This document contains analyses of salary data to examine whether there are differences in pay between male and female employees. Descriptive statistics are presented for overall salary as well as by gender. Tests are conducted to analyze variance and means between male and female salaries, as well as between employees with and without advanced degrees. The analyses find no significant differences in variance or means between male and female salaries or between those with and without degrees, suggesting equal pay for equal work.

1. Sheet3Sum of SalaryColumn LabelsRow LabelsABCDEFGrand
TotalF280.7143.183.2160.9131.2146.7945.8M71.283.313596.76
10.6301.61298.4Grand
Total351.9226.4218.2257.6741.8448.32244.2Row
LabelsABCDEFF280.7143.183.2160.9131.2146.7M71.283.3135
96.7610.6301.6
DataIDSalaryCompa-ratioMidpoint AgePerformance
RatingServiceGenderRaiseDegreeGender1GradeDo not
manipuilate Data set on this page, copy to another page to make
changes154.50.956573485805.70METhe 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)?
228.30.913315280703.90MBNote: to simplfy the analysis, we
will assume that jobs within each grade comprise equal
work.822.80.992233290915.81FA1023.31.014233080714.71FA
The column labels in the table
mean:1124.31.05723411001914.81FAID – Employee sample
number Salary – Salary in thousands
14251.08523329012161FAAge – Age in yearsPerformance
Rating - Appraisal rating (employee evaluation
score)1522.60.983233280814.91FAService – Years of service
(rounded)Gender – 0 = male, 1 = female
1923.91.039233285104.61MAMidpoint – salary grade midpoint
Raise – percent of last raise3122.90.995232960413.91FAGrade
– job/pay gradeDegree (0= BSBA 1 =
MS)4224.41.0592332100815.71FAGender1 (Male or
Female)Compa-ratio - salary divided by
midpoint334.11.100313075513.61FB1259.71.0475752952204.50
ME1341.81.0444030100214.70FC3426.90.869312680204.91MB
741.41.0344032100815.71FC1648.51.213404490405.70MC2746
.21.156403580703.91MC1836.21.1673131801115.60FB549.21.0
254836901605.71MD2035.51.1443144701614.80FB2257.61.199
484865613.81FD4549.91.040483695815.21FD2322.20.9642336
65613.30FA2453.41.112483075913.80FD2523.61.02823417040

2. 40MA2622.30.971232295216.20FA460.91.06857421001605.51
ME2874.41.111674495914.40FF2975.61.129675295505.40MF3
047.50.9894845901804.30MD1763.11.1075727553131FE3228.1
0.906312595405.60MB3363.71.117573590905.51ME4465.91.15
65745901605.21ME3522.70.987232390415.30FA3624.41.05923
2775314.30FA3723.81.034232295216.20FA3864.61.133574595
1104.50ME3937.31.202312790615.50FB4023.71.031232490206
.30MA4140.31.008402580504.30MC4657.41.0075739752003.91
ME4372.31.0796742952015.50FF47560.982573795505.51ME48
68.11.1955734901115.31FE674.11.1066736701204.51MF9731.0
89674910010041MF2178.91.1786743951306.31MF4966.21.161
5741952106.60ME5061.71.0835738801204.60ME
Sheet2Compa-ratioGenderMale
Female1.06801.0681.1001.02501.0251.0341.10601.1060.9921.0
8901.0891.0141.03901.0391.0571.17801.1781.0851.15601.1560
.9831.11701.1171.1070.86900.8691.1991.15601.1560.9951.007
01.0071.0590.98200.9821.0401.10011.1951.03410.99211.01411
.05711.08510.98311.10711.19910.99511.05911.04011.1951
Week 1Week 1: Descriptive Statistics, including
ProbabilityGender1SalaryWhile the lectures will examine our
equal pay question from the compa-ratio viewpoint, our weekly
assignments will focus onF22.8examining the issue using the
salary measure.F22.6M23.9The purpose of this assignmnent is
two fold:F24.41. Demonstrate mastery with Excel tools.F34.12.
Develop descriptive statistics to help examine the
question.F41.83. Interpret descriptive outcomesM26.9F41.4The
first issue in examining salary data to determine if we - as a
company - are paying males and females equally for doing equal
work is to develop someM46.2descriptive statistics to give us
something to make a preliminary decision on whether we have
an issue or not.F36.2F35.51Descriptive Statistics: Develop
basic descriptive statistics for SalaryF49.9The first step in
analyzing data sets is to find some summary descriptive
statistics for key variables. F22.2Suggestion: Copy the gender1
and salary columns from the Data tab to columns T and U at the
right.F53.4Then use Data Sort (by gender1) to get all the male

3. and female salary values grouped together.F22.3F74.4a. Use the
Descriptive Statistics function in the Data Analysis tab Place
Excel outcome in Cell K19F63.1to develop the descriptive
statistics summary for the overall Column1F22.7group's overall
salary. (Place K19 in output range.)F24.4Highlight the mean,
sample standard deviation, and range.Mean44.884F23.8Standard
Error2.6698167177F37.3Median44M57.4Mode24.4F72.3b.Using
Fx (or formula) functions find the following (be sure to show
the formula 18.8784550561Standard
Deviation18.8784550561F68.1and not just the value in each
cell) asked for salary statistics for each gender:Sample
Variance356.3960653061M78.9MaleFemaleKurtosis-
1.4241756975M54.5Mean:48.72841.04Skewness0.2096662654
M28.3Sample Standard
Deviation:18.573940526118.7581093575Range56.7F23.3Range:
52.756.7Minimum22.2F24.3Maximum78.9F25Sum2244.2F22.9
Count50M59.7M48.52Develop a 5-number summary for the
overall, male, and female SALARY variable.M49.2For full
credit, use the excel formulas in each cell rather than simply the
numerical
answer.F57.6OverallMalesFemalesM23.6Max78.975.678.9M60.
93rd Q61.563.753.4M75.6Midpoint4454.536.2M47.51st
Q24.428.123.9M28.1Min44.422.922.20M63.70.41M65.93Locati
on Measures: comparing Male and Female midpoints to the
overall Salary data range.M64.6For full credit, show the excel
formulas in each cell rather than simply the numerical
answer.M23.7Using the entire Salary range and the M and F
midpoints found in Q2MaleFemaleM40.3a. What would each
midpoint's percentile rank be in the overall range?0.640.37Use
Excel's =PERCENTRANK.EXC functionM56b. What is the
normal curve z value for each midpoint within overall
range?0.5888-0.5712Use Excel's =STANDARDIZE function
M74.1M734Probability Measures: comparing Male and Female
midpoints to the overall Salary data range0.62M66.2For full
credit, show the excel formulas in each cell rather than simply
the numerical answer.M61.7Using the entire Salary range and

4. the M and F midpoints found in Q2, findMaleFemalea. The
Empirical Probability of equaling or exceeding (=>) that value
for0.360.64Show the calculation formula = value/50 or
=countif(range,">="&cell)/50b. The Normal curve Prob of =>
that value for each group0.35942356680.2610862997Use "=1-
NORM.S.DIST" functionNote: be sure to use the ENTIRE salary
range for part a when finding the probability.5Conclusions:
What do you make of these results?Be sure to include findings
from this week's lectures as well.In comparing the overall,
male, and female outcomes, what relationship(s) see, to exist
between the data sets? Your findings:it is my finding that half
the males make more than the females and the other half of the
males make less than the females. Of the 56 compared to the
statistical prediction The lecture's related findings:the lecture
makes that argument that all the males make a larger salary but
that is not the case.Overall conclusion:is after you range and
midpoint to figure out the number of people that make larger
salary and how many make less than female in the
companyWhat does this suggest about our equal pay for equal
work question?
Week 2Week 2: Identifying Significant Differences - part
1SalaryCompa-ratioMaleFemaelMaleFemaleTo Ensure full
credit for each question, you need to show how you got your
results. This involves either showing where the data you used
is located 54.534.10.9561.100or showing the excel formula in
each cell.Be sure to copy the appropriate data columns from the
data tab to the right for your use this
week.28.341.40.9131.03460.922.81.0680.992As with our
examination of compa-ratio in the lecture, the first question we
have about salary between the genders involves equality - are
they the same or different?49.223.31.0251.014What we do,
depends upon our
findings.74.124.31.1061.0577341.81.0891.0441As with the
compa-ratio lecture example, we want to examine salary
variation within the groups - are they equal?Use Cell K10 for
the Excel test outcome location.59.7251.0471.085aWhat is the

5. data input ranged used for this question:F-Test Two-Sample for
Variances48.522.61.2130.983Q3:Q27andR3:R2723.963.11.0391.
107b Which is needed for this question: a one- or two-tail
hypothesis statement and test
?MaleFemale78.936.21.1781.167Answer:I would pick the one
tail hypothesisMean1.055561.0693623.635.51.0281.144Why:We
are only trying to prove that they are not
equalVariance0.00813842330.005229323346.257.61.1561.199O
bservations252575.622.21.1290.964c. Step 1:Ho:Male compa-
ratio variance equl Female compa-ratio variance
df242447.553.40.9891.112Ha:Male compa-ratio variance not
equl Female compa-ratio variance
F1.556305245428.122.30.9060.971Step 2:Significance
(Alpha):0.05P(F<=f) one-
tail0.142772879763.774.41.1171.111Step 3:Test Statistic and
test:F statistic and F-test for VarianceF Critical one-
tail1.983759568526.922.90.8690.995Why this test?Need to
determine the mean and significantly
different64.622.71.1330.987Step 4:Decision rule:Decision rule:
Reject the null hypothesis if the p-value is less than
0.0523.724.41.0311.059Step 5:Conduct the test - place test
function in cell k1040.323.81.0081.03465.937.31.1561.202Step
6:Conclusion and Interpretation57.424.41.0071.059What is the
p-value:0.14277287975672.30.9821.079What is your decision:
REJ or NOT reject the null?Not
Reject66.249.91.1611.040Why?p value is greater than
alpha61.768.11.0831.195What is your conclusion about the
variance in the population for male and female
salaries?Conclude that Male compa-ratio variance equl Female
compa-ratio variance 2Once we know about variance quality, we
can move on to means: Are male and female average salaries
equal?Use Cell K35 for the Excel test outcome
location.(Regardless of the outcome of the above F-test, assume
equal variances for this test.)aWhat is the data input ranged
used for this question:F-Test Two-Sample for
Variances03:027andP3:P27b Does this question need a one or

6. two-tail hypothesis statement and test?two tail
hypothesisMaleFemaelWhy:We trying to prove that they are not
equalMean51.93637.832c. Step 1:Ho:male salary
average=female salary
averageVariance314.8007333333309.2356Ha:male salary
average=/female salary averageObservations2525Step
2:Significance (Alpha):0.05df2424Step 3:Test Statistic and
test:F-testF1.0179964187Why this test?because we are dealing
with two tailP(F<=f) one-tail0.4827562326Step 4:Decision
rule:Reject the null hypothesis if the p-value is less than our
alpha of .05F Critical one-tail1.9837595685Step 5:Conduct the
test - place test function in cell K35P(F<=f) 2-
tail0.9655124652Step 6:Conclusion and InterpretationWhat is
the p-value:0.9655124652What is your decision: REJ or NOT
reject the null?Not RejectWhy?p value is greater than alpha
(0.05)What is your conclusion about the means in the
population for male and female salaries?Cocnclude that male
salary average is equal female salary averageThe means did not
equal at all . The males had a higher variance as well3Education
is often a factor in pay differences.Do employees with an
advanced degree (degree = 1) have higher average salaries?Use
Cell K60 for the Excel test outcome location.DegreeNote:
assume equal variance for the salaries in each degree for this
question.01aWhat is the data input ranged used for this
question:t-Test: Paired Two Sample for MeansSalary 0Salary
1N59:N83andO59:O8354.534.1b Does this question need a one
or two-tail hypothesis statement and test?one tailSalary
0Salary128.360.9Why:We are only looking at who has a degree
or notMean43.54446.22459.749.2c. Step 1:Ho:degree have
equal average
salariesVariance339.2742333333384.626941.874.1Ha:degree
have higher average salariesObservations252548.541.4Step
2:Significance (Alpha):0.05Pearson
Correlation0.104006890436.222.8Step 3:Test Statistic and
test:t-Test: Paired Two Sample for MeansHypothesized Mean
Difference035.573Why this test?because of the infomration in

7. usedf2422.223.3Step 4:Decision rule:Reject the null hypothesis
if the p-value is less than our alpha of .05t Stat-
0.526093969653.424.3Step 5:Conduct the test - place test
function in cell K60P(T<=t) one-tail0.301825506923.625t
Critical one-tail1.710882079922.322.6Step 6:Conclusion and
InterpretationP(T<=t) two-tail0.603651013774.463.1What is the
p-value:0.6036510137t Critical two-tail2.063898561675.623.9Is
the t value in the t-distribution tail indicated by the arrow in the
Ha claim?Yes47.578.928.157.6What is your decision: REJ or
NOT reject the null?Not Reject22.746.2Why?P value is greater
than 0.0524.422.9What is your conclusion about the impact of
education on average salaries?23.863.7conclude that degree
have equal average salaries64.626.937.324.4it was shown that
education had a good impart on salary with the degree mean
being higher than those without23.765.940.349.94Considering
both the compa-ratio information from the lectures and your
salary information, what conclusions can you reach about equal
pay for equal work?72.357.4Your findings:I can conclude that
different factors change the salaries for different people. Those
with degrees make more on average.66.256The lecture's related
findings:This just mean they are getting paid more because of
education and experience. 61.768.1Overall conclusion:Why -
what statistical results support this conclusion?The results from
question 3 supports my claims. More reseach needs to be done
to determine if it is really equal pay for equal work.Or is it that
they are getting rewarded for the education they have on top of
the experience, which plays a factot(sometimes).
Sheet1SalaryMaleFemael54.534.128.341.460.922.849.223.374.1
24.37341.859.72548.522.623.963.178.936.223.635.546.257.675.
622.247.553.428.122.363.774.426.922.964.622.723.724.440.323
.865.937.357.424.45672.366.249.961.768.1
Week 3Week 3: Identifying Significant Differences - part 2Data
Input Table:Salary Range GroupsGroup name:ABCDEFTo
Ensure full credit for each question, you need to show how you
got your results. This involves either showing where the data
you used is located List salaries within each

8. grade22.834.141.449.260.974.1or showing the excel formula in
each cell.Be sure to copy the appropriate data columns from the
data tab to the right for your use this
week.23.326.946.257.663.17324.349.963.778.91A good pay
program will have different average salaries by grade. Is this
the case for our company?2565.9aWhat is the data input ranged
used for this question:Use Cell K08 for the Excel test outcome
location.22.657.4Note: assume equal variances for each grade,
even though this may not be accurate, for purposes of this
question.Anova: Single Factor23.956b. Step 1:Ho:Salaries are
equal 22.968.1Ha:Salaries are not equalSUMMARY24.4Step
2:Significance
(Alpha):0.05GroupsCountSumAverageVarianceStep 3:Test
Statistic and test:single factor Analysis of variance
A8189.223.650.7685714286Why this test?It provides a critical
focus on the existing differences between groups which is being
evaluated in this case.B26130.525.92Step 4:Decision
rule:Reject Ho if p<0.05C287.643.811.52Step 5:Conduct the
test - place test function in cell
K08D3156.752.233333333321.7233333333E7435.162.15714285
7119.1195238095Step 6:Conclusion and
InterpretationF322675.33333333339.8433333333What is the p-
value:0What is your decision: REJ or NOT reject the
null?REJWhy?p<0.05ANOVAWhat is your conclusion about the
means in the population for grade salaries?The means in the
population for grade salaries are not equalSource of
VariationSSdfMSFP-valueF critBetween
Groups9010.375123809551802.0750247619155.160880882502.
7400575417Within
Groups220.67047619051911.614235589Total9231.0456242If
the null hypothesis in question 1 was rejected, which pairs of
means differ?(Use the values from the ANOVA table to
complete the follow table.)Groups ComparedMean Diff.T value
used+/- TermLowto HighDifference Significant?Why?A-B-
6.852.0150.0903175535-6.9403175535-6.7596824465YesZero
not in range A-C-20.152.0150.0903175535-20.2403175535-

9. 20.0596824465YesZero not in range A-D-
28.58333333332.0150.0755650867-28.6588984201-
28.5077682466YesZero not in range A-E-
38.50714285712.0150.0539750619-38.5611179191-
38.4531677952YesZero not in range A-F-
51.68333333332.0150.0755650867-51.7588984201-
51.6077682466YesZero not in range B-C-
13.32.0150.0903175535-13.3903175535-
13.2096824465YesZero not in range B-D-
21.73333333332.0150.0755650867-21.8088984201-
21.6577682466YesZero not in range B-E-
31.65714285712.0150.0539750619-31.7111179191-
31.6031677952YesZero not in range B-F-
44.83333333332.0150.0755650867-44.9088984201-
44.7577682466YesZero not in range C-D-
8.43333333332.0150.0755650867-8.5088984201-
8.3577682466YesZero not in range C-E-
18.35714285712.0150.0539750619-18.4111179191-
18.3031677952YesZero not in range C-F-
31.53333333332.0150.0755650867-31.6088984201-
31.4577682466YesZero not in range YesZero not in range D-E-
9.92380952382.0150.0539750619-9.9777845858-
9.8698344619YesZero not in range D-F-
23.12.0150.0755650867-23.1755650867-
23.0244349133YesZero not in range E-F-
13.17619047622.0150.0755650867-13.2517555629-
13.1006253895YesZero not in range 3One issue in salary is the
grade an employee is in - higher grades have higher
salaries.This suggests that one question to ask is if males and
females are distributed in a similar pattern across the salary
grades?aWhat is the data input ranged used for this
question:Use Cell K54 for the Excel test outcome
location.p2.06060360617515E-103b. Step 1:Ho:There is
strong association between grades and salariesHa:There is no
association between grades and salaries Step 2:Significance
(Alpha):0.05Step 3:Test Statistic and test:Chi-square test for

10. associationPlace the actual distribution in the table below.Why
this test?It provide a critical focus on the underlying association
between variablesABCDEFSumStep 4:Decision rule:Reject
p<0.05Male71.283.313596.7610.6301.61298.4Step 5:Conduct
the test - place test function in cell
K54Female280.7143.183.2160.9131.2146.7945.8Sum:351.9226.
4218.2257.6741.8448.32244.2Step 6:Conclusion and
InterpretationPlace the expected distribution in the table
below.What is the p-value:2.06060360617515E-
103ABCDEFWhat is your decision: REJ or NOT reject the
null?RejectMale203.5945815881130.9855449603126.24136886
2149.0365564566429.1743694858259.36757864721298.4Why?P
<0.05Female148.305418411995.414455039791.958631138108.5
634435434312.6256305142188.9324213528945.8What is your
conclusion about the means in the population for male and
female salaries?There is no association between grades and
salaries Sum:351.9226.4218.2257.6741.8448.32244.24What
implications do this week's analysis have for our equal pay
question?Your findings:There is difference in pay between male
and female employeesThe lecture's related findings:There is no
difference in pay among employeesOverall conclusion:Th
analysis conducted in this case highlight that there is
statistically significant difference in pay between
employeesWhy - what statistical results support this
conclusion?The analysis of variance and chi square analysis
support this conclusion
Week 4Week 4: Identifying relationships - correlations and
regressionSalaryMidpoint AgePerformance
RatingServiceRaiseDegreeGender122.823329095.81FTo Ensure
full credit for each question, you need to show how you got
your results. This involves either showing where the data you
used is located 23.323308074.71For showing the excel formula
in each cell.Be sure to copy the appropriate data columns from
the data tab to the right for your use this
week.24.32341100194.81F252332901261F1What is the
correlation between and among the interval/ratio level variables

11. with salary? (Do not include compa-ratio in this
question.)22.623328084.91Fa. Create the correlation table.Use
Cell K08 for the Excel test outcome
location.23.923328514.61Mi.What is the data input ranged used
for this question:22.923296043.91Fii. Create a correlation table
in cell K08.24.4233210085.71F34.131307553.61Fb.
Technically, we should perform a hypothesis testing on each
correlation to determine 26.931268024.91Mif it is significant or
not. However, we can be faithful to the process and save some
41.4403210085.71Ftime by finding the minimum correlation
that would result in a two tail rejection of the
null.46.240358073.91MWe can then compare each correlation to
this value, and those exceeding it (in either a
49.2483690165.71Mpositive or negative direction) can be
considered statistically significant.57.648486563.81F i. What
is the t-value we would use to cut off the two tails?T
=49.948369585.21F ii. What is the associated correlation
value related to this t-value? r
=60.95742100165.51M63.1572755331Fc. What variable(s)
is(are) significantly correlated to
salary?63.757359095.51M65.9574590165.21Md. Are there any
surprises - correlations you though would be significant and are
not, or non significant correlations you thought would
be?57.4573975203.91M5657379555.51Me. Why does or does
not this information help answer our equal pay
question?68.1573490115.31F74.1673670124.51M736749100104
1M2Perform a regression analysis using salary as the dependent
variable and all of the variables used in Q1. Add
the78.9674395136.31Mtwo dummy variables - gender and
education - to your list of independent variables. Show the
result, and interpret your findings by answering the following
questions.Suggestion: Add the dummy variables values to the
right of the last data columns used for Q1.What is the multiple
regression equation predicting/explaining salary using all of our
possible variables except compa-ratio?a.What is the data input
ranged used for this question:b. Step 1: State the appropriate

12. hypothesis statements:Use Cell M34 for the Excel test outcome
location.Ho:Ha:Step 2:Significance (Alpha):Step 3:Test
Statistic and test:Why this test?Step 4:Decision rule:Step
5:Conduct the test - place test function in cell M34Step
6:Conclusion and InterpretationWhat is the p-value:What is
your decision: REJ or NOT reject the null?Why?What is your
conclusion about the factors influencing the population salary
values?c.If we rejected the null hypothesis, we need to test the
significance of each of the variable coefficients.Step 1: State
the appropriate coefficient hypothesis statements:(Write a
single pair, we will use it for each variable
separately.)Ho:Ha:Step 2:Significance (Alpha):Step 3:Test
Statistic and test:Why this test?Step 4:Decision rule:Step
5:Conduct the testNote, in this case the test has been performed
and is part of the Regression output above.Step 6:Conclusion
and InterpretationPlace the t and p-values in the following
tableIdentify your decision on rejecting the null for each
variable. If you reject the null, place the coefficient in the
table.MidpointAgePerf. Rat.SeniorityRaiseGenderDegreet-
value:P-value:Rejection Decision:If Null is rejected, what is the
variable's coefficient value?Using the intercept coefficient and
only the significant variables, what is the equation?Salary =
d.Is gender a significant factor in salary?e.Regardless of
statistical significance, who gets paid more with all other things
being equal?f.How do we know? 3After considering the compa-
ratio based results in the lectures and your salary based results,
what else would you like to knowbefore answering our question
on equal pay? Why?4Between the lecture results and your
results, what is your answer to the questionof equal pay for
equal work for males and females? Why?Your findings:The
lecture's related findings:Overall conclusion:5What does
regression analysis show us about analyzing complex measures?