RMBA Class Notes Peer Discussion Session SH
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Transcript of RMBA Class Notes Peer Discussion Session SH
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Research
Methodology
and Business
Analytics
Class Exercise-1 & 2
24th& 25thSeptember 2013
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Contents1) Find out the mean compensation, JS and TI score for the entire sample. ........................................... 2
2) Are these 3 variables (Compensation, JS & TI), normally distributed. ................................................. 5
3) Construct a frequency table for gender and department affiliation separately. ................................. 7
4) Find out the mean compensation for male executives. ....................................................................... 9
5) Find out mean compensation of female executives. .......................................................................... 13
6) Compute the number of cases fulfilling the following conditions - .................................................... 15
a) Female executives in marketing department whose job satisfaction score is above 3. .................... 15
b) Is the turnover intention for male executives different from female counterparts in HR
department? ............................................................................................................................................... 17
7) Construct a cross table and display gender along the column & department affiliation across the
rows............................................................................................................................................................. 19
8) Find out the median compensation received. Divide the sample into 2 categories as: below median
which can be given code of 1, and median & above (= 2). Use this recoded value of compensation and
find out the average turnover intention of these executives whose compensation is below median. ..... 21
9) Repeat previous question, this time for male executives only. .......................................................... 27
10) Compute the median JS and TI and classify them on basis of below median; median and above.
Construct a cross tabulation to find out how many execs are falling under median and above on both JS
and TI? Is the mean compensation different for executives found under these four different cells? ....... 29
11) Is there a significant relation between compensation and JS? If it so interpret your result. ......... 31
12) Is there an association between Job Satisfaction and intention to leave the organization? (Aftercontrolling compensation) .......................................................................................................................... 33
Lecture on 25/9/2013 ............................................................................................................................. 35
13) Is there an association between median classified JS and median classified Compensation? If
there is, what is the magnitude of the association between the assignments? ........................................ 35
.................................................................................................................................................................... 36
14) Classify the employees on the basis of mean experience. Level them in the same way that we had
done earlier. And find out: (In all the above cases compute the magnitude of such association.) ........... 39
15) Do you find correlation between JS and turnover intention? ........................................................ 44a. Is this significant? ................................................................................................................................ 44
b. Is the correlation between JS & TI influenced by- .............................................................................. 44
16) Find out the impact of ..................................................................................................................... 47
17) Perform a multiple regression analysis to find out the impact of employees experience,
compensation, JS on TI. .............................................................................................................................. 52
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1)Find out the mean compensation, JS and TI score for the entire
sample.
Steps:Analyze->Descriptive Statistics -> Descriptives -> Select variables -> SelectMean from the checkbox -> click continue.
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2)Are these 3 variables (Compensation, JS & TI), normally
distributed.
(Hint: both Skewness and Kurtosis should be within the range -2 to +2)
Steps:Analyze->Descriptive Statistics -> Descriptives -> Select variables -> SelectSkewness and Kurtosis from the checkbox -> click continue.
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Based on the values given this sample corresponds to normal distribution.
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3)Construct a frequency table for gender and department affiliation
separately.
Steps:Analyze->Descriptive Statistics -> Frequencies -> Select variables -> clickcontinue.
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4)Find out the mean compensation for male executives.
This is the next level of mean calculation. First we will sort the data male executivesusing Select Cases functionality in Data tab and then apply the normal operation to findthe mean.
Steps: Data-> Select Cases -> Select If condition -> Gender = 1.Analyze->Descriptive Statistics -> Descriptives -> Select variables -> Select Mean fromthe checkbox -> click continue.
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5)Find out mean compensation of female executives.
This question is similar to question 4.
Steps: Data-> Select Cases -> Select If condition -> Gender = 2.
Analyze->Descriptive Statistics -> Descriptives -> Select variables -> Select Mean fromthe checkbox -> click continue.
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6)Compute the number of cases fulfilling the following conditions -
a)Female executives in marketing department whose job satisfaction
score is above 3.
We will perform the same function of selecting the sub category first and then usefrequency function (as in question 3). But as shown the number of female executivessatisfying the above criteria is zero therefore we get null output in the frequencyfunction.
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b)Is the turnover intention for male executives different from female
counterparts in HR department?
This question deals with comparing the means of female employees TI v/s male
employees TI. So select the sub category first and then apply the mean for TI. Forfemale it is 3.25 & male it is 3.0.
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Similarly do for male Dept=1 & Gender=1.
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7)Construct a cross table and display gender along the column &
department affiliation across the rows.
Steps:Analyze->Descriptive Statistics -> Crosstabs -> Select variables (each for row &column) -> press OK.
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8)Find out the median compensation received. Divide the sample
into 2 categories as: below median which can be given code of 1,
and median & above (= 2). Use this recoded value of compensation
and find out the average turnover intention of these executives
whose compensation is below median.
Steps:Analyze->Descriptive Statistics -> Frequencies -> select variables ->Statistics ->Median.Transform-> Recode into different variables -> Select Compensation-> FillCompensation_R in Name and its description in Label-> Select the condition{Range value through highest = 4800, value=2-> Click Add} {All other values,Value=1, -> Click Add }Change value variables in the variable view tab.Now the values have been recoded, use the if function to select the sub-section
(Compensation_R=1) and then apply mean for compensation.
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9)Repeat previous question, this time for male executives only.
Use the sheet of question 8 & use the if functionto select the sub-section(Compensation_R=1& Gender = 1) and then apply mean for compensation.
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10) Compute the median JS and TI and classify them on basis of
below median; median and above. Construct a cross tabulation to
find out how many execs are falling under median and above on
both JS and TI? Is the mean compensation different for executives
found under these four different cells?
It is similar to question 8.
Steps:Analyze->Descriptive Statistics -> Frequencies -> select variables ->Statistics ->Median. (Calculate median for both JS = 4 & TI = 3).
Transform-> Recode into different variables -> Select Compensation-> Fill JS_R inName and its description in Label-> Select the condition {Range value throughhighest = 4, value=2-> Click Add} {All other values, Value=1, -> Click Add }Change value variables in the variable view tab.
Similarly do for TI. (Using median value as 3)
Analyze->Descriptive Statistics -> Crosstabs -> Select variables (each for row TI_R &column JS_R) -> press OK.
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Now using if function calculate mean for the following 4 conditions:
JS_R=1, TS_R=1 (Mean=0)
JS_R=2, TS_R=1 (Mean=6120)
JS_R=1, TS_R=2 (Mean=4240)
JS_R=2, TS_R=2 (Mean=3000)
We will find that the mean values are coming to be different in each case.
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11) Is there a significant relation between compensation and JS?
If it so interpret your result.
Steps:Analyze->Correlate-> Bivariate -> select variables -> Pearson-> Press OK.
0 to 0.40 -> Low Correlation0.40 to 0.60 -> Moderate Correlation0.60 to 0.80 -> High Correlation>80 -> Very high correlation1 -> Perfect correlation or singularity
We have got correlation =0.654, so it signifies high correlation; p value (0.029) is below0.05 therefore we can infer that relation exists. (H0= No relation exists between thevariable, H1= Relation exists between the variables.)
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12) Is there an association between Job Satisfaction and intention
to leave the organization? (After controlling compensation)
Steps:Analyze->Correlate-> Partial -> select variables & controlling parameter ->Press OK.
Value of -0.89 depicts very high negative correlation.
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Lecture on 25/9/2013
13) Is there an association between median classified JS and
median classified Compensation? If there is, what is the magnitudeof the association between the assignments?
Here we have to take median classified JS & median classified compensation valuesand find out the Chi-square analysis and magnitude of the same.Chi-square analysis depicts whether is association or not; and Phigives the magnitudeof the association.
Steps:Analyze->Descriptive Statistics-> Crosstabs -> select variables.Statistics button-> select Chi-square and Phi and Cramers V-> Press continue.
We get Chi-square value as 0.782 for 1 degree of freedom {i.e. (r-1)*(c-1)}. Thissignifies high association. Here magnitude of association (Phi-value) is 0.267 which islow.
To summarize:Compensation affects Job Satisfaction with a probability of 0.782 but the magnitude ofthe same is 0.267 i.e. JS= 0.267Compensation.
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14) Classify the employees on the basis of mean experience. Level
them in the same way that we had done earlier. And find out: (In all
the above cases compute the magnitude of such association.)
a. Significance of association between mean classified experience and
mean classified compensation.
Here we have to find chi-square and Phi value for compensation_recoded &
experience_recoded. Obtained values are Ch-square= 0.122 & Phi= 0.467. (Here
compensation is dependent on experience, so putting dependent variable in row &
independent variable in column).
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b. Mean classified experience and mean classified JS
It is similar to part (a). Here JS is dependent on Experience. Obtained values are Ch-
square= 0.122 & Phi= 0.467.
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c. Mean classified experience and mean classified TI.
It is similar to part (a). Here TI is dependent on Experience. Obtained values are Ch-
square= 0.376 & Phi= -0.267.
d. Mean classified experience and gender.
It is similar to part (a). Obtained values are Ch-square= 0.376 & Phi= -0.633.
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15) Do you find correlation between JS and turnover intention?
a. Is this significant?
b. Is the correlation between JS & TI influenced by-
a. Employees compensation
b. Experience
c. Gender
Find out in which case the influence of the third variable is high on the relation
between JS & TI.
Solution:
Correlation between JS & TI is -0.938. No, this is not significant.
1. Employees compensation (Answer: Correlation= -0.89)
2. Experience (Answer: Correlation= -0.900)
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3. Gender (Answer: Correlation=-0.947)
In order to find out unique relation between JS & TI, it was suspected that other variable
of interest may influence the relation between JS & TI. In order to check the
suppressing effect of these variables we performed partial correlation for each of these
variables. Correlation between JS & TI was found to be still significant. We have notobserved any significant change in the magnitude of correlation w.r.t. JS, TI & partial
variables. There is no change in the direction of the variables, therefore relation
between JS & TI exists and maximum difference was also meager. Therefore Job
Satisfaction & Turnover Intention remains insignificant of employee compensation,
experience and gender.
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16) Find out the impact of
a. Experience on JS
b. JS on TI
c. Compensation on JS.
Based on the model that you construct predict the JS score of Mr.M whose
compensation is 5800.
Correlation is to check the overlap between variables. This question requires ANOVA
for impact analysis. Regression is not only for strength between variables and also
impact of one variable on the other, prediction can also be made. The values are:
1. Experience on JS: JS is dependent on experience
2. JS on TI: TI is dependent on JS
3. Compensation on JS: JS is dependent on compensation
Experience on JS
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Total impact R = 0.472it gives the overall effect of all independent variables on
dependent variables
Adjusted R = 0.413- it gives the value adjusted for sample used and the no. of
variables, it gives rough estimate of least likely effect obtainable size.
Gap between R and Adjusted R should be low and between +- 0.1. Our difference is
.059 so our model is robust.
R correlation, gives the overlap area
R Variance, it gives the overall effect of allindependent variables on dependentvariables
Adjusted R it gives the value adjusted for sample usedand the no. of variables, it gives roughestimate of least likely effect obtainable
size, i.e. this is the least possible value forany sample
Std. Error of Estimate It gives the standard deviation of theresidual value. Smaller the value better itis.
Regression and ANOVA are based on the analysis of F-statistic. ANOVA helps in
assessing total impact and significant.
Absolute impactin original measurement, Example- one year increase in experience
is going to impact JS by 0.197 and relative impact is 0.687. Regression modelsignificance of individual parameters is based on Tdistribution, whereas the overall
effect is based on F distribution.
Constant = 2.454 (intercept), its the average JS value. Relative impact = 0.687 is same
as correlation because of single variable.
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JS on TI
R = 0.881 and Adjusted R = 0.867, Total impact = 0.881, Absolute impact = -1.153
and Relative impact = -0.938.
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Compensation on JS
(JS is dependent on compensation.) Based on the model that you construct predict the
JS score of Mr.M whose compensation is 5800.
Y=1.393 + 0.000X.
Therefore for X=5800, Y=1.393 (i.e. for compensation=5800, job satisfaction is 1.393).
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17) Perform a multiple regression analysis to find out the impact
of employees experience, compensation, JS on TI.
R = 0.921, implying the variance is 92.1%. We can infer that TI is not influenced by
compensation or experience; and JS has very strong and is significant on TI. (Value of
beta are compensation= -0.349, experience= 0.295, job satisfaction= -0.913
respectively)
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