Decision Science Lab – 2
Department of Management Studies
SUBMITTED TO
Dr. Richa Joshi
SUBMITTED BY
Abhinav Puri
22MMB006
Mba (semester 2)
Table of content:
Experiments:
1.
2.
3.
4.
5.
Missing values
Normality
Reliability
a) Unreliable data
b) Perfectly reliable data
T-test
a) Student t- test
b) Paired t- test
c) Independent t- test
One way Anova.
a) Simple one way anova
b) Welch test
c) Tuckey test
d) Games howell test
Experiment 1
Objective: handling unengaged values
Here we can see there are some missing values in above screenshot.
Editing missing values:
(Treatment of Unsatisfactory Results)
 Returning to the Field – The questionnaires with unsatisfactory
responses may be returned to the field, where the interviewers
re-contact the respondents.
 Assigning Missing Values – If returning the Questionnaire to the
field is not feasible, the editor may assign missing values to
unsatisfactory responses.
 Discarding Unsatisfactory Respondents – In this approach, the
respondents with unsatisfactory responses are simply discarded.
Data Cleaning Treatment of Missing
Responses:
 Substitute a Neutral Value – A neutral value, typically the mean
response to the variable, is substituted for the missing responses.
 Substitute an Imputed Response – The respondent’s pattern of
responses to other questions are used to impute or calculate a
suitable response to the missing questions.
 In case wise deletion, cases, or respondents, with any missing
responses are discarded from the analysis.
 In pairwise deletion, instead of discarding all cases with any
missing values, the researcher uses only the cases or respondents
with complete responses for each calculation.
Here we can see the no. of missing counts in above picture
Filling missing data:
Step: Analyse >missing value analysis> EM> save completed data> give
dataset name> ok
Experiment-2
Objective: Test for normality.
Q: why perform normality?
Ans: For a continuous data, test of the normality is an
important step for deciding the measures of central tendency
and statistical methods for data analysis.
Steps for perform test of normality:
1. Go to Analyse.
2. Go to descriptive analysis
3. Go to explore
4. Go to plots
5. Go to test for normality
6. Click ok
RESULTS:
Hypothesis: To find a relation between sensex and rupee
rate
Null hypothesis: There is no significant relation between
sensex and rupee rate.
Alternate hypothesis: there exists significant relation
between sensex and rupee rate
 Here significant value/ P- value is greater than
(.05) i.e. (.118) therefore null hypothesis accepted.
Experiment-3
Objective: performing reliability test
Q: why do we perform test for reliability?
Ans: Reliability allows us to study the properties of
measurement scales and items that compose the scales. The
reliability analysis procedure calculates a number of commonly
used measures of scale reliability and also provides information
about the relationship between individual items in the scale.
Steps for checking reliability:
1. Go to analyse
2. Go to Scale
3. Go to reliability analysis.
UNRELIABLE DATA:
Output:
Interpretation:
Here Cronbach Alpha is (.361) which is less than (.7)
For a data to be perfect reliable the cronbach alpha should be greater
than .7
Perfect Reliable data:
Output:
Interpretation:
Here cronbach alpha is 1 which is greater than .7 therefore we have
perfect reliable data.
Experiment -4
Objective: performing T-Test
Q: why perform t- test?
Ans: A t- test is a statistical test that is used to compare
the means of two groups.it is often used in hypothesis
testing to determine whether a process or treatment
actually has an effect on the population of interest, or
whether two groups are different from one another.
Steps to perform T-Test:
1. Go to Analyse
2. Go to compare means
3. Go to one-sample t test
1)
Student T-Test
Let us perform T test on student’s height data
1) For mean height = 155cm
Alternate hypothesis: Let us suppose that mean height
of class is not 155 cm
Null hypothesis: mean height of class is 155 cm.
Here sig. value for test vale 155 cm is .00 which is less
than .05 therefore alternate hypothesis accepted.
 Now testing hypothesis for mean height 165 cm
 Null hypothesis: Let us suppose mean height of
class is 165 cm
 Alternate hypothesis: Let us suppose mean
height of class is not 165 cm.
Output: Here sig value is .530 which is greater than 0.05
therefore null hypothesis accepted therefore we can sy
that mean height of class is around 165 cm.
2)
Pared T – Test.
Q: To find relation if there is effect on consumption of
product or sales before and after celebrity
endorsement.
Null hypothesis: there is no significant relation on sales
before celebrity endorsement and after celebrity
endorsement.
Alternate hypothesis: There is a significant relation on
sales on sales before celebrity endorsement and after
celebrity endorsement.
Outcome: here sig. value/ p value is .00 which means
that alternate hypothesis is accepted
Therefore we can say that there is effect on sales before
celebrity endorsement and after celebrity endorsement.
3)
Independent T-test.
Hypothesis: To determine the difference in marks in IQ
test of parents and children with respect to gender.
Null hypothesis: There is no significant relation in IQ
w.r.t gender
Alternate hypothesis: There is a significant relation in IQ
w.r.t. gender
Output:
Here significant value comes out to be .454 which is less
than .005
Therefore null hypothesis accepted.
Therefore we can say that there is no sig difference in IQ
with respect to gender
Experiment-5
Objective: performing one way anova
1)
One way Anova
Hypothesis: to find the significant difference in IQ of
parent w.r.t. income.
Null hypothesis: There is no significant difference in IQ of
parents with respect to their income
Alternate hypothesis: There is a significant relation in IQ
of parents with respect to their income.
Steps for anova test:
1) Go to analyse
2) Go to compare means
3) Go to one way anova
4) Go to options, click homogeneity of varience.
5) Click ok.
Output:
Step 1 : check sig. value for homogeneity of variance. If
sig. value is > .05 only then go for anova table
Step 2 : check sig. value of anova table
 Here in homogeneity of variance sig. value is
.199 which is > to .05. so we can go to anova table
 Here in anova table sig. value is .314 which is
greater than .05 therefore null hypothesis
accepted
Therefore we can say that there is no significant relation
in IQ of parents with respect to income of parents.
2)
Welch test
Hypothesis: To find the sig. relation that if there exists a
relation in IQ of parents with respect to their education
Null hypothesis: there is a no sig. relation in IQ of parent
w.r.t. to their education.
Alternate hypothesis: there is a sig. relation in IQ of
parent w.r.t. to their education.
Steps for welch test:
1) Go to analyse
2) Go to compare means
3) Go to one way anova
4) Go to options, click homogeneity of variance.
5) If sig. value of homogeneity of variance is less than
.05 then go for welch test
6) Click welch test
7) Click ok.
Output:
Here we can see that sig. value in homogeneity of
variance is .023 < .05 therefore we went to welch test
In welch table we can see that we have sig value .061 >
0.05 therefore null hypothesis accepted.
Therefore we can say that there is no significant relation
in marks of IQ w.r.t. education
3)
Tuckey Test ( Post hoc analysis)
Hypothesis/ objective: Testing if there is any difference
in different machines with respect to the number of
defective pieces found.
Null Hypothesis: there is no significant relation in
different machines w.r.t. no. of defective pieces found.
Alternate Hypothesis: There is a significant relation in
different machines w.r.t no. of defective pieces found
Steps for tuckey test:
1) Go to analyse
2) Go to compare means
3) Go to one way anova
4) Go to options, click homogeneity of variance.
5) If sig. value of homogeneity of variance is greater
than .05 then go for Anova test
6) In Anova table is sig. value is <.05 go for tuckey test
7) Click post hoc test
8) Click tuckey test
9) Click ok.
Output:
Here in Tuckey test we have sig. value .062 which is
greater than .05 Therefore null hypothesis accepted.
Therefore we can say that there is no sig. relation in
different machines w.r.t. defected pieces found.
4) Games Howell Test
Objective/hypothesis: to test whether sales (in
thousand) has any relation with the types of vehicle
Null hypothesis: There is no significant relation between
type of vehicle and sales in thousands
Alternate hypothesis: There is a significant relation
between type of vehicle and sales in thousand.
Steps for games howell test:
1) Go to analyse
2) Go to compare means
3) Go to one way anova
4) Go to options, click homogeneity of variance.
5) If sig. value of homogeneity of variance is less than
.05 then go for welch test
6) If in welch test if sig. value is less than .05 go to
games howell test
7) Click on post hoc
8) Click on games howell test
9) Click ok.
Output:
Here in homogeneity of variance the sig. value is .006 <
0.5 therefore we go to welch test
Here in welch test sig. value is .013 < .05 therefore we
will go to games howell test.
Let
u1= car
u2=truck
u3=Jeep
Here, u1 is no equal to u2 so we can say that
Therefore alternative hypothesis accepted.