Spps Solution

8/2/2019 Spps Solution

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Q1. Collect the data from your class mates regarding height and classify the gender

by assigning values to it. Also collect the marks of all the students in the class and

sort them as to know who stands first in the class?

SOLUTION WITH GRAPH-:

gender

Frequency Percent Valid Percent

Cumulative

Percent

Valid male 3390 65.2 65.2 65.2

female 1813 34.8 34.8 100.0

Total 5203 100.0 100.0



Statistics

height

N Valid 5203

Missing 0

Mean 163.22



Q2. Calculate co-efficient of correlation X series & Y series using karl pearson’smethod?

X series Y series

14 13

12 11

14 10

16 15

16 15

17 9

16 14

15 17

SOLUTION-:

Correlations

x series y series

x series Pearson Correlation 1 .192

Sig. (2-tailed) .648

N 8 8

y series Pearson Correlation .192 1


N 8 8



Q3. Calculate co-efficient of correlation of karl pearson’s method from thefollowing:

Marks in Economics Marks in statistics65 67

66 68

67 65

67 68

68 72

69 72

70 69

72 71

SOLUTION-:

Correlations

marks in eco marks in statics

marks in eco Pearson Correlation 1 .603


N 8 8

marks in statics Pearson Correlation .603 1


N 8 8



Q4. Manufacturing & marketing electrics motors would like to build a

regression model consisting of 5 or 6 independent variables. Build the

regression model & recommend whether or not it should be used by the

company.

Dependent variable: market potential interiorSales in lakhs (rs.): No. of dealers of the company in a territory

No. of sales people in the territory

Index of competitor activity in a territory

The data consisting 15 observations from 15 different sales territories

are as follows:

Srno.

sales potential dealers people competitors service Customers

1 5 25 1 6 5 2 20

2 60 150 12 30 4 5 50

3 20 45 5 15 3 2 25

4 11 30 2 10 3 2 20

5 45 75 12 20 2 4 30

6 6 10 3 8 2 3 16

7 15 29 5 18 4 5 308 22 43 7 16 3 6 40

9 29 70 4 15 2 5 39

10 3 40 1 6 5 2 5

11 16 40 4 11 4 2 17

12 8 25 2 9 3 3 10

13 18 32 7 14 3 4 31

14 23 73 10 10 4 3 43

15 81 150 15 35 4 7 70



SOLUTION-:

Coefficientsa

Model

Unstandardized

Coefficients

Standardize

d

Coefficients

B Std. Error Beta t Sig.

1 (Constant) -3.173 5.813 -.546 .600

Potential in territory

(Rs.in lacs)

.227 .075 .439 3.040 .016

No of dealers .819 .631 .164 1.298 .230

No of sales people 1.091 .418 .414 2.609 .031

Index of competitors

activity

-1.893 1.340 -.085 -1.413 .195

No of service people -.549 1.568 -.041 -.350 .735

No of existing

customers

.066 .195 .050 .338 .744

a. Dependent Variable: Sales in territory(Rs.lacs)



Q5. M ltd, a bio-tech ltd specializing in the development of genetically modified

seeds wanted to know the degree to which yield of the wheat per hectare is

influenced by the variety of seeds used for a crop & few other variables. Hence itdeputed an employee in its Bangalore head office to conduct a market research to

study the effect that 6 different conditions ( independent variable) have on the yield

per hector (dependent variable) for a crop of wheat. The research was conducted by

accumulating data from 15 major states in India. The 6 independent variable are:

X1 – rainfall(cms)

X2 - soil type (1-low quality 5- high quality)

X3 – quantity of fertilizers (in quintals per sq.km of land)

X4 - % of land being irrigated by the stste agriculture department

X5 – seed quality (1-low 5-high level of genetically modified quality)X6 - % of automation in cultivation process

The dependent variable is Y yield per heactor (in quintal)

Sr no State

code

yeild Rain

fall

Soil

type

fertilizer Area

irrigated

Seed

type

Automation

1 1 40 23 3 8 78 2 45

2 2 30 12 1 3 47 1 26

3 3 42 28 3 9 82 2 39

4 4 35 15 2 5 51 2 255 5 44 30 3 4 80 3 35

6 6 48 31 4 7 86 3 40

7 7 55 37 5 10 90 4 42

8 8 34 15 2 5 50 2 25

9 9 28 64 1 4 42 1 21

10 10 60 44 5 9 92 5 52

11 11 35 16 2 7 49 2 25

12 12 48 30 1 4 82 2 38

13 13 29 65 1 3 41 1 2014 14 46 30 3 8 80 3 36

15 15 58 42 4 8 89 4 49



SOLUTION-:

Correlations

yeild produced

Fertilizer in

quintal

Fertilizers used

in the state

yeild produced Pearson Correlation 1 .848**

.701**

Sig. (2-tailed) .000 .004

N 15 15 15

Fertilizer in quintal Pearson Correlation .848

**

1 .847

**

Sig. (2-tailed) .000 .000

N 15 15 15

Fertilizers used in the

state

Pearson Correlation .701**

.847**

1

Sig. (2-tailed) .004 .000

N 15 15 15

irrigation carried in the

area

Pearson Correlation .930**

.813**

.707**

Sig. (2-tailed) .000 .000 .003

N 15 15 15

seed quality Pearson Correlation .937**

.919**

.713**

Sig. (2-tailed) .000 .000 .003

N 15 15 15

Automation Pearson Correlation .916**

.818**

.736**

Sig. (2-tailed) .000 .000 .002

N 15 15 15

Rain fall in cm Pearson Correlation .067 .024 -.067

Sig. (2-tailed) .813 .932 .811

N 15 15 15

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



Q6. Data is available for 15 customers of the brand each in Mumbai &

delhi. The customer were asked to rate the brand on a 7 point scale. The

response of all 30 customers is in column labeled rating in the table

The column city indicates – city from where the rating came

1 – Mumbai

2 - Delhi

Present the output from the independent sample ‘t’ test performed on the

data.

Sr no. Ratings City

1 2 12 3 1

3 3 1

4 4 1

5 5 1

6 4 1

7 4 1

8 5 1

9 3 110 4 1

11 5 1

12 4 1

13 3 1

14 3 1

15 4 1

16 3 2

17 4 218 5 2

19 6 2

20 5 2

21 5 2

22 5 2



23 4 2

24 3 2

25 3 2

26 5 2

27 6 228 6 2

29 6 2

30 5 2

SOLUTION-:

Independent Samples Test

t-test for Equality of Means

df Sig. (2-tailed) Mean Difference

Ratings by customers Equal variances assumed 28 .010 -1.000

Equal variances not assumed 26.759 .011 -1.000



Q7. Assume that the sample of 18 respondents whom were asked to rate on a 10

point interval scale. Their attitude towards a brand of garments before and after an

ad campaign was released for the brand.

Ratings:

1 – brand highly disliked

10 – brand is highly disliked

With other ratings having appropriate meaning, the first column contains ratings

given by respondent before this of ad campaign & column 2 represents the rating

after they saw the ad campaign.

Sr no Before ad campaign After ad campaign

1 3 5

2 4 6

3 2 6

4 5 75 3 8

6 4 4

7 5 6

8 3 7

9 4 5

10 2 4

11 2 6

12 4 7

13 1 414 3 6

15 6 8

16 3 4

17 2 5

18 3 6



SOLUTION-:

Paired Samples Test

Paired Differences

t df Sig. (2-tailed)Mean

Std.

Deviation

Std. Error

Mean

95% Confidence Interval of

the Difference

Lower Upper

Pair 1 preference

before campaign

- preference

after campaign

-2.500 1.295 .305 -3.144 -1.856 -8.192 17 .0



Q8. A sample of 18 observations assumes that the magazines in which

the ad copy appears may have an impact on ratings. The hypothesis to be

tested using the data the variables bring rating as the dependent variable,

add copy as a factor and magazine as the block, the magazine appears

could be influenced the ratings given to the ad copy by the respondent.

Sr no. Ad copy ratings

1 1 6.00

2 1 7.00

3 1 5.00

4 1 8.00

5 1 8.00

6 1 8.00

7 2 4.00

8 2 4.00

9 2 5.00

10 2 7.00

11 2 7.00

12 2 6.00

13 3 5.00

14 3 5.00

15 3 4.00

16 3 7.00

17 3 8.00

18 3 7.00



Versions of the ad copy are of 3 different versions which has been

created by an advertising agency for the campaign the ad agency wants

to test which of these 3 version of this advertising copy is preferred bythis target population before the campaign at random. 18 respondents are

assigned to the 3 versions of which are shown to 6 of the respondent.

The respondents are asked to rate the like to show to their on preference

1 – do not like at all

10 - like a lot

SOLUTION-:

ratings by customers

Sum of Squares df Mean Square F Sig.

Between Groups 7.000 2 3.500 1.780 .203

Within Groups 29.500 15 1.967

Total 36.500 17

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