dabm exercise. (1)

8/10/2019 DABM Exercise. (1)

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DESCRIPTIVE STATISTICS USING SPSS

The following are the heights of students belonging to 10thstd of a government school.

Heights (In cms)

140 145 156 142 160 168 150 154 142 156

162 175 172 145 150 156 140 142 160 167

142 155 178 168 172 143 151 146 140 167

(a)Mean.

(b)Median

(c)Mode.

(d)Standard deviation.

(e)Variance.

(f) Range.(g)Maximum & Minimum.

(h)Skewness

(i) Kurtosis

(j) Steam and leaf.

(k)Histogram.


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Ex.no : 1 DESCRIPTIVE STATISTICS USING SPSS

Aim:

To conduct the descriptive statistics for the following data.

Algorithm:

1. Click start ->Programmes->SPSS. SPSS data editor appears.

2. Click Data View Enter thedata.

3. Click Variable View and fill the requirements.

4. Click on Analyze -> Descriptive statistics -> Frequencies. Frequency box

appears. Select height to the variable box . Click Statistics Button and select

Mean, Median, Mode. Clicks continue.

5. Click Analysis-> Descriptives. Descriptive box appears. Select Height to

dependent variable box. Click Option button. Select Standard deviation, Variance,

Range, Minimum, Maximum, Kurtosis, Skewness. Click Continue.

6. Select Analyze-> Descriptives -> Explore. Explore box appears. Send Height to

Dependent list box.

7. Click plots button , select steam and leaf and histogram. Click continue. Click ok.

8. The output appears.


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Output

Statistics Height (In cms)

N Valid 30

Missing 0

Mean 1.5483E2

Median 1.5450E2

Mode 142.00

Frequency Distribution:

Height (In cms)Frequency Percent Valid Percent Cumulative Percent

140 3 10.0 10.0 10.0

142 4 13.3 13.3 23.3

143 1 3.3 3.3 26.7

145 2 6.7 6.7 33.3

146 1 3.3 3.3 36.7

150 2 6.7 6.7 43.3

151 1 3.3 3.3 46.7

154 1 3.3 3.3 50.0

155 1 3.3 3.3 53.3

156 3 10.0 10.0 63.3

160 2 6.7 6.7 70.0

162 1 3.3 3.3 73.3

167 2 6.7 6.7 80.0

168 1 3.3 3.3 83.3

169 1 3.3 3.3 86.7

172 2 6.7 6.7 93.3

175 1 3.3 3.3 96.7

178 1 3.3 3.3 100.0

Total 30 100.0 100.0


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Descriptive Statistics

N Range Minimum Maximum Std. Deviation Variance Skewness Kurtosis

Statistic Statistic Statistic Statistic Statistic Statistic Statistic Std. Error Statistic Std. Error

Height (In cms) 30 38.00 140.00 178.00 11.89634 141.523 .381 .427 -1.125 .833

Valid N (listwise) 30


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Height (In cms) Stem-and-Leaf Plot

Frequency Stem & Leaf

8.00 14 . 00022223

3.00 14 . 556

4.00 15 . 0014

4.00 15 . 5666

3.00 16 . 002

4.00 16 . 7789

2.00 17 . 222.00 17 . 58

Stem width: 10.00

Each leaf: 1 case(s)

Box plot


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Inference:

Mean = 154.8

Median=154.5

Mode =142

Standard deviation = 11.89

Variance= 141.5

Maximum= 178

Minimum=140

Skewness = 0.381

Kurtosis= -1.125

Result :

Hence the Frequency , Descriptive and explore statistics was found out using SPSS.


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PARAMETRIC TEST

CHI SQUARE USING SPSS

Use chi square to test the relationship between sources of information of a product A and

Experience of the respondents in his work life using product. Give your inference.

Source

Experience

Friends/

relativesAgent Advt Exhibition Total

Up to 5 years 8 4 23 9 44

6 to 10 years 18 4 12 12 46

11 to 15 years 3 3 24 12 42

16 to 20 years 2 3 6 4 15

21 to 25 years 8 3 6 14 31

Above 25 years 1 1 6 14 22

Total 40 18 77 65 200


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PARAMETRIC TEST

Ex.no : 2 CHI SQUARE USING SPSS

AIM:

To conduct the chi-square test to test the following hypothesis using SPSS

H0: There is no significant relationship between sources of information of a product and

experience of the respondents

H1: There is a significant relationship between sources of information of a product and

experience of the respondents

Algorithm:

1. Open SPSS new document

2. Enter the data in SPSS Data Editor

3. Click Data then Weight Cases

4. Transfer data to Frequency Variable box. Click Ok

5. Click on analyze then descriptive statistics then click on cross tabs

6. In the rows select the independent variable (Experience) and in the column select the

dependent variable (Sources of Information).

7. Then click on statistics in the same window and select chi-square

8. Then click on continue

9. Finally click OK


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OUTPUT

Chi-square Test

Chi-Square Tests

Value df Asymp. Sig. (2-sided)

Pearson Chi-Square 40.604a 15 .000

Likelihood Ratio 39.723 15 .000

Linear-by-Linear Association 7.663 1 .006

N of Valid Cases 200

a. 9 cells (37.5%) have expected count less than 5. The minimum expected count is 1.35.

Source * Experience Crosstabulation

Count

Experience

Total1 2 3 4

Source 1 8 4 23 9 44

2 18 4 12 12 46

3 3 3 24 12 42

4 2 3 6 4 15

5 8 3 6 14 31

6 1 1 6 14 22

Total 40 18 77 65 200


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Inference:

Chi-square value = 40.604

Since P


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PARAMETRIC TEST

ONE WAY ANOVA USING SPSS

To determine whether different income groups have different purchasing habits concerning a certain

brand, a marketing researcher asked four income groups: Do you always purchase the brand, never

purchase it or sometimes purchase it? The results of the survey were:

Income

Group/

Frequency

Rs.2,000 Rs.2,000

to 2,999

Rs.3,000

to 3,999

Rs.4,000+ Total

Always 25 40 46 45 156

Never 68 40 75 38 251

Sometimes 37 30 19 37 123

TOTAL 130 120 140 140 530

Is there any association between income level and purchasing habits?


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Ex.no: 3 PARAMETRIC TEST

ONE WAY ANOVA USING SPSS

AIM:

To conduct the one way ANOVA, using SPSS to test the given hypothesis.

H0: There is no association between income level of the respondents and their purchasing habits.

H1: There is association between income level of the respondents and their purchasing habits.

Algorithm:



3. Name the Variables, Values and Labels on the data editor



6. Click Analyze, then Compare Means, then One Way Anova. One way Anova

Dialogue Box appears.

7. Transfer the dependent variable (Frequency) into the Dependent List Box and the

Independent variable (Income Group) in to the Factor Box.

8. Open Contrast and select Polynomial Option.

9. Open Options Button and select descriptive, Homogeneity of Variance Test and

Welch and Continue

10.Finally click OK


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OUTPUT

One way ANOVA

Descriptives

Frequency N Mean

Std.

Deviation

Std.

Error

95% Confidence

Interval for Mean

Minimum Maximum

Lower

Bound

Upper

Bound

Rs 2000 130 2.09 .687 .060 1.97 2.21 1 3

Rs 2000 - 2999 110 1.91 .796 .076 1.76 2.06 1 3

Rs 3000 - 3999 140 1.81 .656 .055 1.70 1.92 1 3

Rs 4000 and

above120 1.93 .827 .076 1.78 2.08 1 3

Total 500 1.93 .745 .033 1.87 2.00 1 3

Test of Homogeneity of Variances

Frequency

Levene Statistic df1 df2 Sig.

5.834 3 496 .001

ANOVA

Frequency

Sum of

Squaresdf Mean Square F Sig.

Between Groups (Combined) 5.579 3 1.860 3.401 .018Linear Term Unweighted 2.088 1 2.088 3.819 .051

Weighted 2.372 1 2.372 4.337 .038

Deviation 3.208 2 1.604 2.933 .054

Within Groups 271.243 496 .547

Total 276.822 499


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Robust Tests of Equality of Means

Frequency

Statistica df1 df2 Sig.

Welch 4.061 3 266.932 .008

a. Asymptotically F distributed.

Homogeneous Subsets

Frequency

Income N

Subset for alpha = 0.05

1 2

TukeyHSDa Rs 3000 - 3999 140 1.81

Rs 2000 - 2999 110 1.91 1.91

Rs 4000 and above 120 1.93 1.93

Rs 2000 130 2.09

Sig. .536 .208

Means for groups in homogeneous subsets are displayed.

a. Uses Harmonic Mean Sample Size = 123.995.


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Inference:

F= 3.401, since p


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PARAMETRIC TEST

TWO WAY ANOVA USING SPSS

Consider the following two factor experiment based on a social psychologist interested in

the effect of a type of crime with 3 levels.

A1= Brake and enter.

A2= Sexual Assault.

A3= Mans laughter.

Age:

B1 = 20 years.

B2 = 21 to 30 years.

B3 = 31 to 40 years.

B4 = Above 40 years.

Crime Age Sentence

Break and enter 19 49

Break and enter 20 39Break and enter 23 50






Sexual Assault 20 55




Mans laughter 41 74Mans laughter 26 72



Mans laughter 23 64

Mans laughter 19 70

Mans laughter 24 61


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Ex.no: 4 PARAMETRIC TEST

TWO WAY ANOVA USING SPSS

AIM:

To conduct the Two Way ANOVA Test, using SPSS to test the given hypothesis.

H0: There is no association between age and social psychologist interested in the effect of a type

of crime with 3 levels.

H1: There is an association between age and social psychologist interested in the effect of a type

of crime with 3 levels.

Algorithm:






6. Click on Analyze, then General Linear Model, then Univariate

7. Transfer the Dependent Variable (Sentence ) to the Dependent Variable Box

8. Transfer both Independent Variables( Age and crime) in to the Fixed Factor box.

9. Click on the Plots button, the Univariate Profile Plots Dialogue box opens.

10.Transfer the Independent Variable (Age) from the Factors Box into the Horizontal

Axis Box and transfer the Variable (Crime) in to the Separate Lines box. Click the

Add Button and Click Continue.


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OUTPUT

TWO WAY ANOVA


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Inference:

From Levenes Test of Equality of Error Variances, we can see that we do not have homogeneity

of variances of the dependent variable across group since the sig value is less 0.05.

Since F = 6.609 and p0.05, there is no association between Gender of the respondents and

frequency of visit to the bank.

Since F = 10.632 and p


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NON - PARAMETRIC TEST

KOLMOGOROV-SMIRNOV TEST USING SPSS

Use Kolmogorov-Smirnov Test to study the relationship between rank and factors influencing the

purchase of machinery among the respondents.

Rank /

Factors

Influencing

Purchase

1 2 3 4 5 6 Total

Score

Weighted

average

Price 252 190 72 126 44 38 722 3.61

Quality 228 100 160 96 70 35 689 3.45

Brand name 168 160 170 90 82 25 701 3.51

Warranty 192 195 112 132 64 25 720 3.60


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Ex.no: 5 NON - PARAMETRIC TEST

KOLMOGOROV-SMIRNOV TEST USING SPSS

Aim:

To conduct Kolmogorov-Smirnov Test Using SPSS to test the following Hypothesis.

Ho: There is no normality relationship between brand and factors influencing the purchase of

machinery among the respondents.

H1: There is a normality relationship between brand and factors influencing the purchase of

machinery among the respondents.

Algorithm:

Test for Normality:






6. Click Analyze , then Descriptive Statistics, then Explore

7. Transfer the Rank that needs to be tested for normality into the "Dependent List" box

8. Transfer the independent variable(Factors influencing Purchase) to the "Factor List" box

9. Click the statistics button, then descriptive and continue

10.Click the plot button, then normality plot with test and continue

11.Click ok


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Kolmogorov-Smirnov Test

DescriptivesFactors Statistic Std. Error

Rank Price Mean 2.4931 .05641

95% Confidence

Interval for Mean

Lower Bound 2.3823

Upper Bound 2.6038

5% Trimmed Mean 2.3812

Median 2.0000

Variance 2.297

Std. Deviation 1.51574Minimum 1.00

Maximum 6.00

Range 5.00

Interquartile Range 3.00

Skewness .763 .091

Kurtosis -.521 .182

Quality Mean 2.6880 .05875

95% ConfidenceInterval for Mean

Lower Bound 2.5726

Upper Bound 2.8033


Median 3.0000

Variance 2.378

Std. Deviation 1.54200

Case Processing Summary

Factors Cases

Valid Missing Total

N Percent N Percent N Percent

Rank Price 722 100.0% 0 .0% 722 100.0%

Quality 689 100.0% 0 .0% 689 100.0%

Brand Name 695 100.0% 0 .0% 695 100.0%

Warranty 720 100.0% 0 .0% 720 100.0%


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Minimum 1.00

Maximum 6.00

Range 5.00


Skewness .489 .093

Kurtosis -.842 .186

Brand Name Mean 2.7597 .05454

95% Confidence

Interval for Mean

Lower Bound 2.6526

Upper Bound 2.8668


Median 3.0000

Variance 2.068


Minimum 1.00

Maximum 6.00

Range 5.00


Skewness .467 .093

Kurtosis -.735 .185

Warranty Mean 2.6611 .0537195% Confidence

Interval for Mean

Lower Bound 2.5557

Upper Bound 2.7666


Median 2.0000

Variance 2.077


Minimum 1.00

Maximum 6.00Range 5.00


Skewness .529 .091

Kurtosis -.754 .182


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Tests of Normality

Factors Kolmogorov-Smirnova Shapiro-Wilk

Statistic df Sig. Statistic df Sig.

Rank Price .240 722 .000 .847 722 .000

Quality .194 689 .000 .878 689 .000

Brand Name .173 695 .000 .902 695 .000

Warranty .214 720 .000 .889 720 .000

a. Lilliefors Significance Correction

Normal Q-Q Plots


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Normal Q-Q Plots


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Detrended Normal Q-Q Plots


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Inference:

If the Sig.value of the Shapiro-Wilk Test is less than 0.05 then the data significantly deviate

from a normal distribution.

Result:

Thus the normality was checked with Kolmogorov-Smirnov Test Using SPSS to test the

following Hypothesis.


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NON - PARAMETRIC TEST

KRUSKAL WALLIS TEST

A study compared the effects of four 1 month point of purchase promotions on sales. The unit

sales for five stores using all four promotions in different months follows.

Sales

promotion

Types

Store

1 2 3 4 5

Free sample 78 87 81 89 85

One pack gift 94 91 87 90 88

Cents off 73 78 69 83 76

Refund by

Mail

79 83 78 69 81

Use the kruskal wallis test to determine whether all five stores data are come

from different populations. ( = 0.01)


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Ex.no: 6 NON - PARAMETRIC TEST

KRUSKAL WALLIS TEST USING SPSS

Aim:

To conduct Kruskal Wallis test using SPSS to test the following hypothesis.

H0: All 5 stores data do not come from different population.

H1: All 5 stores data come from different population.

Algorithm:






6. Click Analyze, then Non Parametric Test then K Independent Samples

7. Transfer the Dependent Variable (Sales) into the Test Variable List Box and Stores into

the Grouping Variable Box. Click Kruskal Wallis Test.

8. Click Define Range and type 1 into the minimum box and 5 into the maximum

box.

9. Click Options then select Descriptive.

10.Click Continue, the finally OK


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Kruskal-Wallis Test


N Mean Std. Deviation Minimum Maximum

Sales promotion types1639 2.45 1.112 1 4

Stores 1639 3.0024 1.41637 1.00 5.00

Ranks

Stores N Mean Rank

Sales promotion

types

1 324 828.99

2 339 824.86

3 315 822.29

4 331 799.03

5 330 825.03

Total 1639

Test Statisticsa,b

Sales promotion types

Chi-Square .904

df 4

Asymp. Sig. .924

a. Kruskal Wallis Testb. Grouping Variable: Stores


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Inference:

H (2) = 0.904, since p > 0.05, All 5 stores data came from different population with the mean

Rank of Store1= 828.99, store2= 824.86, store3 = 822.29, store4 = 799.03 and store 5 = 825.03.

Result:

Thus Kruskal Wallis test was conducted using SPSS


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MANNWHITNEY U TEST

To increase sales during heavy shopping days, a chain of stores selling cheese in shopping malls

gives away samples at the stores entrance. The chains management defines the heavy shopping

days and randomly selects the days for sampling. From a sample of days that were considered

heavy shopping days the following data give one stores sales on days when cheese sampling

was done and on days when it was not done.

(Sales in hundreds)

Promotion days: 18 21 23 15 19 26 17 18 22 20

Regular days: 22 17 15 23 25 20 26 24 16 17

Promotion days: 18 21 27

Regular days: 23 21


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Ex.no : 7 MANNWHITNEY U TEST

Aim:

To compare the difference between two independent groups.

Ho : There is no significance difference between the mean sales of promotion day and regular

day.

H1: There is significance difference between the mean sales of promotion day and regular day.

Algorithm:



3. Name the Variables, Values and Labels on the data editor.

4. Click Analyze -> Non- Parametric test -> Two independent samples. Two independent

samples test box appears.

5. Click the variable Sales into Test variable list box.

6. Click the variable Days into Grouping variable box.

7. Click the Define groups button and type 1 in the group 1 box and 2 in group 2 box.

Click Continue.

8. Click Option button and select descriptive and Quartiles. Click continue.

9. Select MannWhitney U test. Click ok.

10.The result appears.


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OUTPUT:


N Mean Std. Deviation Minimum Maximum

Percentiles

25th 50th (Median) 75th

Sales 25 20.5600 3.52467 15.00 27.00 17.5000 21.0000 23.0000

Base 25 1.4800 .50990 1.00 2.00 1.0000 1.0000 2.0000

Ranks

Base N Mean Rank Sum of Ranks

Sales Promtion base 13 12.62 164.00

Regular base 12 13.42 161.00

Total 25

Test Statistics

Sales

Mann-Whitney U 73.000

Wilcoxon W 164.000

Z -.273

Asymp. Sig. (2-tailed) .785

Exact Sig. [2*(1-tailed Sig.)] .810a

a. Not corrected for ties.

b. Grouping Variable: Base


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Inference:

From table it was inferred that u= 73. R1= 164, R2=161, P= 0.805/2 = 0.4025.

Since P>0.05 we reject H1 and accept H0.

Therefore there is significance between the mean sales of promotion day and

regular day.

Result

Hence Mann Whitney u test was conducted using SPSS.


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SPEARMAN RANK CORRLATION CO-EFFICIENT

The following data are random sample of consumers Income and expenditures on certain

luxury items. Compute the spearman rank correlation coefficient and test for the

existences of a population correlation.

Income ($1000s/year): 23 17 34 56 49 31 28 80 65

Luxury Item spending: 10 50 120 225 90 60 55 340 170

($/month)

Income ($1000s/year): 49 26

Luxury Item spending: 25 80

($/month)


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Ex.no: 8 SPEARMAN RANK CORRLATION CO-EFFICIENT

Aim

To conduct the test of correlation by computing spearman rank correlation.

Algorithm




4. Click Analyze-> correlate->bivarite. Bivariate correlation appears. Click two

variables, Income and Luxury item spending into variable box.

5. Select Spearman in correlation coefficient.

6. Click two tailed in test of significance.

7. Click option button, select Exclude cases pair wise in missing value -> Click

continue -> Ok.

8. The result appears.


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OUTPUT

Correlations

Income Luxury item spending

Spearman's rho Income Correlation Coefficient 1.000 .770**

Sig. (2-tailed) . .006

N 11 11

Luxury item spending Correlation Coefficient .770** 1.000

Sig. (2-tailed) .006 .

N 11 11

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


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Inference:

From the output it is inferred that,

Spearman Rank correlation co efficient ( rs)for Luxury item spending = 0.770.

P = 0.006, Since P 0.05, we reject null hypothesis and accept alternate

hypothesis.

Result:

Hence, we conclude that there is significance between the income and the luxury items

spending of the sample.


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SIGN TEST

Use the sign test to see if there is a difference between the number of days until

collections of amount receivables before and after collection policy. Use 0.05

significance level.

Before:30 28 34 35 40 42 33 38 34 45 28

After:34 29 33 32 47 43 40 42 37 44 27

Before:27 25 41 36

After: 33 30 38 36


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Ex.no : 9 SIGN TEST

Aim

To conduct the sign test between two paired data using SPSS.

H0 : There is no difference between the number of days until collection of an amount

receivable before and after a collection policy.

H1: There is difference between the number of days until collection of an amount

receivable before and after a collection policy.

Algorithm




4. Click Analyze - > Non parametric test -> Two related samples. Two related

sample test box appears.

5. Click before into variable 1 and after to variable 2.

6. Select Sign in the test type.

7. Click Options and select Descriptive and quartiles in statistics. Click

continue. Click ok

8. The result appears.


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Inference

Number of positive difference = 9

Number of negative difference = 5

Ties = 1

P = 0.424. Since P 0.05 , we accept null hypothesis.

Result

Hence the sign test was conducted using SPSS.


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CORRELATION & REGRESSION

The heights (in cms) and weight (Kilograms) of 10 basketball players on a team

as follows:

Height (x) : 186 189 190 192 193 193 198 201 203 205

Weight (y): 85 85 86 90 87 91 93 103 100 101

Conduct correlation and regression analysis and the lines of regression and coefficient of

correlation using SPSS.


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Ex.no : 10 CORRELATION & REGRESSION

Aim

To conduct the correlation and regression analysis for the two given variables

using SPSS.

Algorithm:




4. To conduct correlation analysis

(i) Click Analyze - > Correlate -> Bivariate. Bivariate correlation box

appears.

(ii) Send the two variables height and weight into variable box.

(iii) Click Pearson in Correlation coefficient.

(iv) Click Two tailed intest of significance.

(v) Tick the box Flag significance correlations.

(vi) Click Ok

(vii) The out put appears.

5. To conduct regression analysis

(i) Select regression -> Linear.

(ii) Send the variable Height into dependent box and the variable Weight

into independent box.

(iii) Click statistics button and select estimates, model fit and case wise

diagnostic. Clicks continue. Click ok.

(iv) The output appears.


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OUTPUT:

Correlations

Height Weight

Height Pearson Correlation 1 .944**

Sig. (2-tailed) .000

N 10 10

Weight Pearson Correlation .944** 1

Sig. (2-tailed) .000

N 10 10

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

Model Summary

Model R R Square Adjusted R Square Std. Error of the Estimate

1 .944 .892 .878 2.23342

a. Predictors: (Constant), Weight

b. Dependent Variable: Height

ANOVA

Model Sum of Squares df Mean Square F Sig.

1 Regression328.095 1 328.095 65.775 .000

Residual39.905 8 4.988

Total368.000 9

a. Predictors: (Constant), Weight


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Model Sum of Squares df Mean Square F Sig.

1 Regression328.095 1 328.095 65.775 .000

Residual39.905 8 4.988

Total368.000 9

b. Dependent Variable: Height

Coefficients

Model

Unstandardized Coefficients Standardized Coefficients

t Sig.B Std. Error Beta

1 (Constant) 114.634 9.934 11.539 .000

Weight .873 .108 .944 8.110 .000

a. Dependent Variable: Height

Residuals Statistics

Minimum Maximum Mean Std. Deviation N

Predicted Value 188.8046 204.5113 1.9500E2 6.03779 10

Residual -3.51126 2.45022 .00000 2.10569 10

Std. Predicted Value -1.026 1.575 .000 1.000 10

Std. Residual -1.572 1.097 .000 .943 10

a. Dependent Variable: Height


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Inference:

(i) Correlation:

Correlation coefficient = 0.944

P = 0.000, Since P

dabm exercise. (1)

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