors are clustered into three components

DISCRIMINANT ANALYSIS

OBJECTIVE:

To conduct Discriminant Analysis for the given data using SPSS software

PROBLEM:

Conduct Discriminant Analysis that predicts membership of two groups based on the

dependent variable category and creating the discriminant equation with inclusion of 17

independent variables selected by a step-wise procedure based on minimization of Wilk’s

Lambda at each step.

NULL HYPOTHESIS

There is no discrimination in membership of two groups.


There is discrimination in membership of two groups.

PROCEDURE:

1. Select Analyze from the menu Classify Discriminant.

2. Select grouping variable (1,2).

3. Define the range – min:1 & max: 2.

4. Select all the variables as independent variables select ‘Use stepwise method’.

5. Click Statistics check Means, Univariate ANOVA, Box’s M and Under

standardized Continue.

6. Click select Wilk’s Lambda method and enter F value – Entry: 1.15 and Exit: 1

Continue.

7. Click Classify Check All groups equal, Case wise results, Summary table,

Combined-groups Continue.

8. Click OK.

OUTPUT:

Analysis Case Processing Summary

50 100.0

0 .0

0 .0

0 .0

0 .0

50 100.0

Unweighted CasesValid

Missing or out-of-rangegroup codes

At least one missingdiscriminating variable

Both missing orout-of-range group codesand at least one missingdiscriminating variable

Total

Excluded

Total

N Percent

Box's Test of Equality of Covariance Matrices

Log Determinants

6 3.031

6 2.918

6 3.800

1=COMPLETEDPHD, 2=DID NOTCOMPLETE PHDFINISH

NOT FINISH

Pooled within-groups

RankLog

Determinant

The ranks and natural logarithms of determinantsprinted are those of the group covariance matrices.

Test Results

39.633

1.633

21

8474.108

.034

Box's M

Approx.

df1

df2

Sig.

F

Tests null hypothesis of equal population covariance matrices.

Tests of Equality of Group Means

.795 12.356 1 48 .001

.951 2.493 1 48 .121

.998 .113 1 48 .738

.628 28.393 1 48 .000

.974 1.283 1 48 .263

.650 25.889 1 48 .000

.756 15.476 1 48 .000

.534 41.969 1 48 .000

.679 22.722 1 48 .000

1.000 .007 1 48 .934

.993 .337 1 48 .564

.768 14.526 1 48 .000

.904 5.079 1 48 .029

.787 13.017 1 48 .001

.972 1.378 1 48 .246

OVERALL COLLEGE GPA

MAJOR AREA GPA

GRE SCORE ONSPECIALITY EXAM

GRE SCORE ONQUANTATIVE

GRE SCORE ON VERBAL

FIRST LETTER OFRECOMMENDATION

SECOND LETTER OFRECOMMENDATION

THIRD LETTER OFRECOMMENDATION

STUDENTS MOTIVATION

STUDENTS EMOTIONALSTABILITY

FINAICIAL/PERSONALRESOURCES TOCOMPLETE

AGE IN YEARS AT ENTRY

ABILITY TO INTERACTEASILY

RATING OF STUDENTHOSTILITY

MEAN RATING OFSELECTORSIMPRESSION OFAPPLICANT

Wilks'Lambda F df1 df2 Sig.

Stepwise Statistics:

Variables Entered/Removeda,b,c,d

THIRDLETTEROFRECOMMENDATION

.534 1 1 48.000 41.969 1 48.000 .000

STUDENTSMOTIVATION

.451 2 1 48.000 28.638 2 47.000 .000

FIRSTLETTEROFRECOMMENDATION

.415 3 1 48.000 21.602 3 46.000 .000

AGE INYEARSATENTRY

.391 4 1 48.000 17.495 4 45.000 .000

MEANRATINGOFSELECTORSIMPRESSION OFAPPLICANT

.367 5 1 48.000 15.188 5 44.000 .000


.348 6 1 48.000 13.446 6 43.000 .000

Step1

2

3

4

5

6

Entered Statistic df1 df2 df3 Statistic df1 df2 Sig.

Exact F

Wilks' Lambda

At each step, the variable that minimizes the overall Wilks' Lambda is entered.

Maximum number of steps is 30.a.

Minimum partial F to enter is 1.15.b.

Maximum partial F to remove is 1.c.

F level, tolerance, or VIN insufficient for further computation.d.

Wilks' Lambda

1 .534 1 1 48 41.969 1 48.000 .000

2 .451 2 1 48 28.638 2 47.000 .000

3 .415 3 1 48 21.602 3 46.000 .000

4 .391 4 1 48 17.495 4 45.000 .000

5 .367 5 1 48 15.188 5 44.000 .000

6 .348 6 1 48 13.446 6 43.000 .000

Step1

2

3

4

5

6

Number ofVariables Lambda df1 df2 df3 Statistic df1 df2 Sig.

Exact F

Variables in the Analysis

1.000 41.969

.987 23.774 .679

.987 8.633 .534

.957 15.183 .552

.934 4.617 .457

.909 3.943 .451

.955 12.842 .503

.913 3.122 .419

.908 3.418 .421

.969 2.733 .415

.935 13.679 .481

.898 3.642 .397

.897 2.417 .387

.943 3.452 .396

.926 2.941 .391

.935 12.445 .448

.882 4.182 .381

.892 2.580 .369

.885 4.599 .385

.926 2.755 .370

.897 2.372 .367



STUDENTS MOTIVATION


STUDENTS MOTIVATION



STUDENTS MOTIVATION




STUDENTS MOTIVATION





STUDENTS MOTIVATION





Step1

2

3

4

5

6

Tolerance F to RemoveWilks'

Lambda

Standardized Canonical Discriminant Function Coefficientts

.312

.607

.393

.299

.409

.316



STUDENTS MOTIVATION




1

Function

Summary of Canonical Discriminant Functions

Eigenvalues

1.876a 100.0 100.0 .808Function1

Eigenvalue % of Variance Cumulative %CanonicalCorrelation

First 1 canonical discriminant functions were used in theanalysis.

a.

Wilks' Lambda

.348 47.541 6 .000Test of Function(s)1

Wilks'Lambda Chi-square df Sig.

Structure Matrix

.683

.547

.536

.502

.402

-.335

.278

.237

.178

.129

-.126

.124

-.068

.061

-.027


GRE SCORE ONQUANTATIVE

a


STUDENTS MOTIVATION


RATING OF STUDENTHOSTILITY

a

SECOND LETTER OFRECOMMENDATION

a

OVERALL COLLEGE GPAa

ABILITY TO INTERACTEASILY

a

MAJOR AREA GPAa

GRE SCORE ONSPECIALITY EXAM

a


GRE SCORE ON VERBALa


STUDENTS EMOTIONALSTABILITY

a

1

Function

Pooled within-groups correlations between discriminatingvariables and standardized canonical discriminant functions Variables ordered by absolute size of correlation within function.

This variable not used in the analysis.a.

Canonical Discriminant Function Coefficients

.288

.617

.490

.175

.091

.262

-15.564



STUDENTS MOTIVATION




(Constant)

1

Function

Unstandardized coefficients

Functions at Group Centroids

1.342

-1.342


NOT FINISH

1

Function

Unstandardized canonical discriminantfunctions evaluated at group means

Casewise Statistics

1 1 .422 1 .997 .645 2 .003 12.160 2.145

1 1 .698 1 .990 .150 2 .010 9.436 1.730

1 1 .934 1 .967 .007 2 .033 6.766 1.259

1 1 .699 1 .929 .149 2 .071 5.281 .956

2 1** .755 1 .941 .097 2 .059 5.629 1.031

1 1 .178 1 .999 1.812 2 .001 16.244 2.688

1 1 .947 1 .968 .004 2 .032 6.851 1.275

1 1 .765 1 .988 .090 2 .012 8.901 1.641

1 1 .439 1 .997 .599 2 .003 11.958 2.116

1 1 .795 1 .987 .068 2 .013 8.669 1.602

1 1 .345 1 .998 .891 2 .002 13.165 2.286

1 1 .620 1 .906 .246 2 .094 4.788 .846

1 1 .668 1 .991 .184 2 .009 9.693 1.771

2 1** .381 1 .777 .769 2 .223 3.266 .465

1 1 .573 1 .994 .318 2 .006 10.551 1.906

1 1 .239 1 .609 1.386 2 .391 2.271 .165

1 1 .804 1 .986 .062 2 .014 8.601 1.591

1 1 .065 1 1.000 3.393 2 .000 20.487 3.184

1 1 .231 1 .595 1.436 2 .405 2.207 .144

1 1 .970 1 .976 .001 2 .024 7.408 1.380

2 2 .708 1 .931 .141 1 .069 5.332 -.967

1 1 .677 1 .991 .173 2 .009 9.611 1.758

1 1 .583 1 .994 .301 2 .006 10.450 1.891

1 1 .290 1 .998 1.118 2 .002 13.998 2.399

1 1 .980 1 .975 .001 2 .025 7.340 1.367

2 2 .644 1 .992 .214 1 .008 9.902 -1.805

2 2 .854 1 .957 .034 1 .043 6.251 -1.158

2 2 .598 1 .993 .278 1 .007 10.316 -1.870

2 2 .580 1 .893 .306 1 .107 4.541 -.789

1 2** .277 1 .664 1.184 1 .336 2.547 -.254

2 2 .903 1 .964 .015 1 .036 6.564 -1.220

2 2 .555 1 .883 .348 1 .117 4.385 -.752

2 2 .322 1 .998 .982 1 .002 13.508 -2.333

2 2 .999 1 .973 .000 1 .027 7.195 -1.340

2 2 .432 1 .997 .617 1 .003 12.037 -2.127

2 2 .401 1 .997 .706 1 .003 12.422 -2.182

2 2 .345 1 .744 .893 1 .256 3.025 -.397

2 2 .649 1 .992 .207 1 .008 9.852 -1.797

1 2** .800 1 .949 .064 1 .051 5.909 -1.089

2 2 .701 1 .990 .148 1 .010 9.414 -1.726

2 2 .326 1 .724 .966 1 .276 2.894 -.359

2 2 .152 1 .999 2.054 1 .001 16.953 -2.775

2 2 .986 1 .972 .000 1 .028 7.108 -1.324

2 2 .202 1 .999 1.629 1 .001 15.687 -2.619

2 2 .973 1 .976 .001 1 .024 7.385 -1.375

2 2 .709 1 .990 .140 1 .010 9.350 -1.716

1 2** .892 1 .962 .018 1 .038 6.495 -1.206

2 2 .426 1 .812 .634 1 .188 3.565 -.546

2 2 .412 1 .997 .673 1 .003 12.280 -2.162

2 2 .715 1 .990 .134 1 .010 9.301 -1.708

Case Number1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

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20

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22

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50

OriginalActual Group

PredictedGroup p df

P(D>d | G=g)

P(G=g | D=d)

SquaredMahalanobisDistance to

Centroid

Highest Group

Group P(G=g | D=d)

SquaredMahalanobisDistance to

Centroid

Second Highest Group

Function 1

DiscriminantScores

Misclassified case**.

Classification Resultsa

22 3 25

2 23 25

88.0 12.0 100.0

8.0 92.0 100.0


NOT FINISH

FINISH

NOT FINISH

Count

%

OriginalFINISH NOT FINISH

Predicted GroupMembership

Total

90.0% of original grouped cases correctly classified.a.

INFERENCE:

1. The F and significant F values identify for which variables the two groups differ

significantly.

2. The canonical correlation coefficient is .808 which shows strong correlation.

3. The significance values are <0.05

RESULT:

There is discrimination in membership between two groups.

CLUSTER ANALYSIS

OBJECTIVE:

To conduct K-means cluster analysis.

PROBLEM:

Brands of 21 VCRs are given along with their attributes. Determine the hierarchical K-means cluster analysis.

PROCEDURE:

1. Select Analyze Classify K-means cluster2. Select all variables and move into the variables boz.3. Label case as brand4. Enter ‘number of clusters’ as 35. Then select the required options from ‘Save’. Then Continue6. Click Options check initial cluster centre continue ok

OUTPUT:

Quick Cluster

Iteration Historya

52.791 70.267 80.393

14.037 12.547 .000

.000 .000 .000

Iteration1

2

3

1 2 3

Change in Cluster Centers

Convergence achieved due to no or smallchange in cluster centers. The maximumabsolute coordinate change for any center is.000. The current iteration is 3. The minimumdistance between initial centers is 335.404.

a.

Number of Cases in each Cluster

8.000

8.000

5.000

21.000

.000

1

2

3

Cluster

Valid

Missing

Initial Cluster Centers

200 535 380

3 5 2

3 5 2

3 5 2

3 5 2

3 5 2

3 3 3

4 4 3

2 3 3

4 5 4

4 4 4

4 4 4

4 4 4

8 6 4

365 365 30

3 3 4

3 4 4

3 4 4

3 12 6

3 12 6

3 12 6

price

pictur1

pictur2

pictur3

pictur4

pictur5

program

recept1

recept3

audio1

audio2

audio3

features

events

days

remote1

remote2

remote3

extras1

extras2

extras3

1 2 3

Cluster

Final Cluster Centers

239 453 460

3 4 4

3 4 4

3 4 4

3 4 4

3 4 4

3 3 3

4 4 3

2 3 3

4 4 4

4 4 3

4 4 4

4 4 4

8 7 6

365 365 30

3 3 4

3 3 4

3 3 4

3 8 10

3 8 10

3 8 10

price

pictur1

pictur2

pictur3

pictur4

pictur5

program

recept1

recept3

audio1

audio2

audio3

features

events

days

remote1

remote2

remote3

extras1

extras2

extras3

1 2 3

Cluster

INFERENCE:

Three clusters were formed.

RESULT:

Thus cluster analysis was done using SPSS.

REGRESSION

OBJECTIVE:

To find the dependency of the variables with respect to the sales of the company.

PROBLEM:





Dependent Variable:










Dealers (X2)

People (X3)

Competition (X4)

Service (X5)

Customers (X6)

1 5 25 1 6 5 2 202 60 150 12 30 4 5 503 20 45 5 15 3 2 254 11 30 2 10 3 2 205 45 75 12 20 2 4 306 6 10 3 8 2 3 167 15 29 5 18 4 5 308 22 43 7 16 3 6 409 29 70 4 15 2 5 3910 3 40 1 6 5 2 511 16 40 4 11 4 2 1712 8 25 2 9 3 3 1013 18 32 7 14 3 4 3114 23 73 10 10 4 3 4315 81 150 15 35 4 7 70

Null Hypothesis

There is no dependence between the independent variables, market potential, no. of

dealers, no. of sales person, index of the competitor activity, no. of service people, no. of

existing customers and the dependent variable sales at 95% confidence interval.


There is dependence between the independent and dependent variables at 95%

confidence interval.

PROCEDURE:




3. Choose Analyze >Regression > Linear from the main menu.

4. In the linear regression dialogue box enter sales as the dependent variable and all the

other variables as the independent variables.

5. Click the statistics button and click the regression coefficient estimates, model fit and

descriptive check boxes.

6. The output chart is generated and it is analyzed and inference obtained.

OUTPUT:

Model Summary (b)

Model

R R SquareAdjusted R Square

Std. Error of the

Estimate1 .989(a) .977 .960 4.391

a Predictors: (Constant), No. of existing customers in the territory, Index of

competitor activity in the territory, No. of service people in the territory, Market potential in

the territory (in Rs. lakhs), No. of dealers of the company in the territory, No. of sales people

in the territory

b Dependent Variable: Sales in Rs.lakhs in the territory

ANOVA (b)

Model

Sum of Squares

dfMean Square

F Sig.

1Regressi

on6609.48

56 1101.581

57.133

.000(a)

Residual 154.249 8 19.281

Total6763.73

314

a Predictors: (Constant), No. of existing customers in the territory, Index of competitor

activity in the territory, No. of service people in the territory, Market potential in the territory

(in Lakhs), No. of dealers of the company in the territory, No. of sales people in the territory

b Dependent Variable: Sales in Lakhs in the territory

Coefficients (a)

Model Unstandardize

d Coefficients

Standardized

Coefficients

T Sig.

95% Confidence Interval for B

BStd.

ErrorBeta

Lower Bound

Upper Bound

1 (Constant) -3.173 5.813 -.546 .600 -16.579 10.233Market potential in the territory (in Rs. Lakhs)

.227 .075 .439 3.040 .016 .055 .399


.819 .631 .164 1.298 .230 -.636 2.275

No. of sales people in the territory

1.091 .418 .414 2.609 .031 .127 2.055

Index of competitor activity in the territory

-1.893 1.340 -.085 -1.413 .195 -4.982 1.197

No. of service people in the territory

-.549 1.568 -.041 -.350 .735 -4.166 3.067


.066 .195 .050 .338 .744 -.384 .516

a Dependent Variable: Sales in Rs.lakhs in the territory

Therefore the estimating equation is: Y= 0.439X1+0.164X2+0.414X3-0.085X4-0.041X5+0.05X6

INFERENCE:

1. The variables are selected using the enter method.

2. The values of R ranging from 0 to 1 are determined. Larger values indicate stronger

relationship.

3. The significance value .000 arrived through ANOVA is less than 0.05. So null

hypothesis is rejected and alternate hypothesis is accepted. Hence the independent

variables explain the variations of the dependent variable.

4. The t statistics shows the relative importance of each variable with respect to the

regression coefficients where t values below -2 or above +2 are good predictors.

5. The t statistic and its significance value are used to test the null hypothesis that the

regression coefficient is zero (or that there is no linear relationship between the

dependent and independent variable).

6. The significance levels of the market potential (0.016) and no of sales people (0.031)

are less than 0.05. So null hypothesis is rejected and alternate hypothesis is accepted.

Hence the variables are linearly related with sales.

7. The significance level of the number of dealers (0.230), index of competitor (0.195),

no of service people (0.735) and no. of existing customer (0.744) is greater than 0.05.

So null hypothesis is not rejected and alternate hypothesis is not accepted. Hence the

variables are not linearly related.

8. Residuals are estimates of the true errors in the model. The residual statistic gives the

difference between the observed value of the dependent variable and the value

predicted by the model.

9. Since the residual value (154.249) is less than regression value (6609.485) the

estimating equation is the best fit.

10. Since the significance value is less than 0.05 the estimating equation is the best fit.

11. Since the model is appropriate for the data, the residuals follow a normal distribution

as indicated by a histogram.

RESULT

There is dependency between the sales (dependent variable) and the market potential

of the territory and number of sales person in the territory (independent variables). The other

independent variables, number of dealers, number of service people, index of the competitor

activity and number of existing customers in the territory has a non-linear relationship with

sales.The estimating equation is :

Y= 0.439X1+0.164X2+0.414X3-0.085X4-0.041X5+0.05X6

CORRELATION

OBJECTIVE:

To find the interrelationship between the dependent and the independent variables.

PROBLEM:





Dependent Variable:










Dealers (X2)

People (X3)

Competition (X4)

Service (X5)

Customers (X6)

1 5 25 1 6 5 2 202 60 150 12 30 4 5 503 20 45 5 15 3 2 254 11 30 2 10 3 2 205 45 75 12 20 2 4 306 6 10 3 8 2 3 167 15 29 5 18 4 5 308 22 43 7 16 3 6 409 29 70 4 15 2 5 3910 3 40 1 6 5 2 511 16 40 4 11 4 2 1712 8 25 2 9 3 3 1013 18 32 7 14 3 4 3114 23 73 10 10 4 3 4315 81 150 15 35 4 7 70

Null Hypothesis

There is no significant relationship between the independent and the dependent

variables at 95% confidence interval.


There is significant relationship between the independent and dependent variables at

95% confidence interval.

PROCEDURE:




10. Choose Analyze > Correlate > Bivariate from the main menu.

11. In the bivariate correlations dialogue box select all the dependent and independent

variables.

12. Select Pearson’s correlation coefficient with test of significance being one tailed.

13. Also include the statistics for mean and standard deviation.

14. The output chart is generated, analyzed and inference obtained.

OUTPUT:


CHI-SQUARE TEST

MeanStd.

Deviation N

Sales in Rs.lakhs in the territory 24.13 21.980 15Market potential in the territory (in Rs.

lakhs)55.80 42.543 15


6.00 4.408 15

No. of sales people in the territory 14.87 8.340 15Index of competitor activity in the

territory3.40 .986 15

No. of service people in the territory 3.67 1.633 15

No. of existing customers in the territory 29.73 16.829 15

OBJECTIVE:

To find out whether there is a significant association between the income of the individuals

and intention to purchase.

PROBLEM:

A customer survey was conducted for a brand of detergent. One of the questions dealt

with the income category and the other asked the respondent to rate his purchase intention.

These two variables are listed in the table below. Both variables are coded as follows:

INCOME TABLE

CODE INCOME IN Rs./month1 <=50002 5001-100003 10001-200004 >20000

PURCHASE TABLE

Code Intention1 None2 Low3 High4 Very high5 Certain

S. No. Income Code Intent Intent Code1 <5000 1 None 12 <5000 1 Low 23 <5000 1 Low 24 <5000 1 None 15 <5000 1 High 36 5001-10000 2 Low 27 5001-10000 2 High 38 5001-10000 2 Very high 49 5001-10000 2 High 310 5001-10000 2 Low 211 10001-20000 3 High 312 10001-20000 3 Very high 413 10001-20000 3 Certain 514 10001-20000 3 High 315 10001-20000 3 Very high 416 >20000 4 High 317 >20000 4 Certain 518 >20000 4 Very high 419 >20000 4 Certain 520 >20000 4 Certain 5

Null Hypothesis

There is no significant association between income and purchase intention.

Alternate Hypothesis

There is significant association between income and purchase intention.

PROCEDURE:

1. The field names and the corresponding data types are entered in the variable view

with the income and purchase intention in nominal measure.

2. The given data is entered in the data view.

3. Choose Analyze > Descriptive statistics > Cross tabs > statistics from the main menu.

4. Select Chi-square test and the required cells are checked.

5. Select income of the respondent (the independent variable) into the rows option and

the intention of the respondents (the dependent variable) into the columns option.

6. The value and significance column are compared from the output and the inference is

made.

OUTPUT:

CHI-SQUARE TEST

Case Processing Summary

CasesValid Missing Total

N Percent N Percent N PercentIncome of the respondent * Intention to purchase

20 100.0% 0 .0% 20 100.0%

Income of the respondent * Intention to purchase Cross tabulationCount

Intention to purchase Total

None Low HighVery high

Certain None

Income of the

respondent

<5000 2 2 1 0 0 55001-10000 0 2 2 1 0 5

10001-20000

0 0 2 2 1 5

>20000 0 0 1 1 3 5Total 2 4 6 4 4 20

Chi-Square Tests

Value Df Asymp. Sig. (2-sided)

Pearson Chi-Square 18.667(a) 12 .097Likelihood Ratio 21.134 12 .048Linear-by-Linear

Association11.790 1 .001

N of Valid Cases 20

a. 20 cells (100.0%) have expected count less than 5. The minimum expected count is .50.

Symmetric Measures

ValueApprox.

Sig.Nominal by Nominal Phi .966 .097 Cramer's V .558 .097N of Valid Cases 20

a. Not assuming the null hypothesis.b. Using the asymptotic standard error assuming the null hypothesis.

Income of the respondent>2000010001-200005001-10000<5000

Co

un

t

3

2

1

0

Bar Chart

certainvery highhighlownone

Intention to purchase

INFERENCE:

1. The Chi-square tests the hypothesis that the row and column variables in a cross

tabulation are independent.

2. The significance value 0.097 is greater than 0.05

3. So null hypothesis is not rejected and alternate hypothesis not accepted. Hence there

is no association between the two variables, Income and the intention.

4. The nominal directional measures indicate both the strength and significance of the

relationship between the row and column variables of the cross tabulation.

5. The value of each statistic can range from 0 to 1 and indicates the proportional

reduction in error in predicting the value of one variable based on the value of other

variable.

6. Also the significance value is greater than 0.05 indicating that there is no relationship

between the two variables.

7. Hence the two attributes are independent.

RESULT:

There is no association between the income and the purchase intention of the

individual.

spss

Documents

sample t test

sample test test value

sample ttest objective

mean difference

sample t test dialogue

test variables

withadd t

error mean