dabm exercise. (1)
TRANSCRIPT
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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:
1. Open SPSS new document
2. Enter the data in SPSS Data Editor
3. Name the Variables, Values and Labels on the data editor
4. Click Data then Weight Cases
5. Transfer data to Frequency Variable box. Click Ok
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
Break and enter 24 55
Break and enter 33 43
Break and enter 36 38
Break and enter 42 53
Break and enter 44 48
Sexual Assault 20 55
Sexual Assault 20 60
Break and enter 23 46
Break and enter 32 30
Mans laughter 41 74Mans laughter 26 72
Sexual Assault 44 62
Sexual Assault 19 60
Mans laughter 23 64
Mans laughter 19 70
Mans laughter 24 61
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Break and enter 42 42
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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:
1. Open SPSS new document
2. Enter the data in SPSS Data Editor
3. Name the Variables, Values and Labels on the data editor
4. Click Data then Weight Cases
5. Transfer data to Frequency Variable box. Click Ok
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:
1. Open SPSS new document
2. Enter the data in SPSS Data Editor
3. Name the Variables, Values and Labels on the data editor
4. Click Data then Weight Cases
5. Transfer data to Frequency Variable box. Click Ok
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
5% Trimmed Mean 2.5977
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
Interquartile Range 3.00
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
5% Trimmed Mean 2.6930
Median 3.0000
Variance 2.068
Std. Deviation 1.43789
Minimum 1.00
Maximum 6.00
Range 5.00
Interquartile Range 2.00
Skewness .467 .093
Kurtosis -.735 .185
Warranty Mean 2.6611 .0537195% Confidence
Interval for Mean
Lower Bound 2.5557
Upper Bound 2.7666
5% Trimmed Mean 2.5849
Median 2.0000
Variance 2.077
Std. Deviation 1.44116
Minimum 1.00
Maximum 6.00Range 5.00
Interquartile Range 3.00
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:
1. Open SPSS new document
2. Enter the data in SPSS Data Editor
3. Name the Variables, Values and Labels on the data editor
4. Click Data then Weight Cases
5. Transfer data to Frequency Variable box. Click Ok
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
Descriptive Statistics
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:
1. Open SPSS new document
2. Enter the data in SPSS Data Editor
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:
Descriptive Statistics
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
1. Open SPSS new document
2. Enter the data in SPSS Data Editor
3. Name the Variables, Values and Labels on the data editor.
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
1. Open SPSS new document
2. Enter the data in SPSS Data Editor
3. Name the Variables, Values and Labels on the data editor.
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:
1. Open SPSS new document
2. Enter the data in SPSS Data Editor
3. Name the Variables, Values and Labels on the data editor.
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 co- efficient = 0.944
P = 0.000, Since P