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1. Outline the differences between Hoarding power and
Encouraging.
2. Explain about the power of Congruency in Leadership.
DataIDSalaryCompaMidpoint AgePerformance
RatingServiceGenderRaiseDegreeGender1GrCopy Employee
Data set to this page.822.10.962233290915.81FAThe ongoing
question that the weekly assignments will focus on is: Are
males and females paid the same for equal work (under the
Equal Pay Act)? 1522.60.984233280814.91FANote: to simplfy
the analysis, we will assume that jobs within each grade
comprise equal
work.3522.60.984232390415.30FA37230.999232295216.20FAT
he column labels in the table
mean:1023.11.003233080714.71FAID – Employee sample
number Salary – Salary in thousands
2323.11.004233665613.30FAAge – Age in yearsPerformance
Rating – Appraisal rating (Employee evaluation
score)1123.31.01223411001914.81FASERvice – Years of
serviceGender: 0 = male, 1 = female
2623.51.020232295216.20FAMidpoint – salary grade midpoint
Raise – percent of last raise3123.61.028232960413.91FAGrade
– job/pay gradeDegree (0= BSBA 1 =
MS)3623.61.026232775314.30FAGender1 (Male or
Female)Compa-ratio - salary divided by
midpoint4023.81.034232490206.30MA14241.04523329012161F
A4224.21.0512332100815.71FA1924.31.055233285104.61MA2
5251.0872341704040MA3226.50.855312595405.60MB227.70.8
95315280703.90MB3428.60.923312680204.91MB3933.91.0943
12790615.50FB2034.11.1013144701614.80FB1834.51.1133131
801115.60FB335.11.132313075513.61FB1341.11.02740301002
14.70FC741.31.0324032100815.71FC1642.21.054404490405.70
MC4145.81.144402580504.30MC2746.91.172403580703.91MC
548.21.0044836901605.71MD3049.31.0274845901804.30MD24
56.31.173483075913.80FD4556.91.185483695815.21FD4757.21
.003573795505.51ME3357.51.008573590905.51ME4581.01857
421001605.51ME3858.81.0325745951104.50ME5059.61.04657
38801204.60ME4660.21.0575739752003.91ME2260.31.257484
865613.81FD161.61.081573485805.70ME4461.81.08557459016
05.21ME49631.1055741952106.60ME1763.71.1185727553131F
E1264.71.1355752952204.50ME4869.51.2195734901115.31FE9
73.91.103674910010041MF4375.61.1286742952015.50FF2976.
31.139675295505.40MF2177.21.1526743951306.31MF678.11.1
656736701204.51MF2878.31.169674495914.40FF
Week 2This assignment covers the material presented in weeks
1 and 2.Six QuestionsBefore starting this assignment, make sure
the the assignment data from the Employee Salary Data Set file
is copied over to this Assignment file.You can do this either by
a copy and paste of all the columns or by opening the data file,
right clicking on the Data tab, selecting Move or Copy, and
copying the entire sheet to this file(Weekly Assignment Sheet
or whatever you are calling your master assignment file).It is
highly recommended that you copy the data columns (with
labels) and paste them to the right so that whatever you do will
not disrupt the original data values and relationships.To Ensure
full credit for each question, you need to show how you got
your results. For example, Question 1 asks for several data
values. If you obtain them using descriptive statistics,then the
cells should have an "=XX" formula in them, where XX is the
column and row number showing the value in the descriptive
statistics table. If you choose to generate each value using
fxfunctions, then each function should be located in the cell and
the location of the data values should be shown.So, Cell D31 -
as an example - shoud contain something like "=T6" or
"=average(T2:T26)". Having only a numerical value will not
earn full credit.The reason for this is to allow instructors to
provide feedback on Excel tools if the answers are not correct -
we need to see how the results were obtained.In starting the
analysis on a research question, we focus on overall descriptive
statistics and seeing if differences exist. Probing into reasons
and mitigating factors is a follow-up activity.1The first step in
analyzing data sets is to find some summary descriptive
statistics for key variables. Since the assignment problems
willfocus mostly on the compa-ratios, we need to find the mean,
standard deviations, and range for our groups: Males, Females,
and Overall.Sorting the compa-ratios into male and females will
require you copy and paste the Compa-ratio and Gender1
columns, and then sort on Gender1.The values for age,
performance rating, and service are provided for you for future
use, and - if desired - to test your approach to the compa-ratio
answers (see if you can replicate the values).You can use either
the Data Analysis Descriptive Statistics tool or the Fx =average
and =stdev functions. The range can be found using the
difference between the =max and =min functions with Fx
functions or from Descriptive Statistics.Suggestion: Copy and
paste the compa-ratio data to the right (Column T) and gender
data in column U. If you use Descriptive statistics, Place the
output table in row 1 of a column to the right.If you did not use
Descriptive Statistics, make sure your cells show the location of
the data (Example: =average(T2:T51)Compa-ratioAgePerf.
Rat.ServiceOverallMean1.069035.785.99.0Standard
Deviation0.00118.251311.41475.7177Note - remember the data
is a sample from the larger company
populationRange0.402304521FemaleMean1.072332.584.27.9Sta
ndard
Deviation0.01676.913.64.9Range0.29526.045.018.0MaleMean1.
054138.987.610.0Standard
Deviation0.01698.48.76.4Range0.31728.030.021.0A key issue
in comparing data sets is to see if they are distributed/shaped
the same. At this point we can do this by looking at the
probabilities that males and females are distributed in the same
way for a grade levels.2Empirical Probability: What is the
probability for a:Probabilitya. Randomly selected person
being in grade E or above?0.88b. Randomly selected person
being a male in grade E or above? 0.4c. Randomly selected
male being in grade E or above? 0.84d. Why are the results
different?The probability of randomly selecting male takes into
focus whole population less those who got grade F while the
probability of randomly selecting male takes into focus only the
male population of those who scored E grade and above.The
probability of a selected person being grade E or above takes
into account the probability of all grades less grade F3Normal
Curve based probability: For each group (overall, females,
males), what are the values for each question below?:Make sure
your answer cells show the Excel function and cell location of
the data used.AThe probability of being in the top 1/3 of the
compa-ratio distribution..Note, we can find the cutoff value for
the top 1/3 using the fx Large function: =large(range,
value).Value is the number that identifies the x-largest value.
For the top 1/3 value would be the value that starts the top 1/3
of the range,For the overall group, this would be the 50/3 or
17th (rounded), for the gender groups, it would be the 25/3 =
8th (rounded) value.OverallFemaleMaleAll of the functions
below are in the fx statistical list.i.How nany salaries are in the
top 1/3 (rounded to nearest whole number) for each group?
60.641.962.83Use the "=ROUND" function (found in Math or
All list)iiWhat Compa-ratio value starts the top 1/3 of the range
for each group?1.1081.1041.111Use the "=LARGE"
functioniiiWhat is the z-score for this
value?0.766340.1872270.558967Use Excel's STANDARDIZE
function iv.What is the normal curve probability of exceeding
this score?0.60710.16510.1361369Use "=1-NORM.S.DIST"
functionBHow do you interpret the relationship between the
data sets? What does this suggest about our equal pay for equal
work question?Based on Compa results, the value that cuts off
the top third salary value is almost the same. The information
provided in this case is not sufficient to make conclusion
regarding equal pay4Based on our sample data set, can the male
and female compa-ratios in the population be equal to each
other?AFirst, we need to determine if these two groups have
equal variances, in order to decide which t-test to use.What is
the data input ranged used for this question:Step 1:Ho:Male and
female compa ratios in the population are equal to each
otherHa:Male and female compa ratios in the population are not
equal to each otherStep 2:Decision Rule:Reject Ho if
P<0.05Step 3:Statistical test:Two sample t test Why?There is
need to determine if these two groups have equal variances.Step
4:Conduct the test - place cell B77 in the output location box.t-
Test: Two-Sample Assuming Equal
VariancesMaleFemaleMean1.055161.07464Variance0.00657189
0.00655624Observations2525Pooled
Variance0.006564065.Hypothesized Mean Difference0df48t
Stat-0.850075545P(T<=t) one-tail0.1997517452t Critical one-
tail1.6772241961P(T<=t) two-tail0.3995034904t Critical two-
tail2.0106347576Step 5:Conclusion and InterpretationWhat is
the p-value:0.3995Is the P-value < 0.05 (for a one tail test) or
0.025 (for a two tail test)?NoWhat is your decision: REJ or
NOT reject the null?Not Reject the nullWhat does this result
say about our question of variance equality?there exists
statitically significant differenct in male and female compaBAre
male and female average compa-ratios equal?(Regardless of the
outcome of the above F-test, assume equal variances for this
test.)What is the data input ranged used for this question:Step
1:Ho:Male Compa ratio =Female Compa ratioHa:Male Compa
ratio =/=Female Compa ratioStep 2:Decision Rule:Reject Ho if
P<0.05Step 3:Statistical test:Analysis of variance Why?Aims to
determine the existing difference between the male and female
average Step 4:Conduct the test - place cell B109 in the output
location box.Anova: Single
FactorSUMMARYGroupsCountSumAverageVarianceCompa505
3.2451.06490.0065269082Gender50250.50.2551020408ANOVA
Source of VariationSSdfMSFP-valueF critBetween
Groups7.9778002517.9778002560.985607908603.938111078Wi
thin
Groups12.8198185980.1308144745Total20.7976187599Step
5:Conclusion and InterpretationWhat is the p-value:0Is the P-
value < 0.05 (for a one tail test) or 0.025 (for a two tail
test)?YesWhat is your decision: REJ or NOT reject the
null?RejectWhat does your decision on rejecting the null
hypothesis mean?the mean are not equal between male and
female compa ratioIf the null hypothesis was rejected, calculate
the effect size value:0.24If the effect size was calculated, what
doe the result mean in terms of why the null hypothesis was
rejected?The effect of size is less than 0.5 thus indicating there
is a small effect on the sample in identifying the differenceWhat
does the result of this test tell us about our question on salary
equality?There is difference between the male and female
salaries5Is the Female average compa-ratio equal to or less than
the midpoint value of 1.00?This question is the same as: Does
the company, pay its females - on average - at or below the
grade midpoint (which is considered the market
rate)?Suggestion: Use the data column T to the right for your
null hypothesis value.What is the data input ranged used for this
question:Step 1:Ho: Female average compa-ratio equal to or less
than the midpoint value of 1.00Ha: Female average compa-ratio
is not equal to the midpoint value of 1.00Step 2:Decision
Rule:Reject Ho if P<0.05Step 3:Statistical test:One sample t
testWhy?focuses on determining whether female average is less
or equal to 1Step 4:Conduct the test - place cell B162 in the
output location box.t-Test: Two-Sample Assuming Equal
VariancesVariable 1Variable
2Mean1.074641Variance0.006556240Observations2525Pooled
Variance0.00327812Hypothesized Mean Difference0df48t
Stat4.6090796525P(T<=t) one-tail0.0000150462t Critical one-
tail1.6772241961P(T<=t) two-tail0.0000300925t Critical two-
tail2.0106347576Step 5:Conclusion and InterpretationWhat is
the p-value:0.0000150462Is the P-value < 0.05 (for a one tail
test) or 0.025 (for a two tail test)?YesWhat, besides the p-value,
needs to be considered with a one tail test?t statistic Decision:
Reject or do not reject Ho?reject What does your decision on
rejecting the null hypothesis mean? Female average compa-ratio
is not equal to the midpoint value of 1.00If the null hypothesis
was rejected, calculate the effect size value:1.3If the effect size
was calculated, what doe the result mean in terms of why the
null hypothesis was rejected?The effect of size is greater than 1
thus indicating there is a very large effect on the sample in
identifying the differenceWhat does the result of this test tell us
about our question on salary equality?the female compa ratio is
not less or equal to 16Considering both the salary information
in the lectures and your compa-ratio information, what
conclusions can you reach about equal pay for equal
work?Compa shows that the salaries are almost the same and
the difference between them is negligible although there is no
equal pay for equal workWhy - what statistical results support
this conclusion?the two sample test and analysis of variance
conducted shows clearly that there is difference in salaries
between men and women
Week 3Week 3ANOVAThree QuestionsRemember to show how
you got your results in the appropriate cells. For questions
using functions, show the input range when asked.Group
name:G1G2G3G4G5G61One interesting question is are the
average compa-ratios equal across salary ranges of 10K
each.Salary Intervals: 22-2930-3940-4950-5960-6970-79While
compa-ratios remove the impact of grade on salaries, are they
different for different pay levels,Compa-ratio values:
0.9621.0941.0271.1731.0571.103that is are people at different
levels paid differently relative to the midpoint? (Put data
values at
right.)0.9841.1011.0321.1851.2571.1280.9841.1131.0541.0031.
0811.139What is the data input ranged used for this
question:0.9991.1321.1441.0081.0851.152Step 1:Ho:Compa
ratios are equal1.0031.1721.0181.1051.165Ha:Compa ratios are
unequal1.0041.0041.0321.1181.169Step 2:Decision Rule:Reject
Ho if P<0.051.0121.0271.0461.135Step 3:Statistical test:Single
fact Anova1.0201.219Why?Aims to determine the diffrence in
compa variable based on salary groups1.028Step 4:Conduct the
test - place cell b16 in the output location box.1.026Anova:
Single
Factor1.0341.045SUMMARY1.051GroupsCountSumAverageVar
iance1.055G11817.9670.99816666670.00342826471.087G244.4
41.110.00027666670.855G377.461.06571428570.00425090480.
895G477.4651.06642857140.0061336190.923G589.0571.13212
50.0049375536G666.8561.14266666670.0006162667ANOVASo
urce of VariationSSdfMSFP-valueF critBetween
Groups0.160756648850.03215132988.89376364560.000006793
62.4270401198Within
Groups0.1590618512440.0036150421Total0.319818549Step
5:Conclusions and InterpretationWhat is the p-
value?0.0000067936Is P-value < 0.05?YesWhat is your
decision: REJ or NOT reject the null?Reject the null
hypothesisIf the null hypothesis was rejected, what is the effect
size value (eta squared)?0.5026496241If calculated, what does
the effect size value tell us about why the null hypothesis was
rejected?There is a large effect size which shows that the
rejection of null hypothesis was caused by variable
interactionWhat does that decision mean in terms of our equal
pay question?There exist significant difference in pay2If the
null hypothesis in question 1 was rejected, which pairs of means
differ?Why?Groups ComparedDiffT+/- TermLowto
HighDifference Significant?Why?G1 G2-
0.11183333332.0150.0669812904-0.1788146237-
0.044852043yesZero not in rageG1 G3-
1.06571428572.0150.0539750619-1.1196893477-
1.0117392238yesZero not in rageG1 G4-
0.06826190482.0150.0539750619-0.1222369667-
0.0142868428yesZero not in rageG1 G5-
0.13395833332.0150.0514889275-0.1854472608-
0.0824694058yesZero not in rageG1 G6-
0.14452.0150.0571218364-0.2016218364-0.0873781636yesZero
not in rageG2 G30.04428571432.0150.0759496444-
0.03166393010.1202353586noZero in rangeG2
G40.04357142862.0150.0759496444-
0.03237821580.1195210729noZero in rangeG2 G5-
0.0221252.0150.0742034421-
0.09632844210.0520784421noZero in rangeG2 G6-
0.03266666672.0150.0782172958-
0.11088396240.0455506291noZero in rangeG3 G4-
0.00071428572.0150.0647700743-
0.06548436010.0640557886noZero in rangeG3 G5-
0.06641071432.0150.0627133548-0.1291240691-
0.0036973595YesZero not in rangeG3 G6-
0.0769523812.0150.0674148308-0.1443672117-
0.0095375502yesZero not in rangeG4 G5-
0.06569642862.0150.0627133548-0.1284097834-
0.0029830738yesZero not in rangeG4 G6-
0.07623809522.0150.0674148308-0.143652926-
0.0088232645yesZero not in rangeG5 G6-
0.01054166672.0150.0654412847-
0.07598295140.0548996181yesZero not in range3Since compa
is already a measure of pay for equal work, do these results
impact your conclusion on equal pay for equal work? Why or
why not?Yes. These results provides an understanding on the
existing difference in pay across the different salary ranges
considered.The information significantly reflects on the
conclusion regarding equal pay for equal work done. Where
from the analysis there is no equal pay among the groups
Week 4Regression and CorellationFive QuestionsCompa-
ratioMidpointAgePerformance
RatingServiceRaiseDegreeGenderRemember to show how you
got your results in the appropriate cells. For questions using
functions, show the input range when asked.1Create a
correlation table using Compa-ratio and the other interval level
variables, except for Salary.Suggestion, place data in columns T
- Y.What range was placed in the Correlation input range
box:Place C9 in output box.bWhat are the statistically
significant correlations related to Compa-ratio?T =Significant r
=cAre there any surprises - correlations you though would be
significant and are not, or non significant correlations you
thought would be?dWhy does or does not this information help
answer our equal pay question?2Perform a regression analysis
using compa as the dependent variable and the variables used in
Q1 along withincluding the dummy variables. Show the result,
and interpret your findings by answering the following
questions.Suggestion: Place the dummy variables values to the
right of column Y.What range was placed in the Regression
input range box:Note: be sure to include the appropriate
hypothesis statements.Regression hypothesesHo:Ha:Coefficient
hyhpotheses (one to stand for all the separate
variables)Ho:Ha:Place B36 in output box.Interpretation:For the
Regression as a whole:What is the value of the F statistic: What
is the p-value associated with this value: Is the p-value <
0.05?What is your decision: REJ or NOT reject the null?What
does this decision mean? For each of the coefficients:
MidpointAgePerf. Rat.ServiceGenderDegreeWhat is the
coefficient's p-value for each of the variables: Is the p-value <
0.05?Do you reject or not reject each null hypothesis: What are
the coefficients for the significant variables?Using the intercept
coefficient and only the significant variables, what is the
equation?Compa-ratio = Is gender a significant factor in compa-
ratio?Regardless of statistical significance, who gets paid more
with all other things being equal?How do we know? 3What does
regression analysis show us about analyzing complex
measures?4Between the lecture results and your results, what
else would you like to knowbefore answering our question on
equal pay? Why?5Between the lecture results and your results,
what is your answer to the questionof equal pay for equal work
for males and females? Why?

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  • 1. 1. Outline the differences between Hoarding power and Encouraging. 2. Explain about the power of Congruency in Leadership. DataIDSalaryCompaMidpoint AgePerformance RatingServiceGenderRaiseDegreeGender1GrCopy Employee Data set to this page.822.10.962233290915.81FAThe ongoing question that the weekly assignments will focus on is: Are males and females paid the same for equal work (under the Equal Pay Act)? 1522.60.984233280814.91FANote: to simplfy the analysis, we will assume that jobs within each grade comprise equal work.3522.60.984232390415.30FA37230.999232295216.20FAT he column labels in the table mean:1023.11.003233080714.71FAID – Employee sample number Salary – Salary in thousands 2323.11.004233665613.30FAAge – Age in yearsPerformance Rating – Appraisal rating (Employee evaluation score)1123.31.01223411001914.81FASERvice – Years of serviceGender: 0 = male, 1 = female 2623.51.020232295216.20FAMidpoint – salary grade midpoint Raise – percent of last raise3123.61.028232960413.91FAGrade – job/pay gradeDegree (0= BSBA 1 = MS)3623.61.026232775314.30FAGender1 (Male or Female)Compa-ratio - salary divided by midpoint4023.81.034232490206.30MA14241.04523329012161F A4224.21.0512332100815.71FA1924.31.055233285104.61MA2 5251.0872341704040MA3226.50.855312595405.60MB227.70.8 95315280703.90MB3428.60.923312680204.91MB3933.91.0943 12790615.50FB2034.11.1013144701614.80FB1834.51.1133131 801115.60FB335.11.132313075513.61FB1341.11.02740301002 14.70FC741.31.0324032100815.71FC1642.21.054404490405.70 MC4145.81.144402580504.30MC2746.91.172403580703.91MC
  • 2. 548.21.0044836901605.71MD3049.31.0274845901804.30MD24 56.31.173483075913.80FD4556.91.185483695815.21FD4757.21 .003573795505.51ME3357.51.008573590905.51ME4581.01857 421001605.51ME3858.81.0325745951104.50ME5059.61.04657 38801204.60ME4660.21.0575739752003.91ME2260.31.257484 865613.81FD161.61.081573485805.70ME4461.81.08557459016 05.21ME49631.1055741952106.60ME1763.71.1185727553131F E1264.71.1355752952204.50ME4869.51.2195734901115.31FE9 73.91.103674910010041MF4375.61.1286742952015.50FF2976. 31.139675295505.40MF2177.21.1526743951306.31MF678.11.1 656736701204.51MF2878.31.169674495914.40FF Week 2This assignment covers the material presented in weeks 1 and 2.Six QuestionsBefore starting this assignment, make sure the the assignment data from the Employee Salary Data Set file is copied over to this Assignment file.You can do this either by a copy and paste of all the columns or by opening the data file, right clicking on the Data tab, selecting Move or Copy, and copying the entire sheet to this file(Weekly Assignment Sheet or whatever you are calling your master assignment file).It is highly recommended that you copy the data columns (with labels) and paste them to the right so that whatever you do will not disrupt the original data values and relationships.To Ensure full credit for each question, you need to show how you got your results. For example, Question 1 asks for several data values. If you obtain them using descriptive statistics,then the cells should have an "=XX" formula in them, where XX is the column and row number showing the value in the descriptive statistics table. If you choose to generate each value using fxfunctions, then each function should be located in the cell and the location of the data values should be shown.So, Cell D31 - as an example - shoud contain something like "=T6" or "=average(T2:T26)". Having only a numerical value will not earn full credit.The reason for this is to allow instructors to provide feedback on Excel tools if the answers are not correct - we need to see how the results were obtained.In starting the analysis on a research question, we focus on overall descriptive
  • 3. statistics and seeing if differences exist. Probing into reasons and mitigating factors is a follow-up activity.1The first step in analyzing data sets is to find some summary descriptive statistics for key variables. Since the assignment problems willfocus mostly on the compa-ratios, we need to find the mean, standard deviations, and range for our groups: Males, Females, and Overall.Sorting the compa-ratios into male and females will require you copy and paste the Compa-ratio and Gender1 columns, and then sort on Gender1.The values for age, performance rating, and service are provided for you for future use, and - if desired - to test your approach to the compa-ratio answers (see if you can replicate the values).You can use either the Data Analysis Descriptive Statistics tool or the Fx =average and =stdev functions. The range can be found using the difference between the =max and =min functions with Fx functions or from Descriptive Statistics.Suggestion: Copy and paste the compa-ratio data to the right (Column T) and gender data in column U. If you use Descriptive statistics, Place the output table in row 1 of a column to the right.If you did not use Descriptive Statistics, make sure your cells show the location of the data (Example: =average(T2:T51)Compa-ratioAgePerf. Rat.ServiceOverallMean1.069035.785.99.0Standard Deviation0.00118.251311.41475.7177Note - remember the data is a sample from the larger company populationRange0.402304521FemaleMean1.072332.584.27.9Sta ndard Deviation0.01676.913.64.9Range0.29526.045.018.0MaleMean1. 054138.987.610.0Standard Deviation0.01698.48.76.4Range0.31728.030.021.0A key issue in comparing data sets is to see if they are distributed/shaped the same. At this point we can do this by looking at the probabilities that males and females are distributed in the same way for a grade levels.2Empirical Probability: What is the probability for a:Probabilitya. Randomly selected person being in grade E or above?0.88b. Randomly selected person being a male in grade E or above? 0.4c. Randomly selected
  • 4. male being in grade E or above? 0.84d. Why are the results different?The probability of randomly selecting male takes into focus whole population less those who got grade F while the probability of randomly selecting male takes into focus only the male population of those who scored E grade and above.The probability of a selected person being grade E or above takes into account the probability of all grades less grade F3Normal Curve based probability: For each group (overall, females, males), what are the values for each question below?:Make sure your answer cells show the Excel function and cell location of the data used.AThe probability of being in the top 1/3 of the compa-ratio distribution..Note, we can find the cutoff value for the top 1/3 using the fx Large function: =large(range, value).Value is the number that identifies the x-largest value. For the top 1/3 value would be the value that starts the top 1/3 of the range,For the overall group, this would be the 50/3 or 17th (rounded), for the gender groups, it would be the 25/3 = 8th (rounded) value.OverallFemaleMaleAll of the functions below are in the fx statistical list.i.How nany salaries are in the top 1/3 (rounded to nearest whole number) for each group? 60.641.962.83Use the "=ROUND" function (found in Math or All list)iiWhat Compa-ratio value starts the top 1/3 of the range for each group?1.1081.1041.111Use the "=LARGE" functioniiiWhat is the z-score for this value?0.766340.1872270.558967Use Excel's STANDARDIZE function iv.What is the normal curve probability of exceeding this score?0.60710.16510.1361369Use "=1-NORM.S.DIST" functionBHow do you interpret the relationship between the data sets? What does this suggest about our equal pay for equal work question?Based on Compa results, the value that cuts off the top third salary value is almost the same. The information provided in this case is not sufficient to make conclusion regarding equal pay4Based on our sample data set, can the male and female compa-ratios in the population be equal to each other?AFirst, we need to determine if these two groups have equal variances, in order to decide which t-test to use.What is
  • 5. the data input ranged used for this question:Step 1:Ho:Male and female compa ratios in the population are equal to each otherHa:Male and female compa ratios in the population are not equal to each otherStep 2:Decision Rule:Reject Ho if P<0.05Step 3:Statistical test:Two sample t test Why?There is need to determine if these two groups have equal variances.Step 4:Conduct the test - place cell B77 in the output location box.t- Test: Two-Sample Assuming Equal VariancesMaleFemaleMean1.055161.07464Variance0.00657189 0.00655624Observations2525Pooled Variance0.006564065.Hypothesized Mean Difference0df48t Stat-0.850075545P(T<=t) one-tail0.1997517452t Critical one- tail1.6772241961P(T<=t) two-tail0.3995034904t Critical two- tail2.0106347576Step 5:Conclusion and InterpretationWhat is the p-value:0.3995Is the P-value < 0.05 (for a one tail test) or 0.025 (for a two tail test)?NoWhat is your decision: REJ or NOT reject the null?Not Reject the nullWhat does this result say about our question of variance equality?there exists statitically significant differenct in male and female compaBAre male and female average compa-ratios equal?(Regardless of the outcome of the above F-test, assume equal variances for this test.)What is the data input ranged used for this question:Step 1:Ho:Male Compa ratio =Female Compa ratioHa:Male Compa ratio =/=Female Compa ratioStep 2:Decision Rule:Reject Ho if P<0.05Step 3:Statistical test:Analysis of variance Why?Aims to determine the existing difference between the male and female average Step 4:Conduct the test - place cell B109 in the output location box.Anova: Single FactorSUMMARYGroupsCountSumAverageVarianceCompa505 3.2451.06490.0065269082Gender50250.50.2551020408ANOVA Source of VariationSSdfMSFP-valueF critBetween Groups7.9778002517.9778002560.985607908603.938111078Wi thin Groups12.8198185980.1308144745Total20.7976187599Step 5:Conclusion and InterpretationWhat is the p-value:0Is the P- value < 0.05 (for a one tail test) or 0.025 (for a two tail
  • 6. test)?YesWhat is your decision: REJ or NOT reject the null?RejectWhat does your decision on rejecting the null hypothesis mean?the mean are not equal between male and female compa ratioIf the null hypothesis was rejected, calculate the effect size value:0.24If the effect size was calculated, what doe the result mean in terms of why the null hypothesis was rejected?The effect of size is less than 0.5 thus indicating there is a small effect on the sample in identifying the differenceWhat does the result of this test tell us about our question on salary equality?There is difference between the male and female salaries5Is the Female average compa-ratio equal to or less than the midpoint value of 1.00?This question is the same as: Does the company, pay its females - on average - at or below the grade midpoint (which is considered the market rate)?Suggestion: Use the data column T to the right for your null hypothesis value.What is the data input ranged used for this question:Step 1:Ho: Female average compa-ratio equal to or less than the midpoint value of 1.00Ha: Female average compa-ratio is not equal to the midpoint value of 1.00Step 2:Decision Rule:Reject Ho if P<0.05Step 3:Statistical test:One sample t testWhy?focuses on determining whether female average is less or equal to 1Step 4:Conduct the test - place cell B162 in the output location box.t-Test: Two-Sample Assuming Equal VariancesVariable 1Variable 2Mean1.074641Variance0.006556240Observations2525Pooled Variance0.00327812Hypothesized Mean Difference0df48t Stat4.6090796525P(T<=t) one-tail0.0000150462t Critical one- tail1.6772241961P(T<=t) two-tail0.0000300925t Critical two- tail2.0106347576Step 5:Conclusion and InterpretationWhat is the p-value:0.0000150462Is the P-value < 0.05 (for a one tail test) or 0.025 (for a two tail test)?YesWhat, besides the p-value, needs to be considered with a one tail test?t statistic Decision: Reject or do not reject Ho?reject What does your decision on rejecting the null hypothesis mean? Female average compa-ratio is not equal to the midpoint value of 1.00If the null hypothesis was rejected, calculate the effect size value:1.3If the effect size
  • 7. was calculated, what doe the result mean in terms of why the null hypothesis was rejected?The effect of size is greater than 1 thus indicating there is a very large effect on the sample in identifying the differenceWhat does the result of this test tell us about our question on salary equality?the female compa ratio is not less or equal to 16Considering both the salary information in the lectures and your compa-ratio information, what conclusions can you reach about equal pay for equal work?Compa shows that the salaries are almost the same and the difference between them is negligible although there is no equal pay for equal workWhy - what statistical results support this conclusion?the two sample test and analysis of variance conducted shows clearly that there is difference in salaries between men and women Week 3Week 3ANOVAThree QuestionsRemember to show how you got your results in the appropriate cells. For questions using functions, show the input range when asked.Group name:G1G2G3G4G5G61One interesting question is are the average compa-ratios equal across salary ranges of 10K each.Salary Intervals: 22-2930-3940-4950-5960-6970-79While compa-ratios remove the impact of grade on salaries, are they different for different pay levels,Compa-ratio values: 0.9621.0941.0271.1731.0571.103that is are people at different levels paid differently relative to the midpoint? (Put data values at right.)0.9841.1011.0321.1851.2571.1280.9841.1131.0541.0031. 0811.139What is the data input ranged used for this question:0.9991.1321.1441.0081.0851.152Step 1:Ho:Compa ratios are equal1.0031.1721.0181.1051.165Ha:Compa ratios are unequal1.0041.0041.0321.1181.169Step 2:Decision Rule:Reject Ho if P<0.051.0121.0271.0461.135Step 3:Statistical test:Single fact Anova1.0201.219Why?Aims to determine the diffrence in compa variable based on salary groups1.028Step 4:Conduct the test - place cell b16 in the output location box.1.026Anova: Single Factor1.0341.045SUMMARY1.051GroupsCountSumAverageVar
  • 8. iance1.055G11817.9670.99816666670.00342826471.087G244.4 41.110.00027666670.855G377.461.06571428570.00425090480. 895G477.4651.06642857140.0061336190.923G589.0571.13212 50.0049375536G666.8561.14266666670.0006162667ANOVASo urce of VariationSSdfMSFP-valueF critBetween Groups0.160756648850.03215132988.89376364560.000006793 62.4270401198Within Groups0.1590618512440.0036150421Total0.319818549Step 5:Conclusions and InterpretationWhat is the p- value?0.0000067936Is P-value < 0.05?YesWhat is your decision: REJ or NOT reject the null?Reject the null hypothesisIf the null hypothesis was rejected, what is the effect size value (eta squared)?0.5026496241If calculated, what does the effect size value tell us about why the null hypothesis was rejected?There is a large effect size which shows that the rejection of null hypothesis was caused by variable interactionWhat does that decision mean in terms of our equal pay question?There exist significant difference in pay2If the null hypothesis in question 1 was rejected, which pairs of means differ?Why?Groups ComparedDiffT+/- TermLowto HighDifference Significant?Why?G1 G2- 0.11183333332.0150.0669812904-0.1788146237- 0.044852043yesZero not in rageG1 G3- 1.06571428572.0150.0539750619-1.1196893477- 1.0117392238yesZero not in rageG1 G4- 0.06826190482.0150.0539750619-0.1222369667- 0.0142868428yesZero not in rageG1 G5- 0.13395833332.0150.0514889275-0.1854472608- 0.0824694058yesZero not in rageG1 G6- 0.14452.0150.0571218364-0.2016218364-0.0873781636yesZero not in rageG2 G30.04428571432.0150.0759496444- 0.03166393010.1202353586noZero in rangeG2 G40.04357142862.0150.0759496444- 0.03237821580.1195210729noZero in rangeG2 G5- 0.0221252.0150.0742034421- 0.09632844210.0520784421noZero in rangeG2 G6-
  • 9. 0.03266666672.0150.0782172958- 0.11088396240.0455506291noZero in rangeG3 G4- 0.00071428572.0150.0647700743- 0.06548436010.0640557886noZero in rangeG3 G5- 0.06641071432.0150.0627133548-0.1291240691- 0.0036973595YesZero not in rangeG3 G6- 0.0769523812.0150.0674148308-0.1443672117- 0.0095375502yesZero not in rangeG4 G5- 0.06569642862.0150.0627133548-0.1284097834- 0.0029830738yesZero not in rangeG4 G6- 0.07623809522.0150.0674148308-0.143652926- 0.0088232645yesZero not in rangeG5 G6- 0.01054166672.0150.0654412847- 0.07598295140.0548996181yesZero not in range3Since compa is already a measure of pay for equal work, do these results impact your conclusion on equal pay for equal work? Why or why not?Yes. These results provides an understanding on the existing difference in pay across the different salary ranges considered.The information significantly reflects on the conclusion regarding equal pay for equal work done. Where from the analysis there is no equal pay among the groups Week 4Regression and CorellationFive QuestionsCompa- ratioMidpointAgePerformance RatingServiceRaiseDegreeGenderRemember to show how you got your results in the appropriate cells. For questions using functions, show the input range when asked.1Create a correlation table using Compa-ratio and the other interval level variables, except for Salary.Suggestion, place data in columns T - Y.What range was placed in the Correlation input range box:Place C9 in output box.bWhat are the statistically significant correlations related to Compa-ratio?T =Significant r =cAre there any surprises - correlations you though would be significant and are not, or non significant correlations you thought would be?dWhy does or does not this information help answer our equal pay question?2Perform a regression analysis using compa as the dependent variable and the variables used in
  • 10. Q1 along withincluding the dummy variables. Show the result, and interpret your findings by answering the following questions.Suggestion: Place the dummy variables values to the right of column Y.What range was placed in the Regression input range box:Note: be sure to include the appropriate hypothesis statements.Regression hypothesesHo:Ha:Coefficient hyhpotheses (one to stand for all the separate variables)Ho:Ha:Place B36 in output box.Interpretation:For the Regression as a whole:What is the value of the F statistic: What is the p-value associated with this value: Is the p-value < 0.05?What is your decision: REJ or NOT reject the null?What does this decision mean? For each of the coefficients: MidpointAgePerf. Rat.ServiceGenderDegreeWhat is the coefficient's p-value for each of the variables: Is the p-value < 0.05?Do you reject or not reject each null hypothesis: What are the coefficients for the significant variables?Using the intercept coefficient and only the significant variables, what is the equation?Compa-ratio = Is gender a significant factor in compa- ratio?Regardless of statistical significance, who gets paid more with all other things being equal?How do we know? 3What does regression analysis show us about analyzing complex measures?4Between the lecture results and your results, what else would you like to knowbefore answering our question on equal pay? Why?5Between the lecture results and your results, what is your answer to the questionof equal pay for equal work for males and females? Why?