FOR TUTOR VIVIANE - Need help with week 2

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Data

See comments at the right of the data set.
ID Salary Compa Midpoint Age Performance Rating Service Gender Raise Degree Gender1 Grade Overall Ranked Salaries Rank Salaries F/M Z SCORE for ALL DATA SALARIES Z SCORE for ALL FEMALE SALARIES Z SCORE for ALL MALE SALARIES Z SCORE for ALL DATA COMPA Z SCORE for ALL FEMALE COMPA Z SCORE for ALL MALE COMPA TOP 1/3 Salaries - ALL DATA TOP 1/3 Salaries - Z Score - ALL DATA TOP 1/3 Salaries - ALL FEMALE TOP 1/3 Salaries - Z Score - ALL FEMALE TOP 1/3 Salaries - ALL MALE TOP 1/3 Salaries - Z Score - ALL MALE TOP 1/3 Compa - ALL DATA TOP 1/3 Compa - Z Score - ALL DATA TOP 1/3 Compa - ALL FEMALE TOP 1/3 Compa - Z Score - ALL FEMALE TOP 1/3 Compa - ALL MALE TOP 1/3 Compa - Z Score - ALL MALE
8 23 1.000 23 32 90 9 1 5.8 0 F A 45 20 -1.1457496095 -0.8199455817 -0.8132761837 -0.9769037476 1 77 1.6665448865 1 77 2.1318585125 1 77 1.4063598762 1 1.21 1.9202065081 1 1.21 2.0083958303 1 1.175 1.4173687977 The 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)?
10 22 0.956 23 30 80 7 1 4.7 0 F A 49 24 -1.1978291372 -0.8746086205 -1.3860058906 -1.6023950877 2 77 1.6665448865 2 75 2.0225324349 2 76 1.3501054811 2 1.187 1.6208250704 2 1.187 1.6814344479 2 1.157 1.2025436178 Note: to simplfy the analysis, we will assume that jobs within each grade comprise equal work.
11 23 1.000 23 41 100 19 1 4.8 0 F A 45 20 -1.1457496095 -0.8199455817 -0.8132761837 -0.9769037476 3 76 1.6144653588 3 69 1.6945542022 3 76 1.3501054811 3 1.175 1.4646260595 3 1.161 1.3118259288 3 1.157 1.2025436178
14 24 1.043 23 32 90 12 1 6 0 F A 37 14 -1.0936700818 -0.7652825429 -0.2535630611 -0.3656281197 4 76 1.6144653588 4 65 1.4759020471 4 72 1.1250879009 4 1.161 1.28239388 4 1.149 1.1412373815 4 1.149 1.1070657601 The column labels in the table mean:
15 24 1.043 23 32 80 8 1 4.9 0 F A 37 14 -1.0936700818 -0.7652825429 -0.2535630611 -0.3656281197 5 75 1.5623858311 5 57 1.0385977368 5 66 0.7875615306 5 1.157 1.230327543 5 1.145 1.0843745324 5 1.14 0.9996531702 ID – Employee sample number Salary – Salary in thousands
23 23 1.000 23 36 65 6 1 3.3 1 F A 45 20 -1.1457496095 -0.8199455817 -0.8132761837 -0.9769037476 6 72 1.406147248 6 55 0.9292716593 6 66 0.7875615306 6 1.157 1.230327543 6 1.14 1.013295971 6 1.134 0.9280447769 Age – Age in years Performance Rating – Appraisal rating (Employee evaluation score)
26 24 1.043 23 22 95 2 1 6.2 1 F A 37 14 -1.0936700818 -0.7652825429 -0.2535630611 -0.3656281197 7 69 1.2499086649 7 50 0.6559564654 7 65 0.7313071356 7 1.149 1.126194869 7 1.129 0.856923136 7 1.134 0.9280447769 Service – Years of service (rounded) Gender: 0 = male, 1 = female
31 24 1.043 23 29 60 4 1 3.9 0 F A 37 14 -1.0936700818 -0.7652825429 -0.2535630611 -0.3656281197 8 66 1.0936700818 8 42 0.2186521551 8 64 0.6750527406 8 1.149 1.126194869 8 1.119 0.7147660132 8 1.122 0.7848279904 Midpoint – salary grade midpoint Raise – percent of last raise
35 24 1.043 23 23 90 4 1 5.3 1 F A 37 14 -1.0936700818 -0.7652825429 -0.2535630611 -0.3656281197 9 66 1.0936700818 9 41 0.1639891163 9 62 0.5625439505 9 1.145 1.0741285321 9 1.096 0.3878046309 9 1.087 0.3671123629 Grade – job/pay grade Degree (0= BS\BA 1 = MS)
36 23 1.000 23 27 75 3 1 4.3 1 F A 45 20 -1.1457496095 -0.8199455817 -0.8132761837 -0.9769037476 10 65 1.0415905541 10 1.14 1.0090456108 Gender1 (Male or Female) Compa - salary divided by midpoint
37 22 0.956 23 22 95 2 1 6.2 1 F A 49 24 -1.1978291372 -0.8746086205 -1.3860058906 -1.6023950877 11 65 1.0415905541 11 1.14 1.0090456108
42 24 1.043 23 32 100 8 1 5.7 0 F A 37 14 -1.0936700818 -0.7652825429 -0.2535630611 -0.3656281197 12 64 0.9895110264 12 1.134 0.9309461053
3 34 1.096 31 30 75 5 1 3.6 0 F B 31 12 -0.5728748047 -0.2186521551 0.436315904 0.3878046309 13 62 0.885351971 13 1.134 0.9309461053
18 36 1.161 31 31 80 11 1 5.6 1 F B 29 10 -0.4687157493 -0.1093260776 1.28239388 1.3118259288 14 60 0.7811929155 14 1.129 0.8658631841
20 34 1.096 31 44 70 16 1 4.8 1 F B 31 12 -0.5728748047 -0.2186521551 0.436315904 0.3878046309 15 60 0.7811929155 15 1.122 0.7747470944
39 35 1.129 31 27 90 6 1 5.5 1 F B 30 11 -0.520795277 -0.1639891163 0.8658631841 0.856923136 16 60 0.7811929155 16 1.119 0.7356973416
7 41 1.025 40 32 100 8 1 5.7 0 F C 27 9 -0.2083181108 0.1639891163 -0.4878615776 -0.6215109407 17 58 0.6770338601 17 1.096 0.436315904
13 42 1.050 40 30 100 2 1 4.7 1 F C 26 8 -0.1562385831 0.2186521551 -0.1624469714 -0.2661181338
22 57 1.187 48 48 65 6 1 3.8 0 F D 18 5 0.6249543324 1.0385977368 1.6208250704 1.6814344479
24 50 1.041 48 30 75 9 1 3.8 1 F D 21 7 0.2603976385 0.6559564654 -0.2795962296 -0.3940595443
45 55 1.145 48 36 95 8 1 5.2 0 F D 20 6 0.520795277 0.9292716593 1.0741285321 1.0843745324
17 69 1.210 57 27 55 3 1 3 0 F E 7 3 1.2499086649 1.6945542022 1.9202065081 2.0083958303
48 65 1.140 57 34 90 11 1 5.3 1 F E 10 4 1.0415905541 1.4759020471 1.0090456108 1.013295971
28 75 1.119 67 44 95 9 1 4.4 1 F F 5 2 1.5623858311 2.0225324349 0.7356973416 0.7147660132
43 77 1.149 67 42 95 20 1 5.5 1 F F 1 1 1.6665448865 2.1318585125 1.126194869 1.1412373815
19 24 1.043 23 32 85 1 0 4.6 1 M A 37 24 -1.0936700818 -1.5751230613 -0.2535630611 -0.1580158545
25 24 1.043 23 41 70 4 0 4 0 M A 37 24 -1.0936700818 -1.5751230613 -0.2535630611 -0.1580158545
40 25 1.086 23 24 90 2 0 6.3 0 M A 36 23 -1.0415905541 -1.5188686663 0.3061500615 0.3551776307
2 27 0.870 31 52 80 7 0 3.9 0 M B 35 22 -0.9374314987 -1.4063598762 -2.5054321358 -2.2227245274
32 28 0.903 31 25 95 4 0 5.6 0 M B 33 20 -0.885351971 -1.3501054811 -2.0758848557 -1.8288783644
34 28 0.903 31 26 80 2 0 4.9 1 M B 33 20 -0.885351971 -1.3501054811 -2.0758848557 -1.8288783644
16 47 1.175 40 44 90 4 0 5.7 0 M C 23 16 0.1041590554 -0.2812719752 1.4646260595 1.4173687977
27 40 1.000 40 35 80 7 0 3.9 1 M C 28 19 -0.2603976385 -0.6750527406 -0.8132761837 -0.6712093397
41 43 1.075 40 25 80 5 0 4.3 0 M C 25 18 -0.1041590554 -0.5062895554 0.1629676348 0.2238955763
5 47 0.979 48 36 90 16 0 5.7 1 M D 23 16 0.1041590554 -0.2812719752 -1.0866244529 -0.9218387162
30 49 1.020 48 45 90 18 0 4.3 0 M D 22 15 0.2083181108 -0.1687631851 -0.5529444988 -0.4325146954
1 58 1.017 57 34 85 8 0 5.7 0 M E 17 13 0.6770338601 0.3375263703 -0.5919942515 -0.4683188921
4 66 1.157 57 42 100 16 0 5.5 1 M E 8 5 1.0936700818 0.7875615306 1.230327543 1.2025436178
12 60 1.052 57 52 95 22 0 4.5 0 M E 14 10 0.7811929155 0.4500351604 -0.1364138029 -0.0506032646
33 64 1.122 57 35 90 9 0 5.5 1 M E 12 8 0.9895110264 0.6750527406 0.7747470944 0.7848279904
38 56 0.982 57 45 95 11 0 4.5 0 M E 19 14 0.5728748047 0.2250175802 -1.0475747002 -0.8860345195
44 60 1.052 57 45 90 16 0 5.2 1 M E 14 10 0.7811929155 0.4500351604 -0.1364138029 -0.0506032646
46 65 1.140 57 39 75 20 0 3.9 1 M E 10 7 1.0415905541 0.7313071356 1.0090456108 0.9996531702
47 62 1.087 57 37 95 5 0 5.5 1 M E 13 9 0.885351971 0.5625439505 0.3191666457 0.3671123629
49 60 1.052 57 41 95 21 0 6.6 0 M E 14 10 0.7811929155 0.4500351604 -0.1364138029 -0.0506032646
50 66 1.157 57 38 80 12 0 4.6 0 M E 8 5 1.0936700818 0.7875615306 1.230327543 1.2025436178
6 76 1.134 67 36 70 12 0 4.5 1 M F 3 2 1.6144653588 1.3501054811 0.9309461053 0.9280447769
9 77 1.149 67 49 100 10 0 4 1 M F 1 1 1.6665448865 1.4063598762 1.126194869 1.1070657601
21 76 1.134 67 43 95 13 0 6.3 1 M F 3 2 1.6144653588 1.3501054811 0.9309461053 0.9280447769
29 72 1.074 67 52 95 5 0 5.4 0 M F 6 4 1.406147248 1.1250879009 0.1499510505 0.2119608441

Week 1

Week 1. Measurement and Description - chapters 1 and 2
1 Measurement issues. Data, even numerically coded variables, can be one of 4 levels -
nominal, ordinal, interval, or ratio. It is important to identify which level a variable is, as
this impact the kind of analysis we can do with the data. For example, descriptive statistics
such as means can only be done on interval or ratio level data.
Please list under each label, the variables in our data set that belong in each group.
Nominal Ordinal Interval Ratio
Gender Grade I.D. Salary
Degree I.D. Age Service
Gender 1 Performance Rating Age
Grade
b. For each variable that you did not call ratio, why did you make that decision?
Everything in the Nominal Data category are that of non-numeric group lables. Ordinal data such as grade and I.D. can be ranked in some numerical way to provide meaning. Interval data shows a difference between two values that are meaningful such as 1 to 10 and 50 to 60. Ratio data must be a positive number and if the number reaches zero then it ceases to exsist as a ratio.
2 The first step in analyzing data sets is to find some summary descriptive statistics for key variables.
For salary, compa, age, performance rating, and service; find the mean, standard deviation, and range for 3 groups: overall sample, Females, and Males.
You can use either the Data Analysis Descriptive Statistics tool or the Fx =average and =stdev functions.
(the range must be found using the difference between the =max and =min functions with Fx) functions.
Note: Place data to the right, if you use Descriptive statistics, place that to the right as well.
Salary Compa Age Perf. Rat. Service
Overall Mean 45.000 1.062 35.720 85.900 8.960
Standard Deviation 19.201 0.077 8.251 11.415 5.718
Range 55.000 0.340 30.000 45.000 21.000
Female Mean 38.000 1.069 32.520 84.200 7.920
Standard Deviation 18.294 0.070 6.881 13.592 4.907
Range 55.000 0.254 26.000 45.000 18.000
Male Mean 52.000 1.056 38.920 87.600 10.000
Standard Deviation 17.776 0.084 8.386 8.675 6.357
Range 53.000 0.305 28.000 30.000 21.000
3 What is the probability for a: Probability
a.       Randomly selected person being a male in grade E? 20.00%
b.      Randomly selected male being in grade E? 40.00%
Note part b is the same as given a female, what is probabilty of being in grade E? 8.00%
c. Why are the results different?
The results are different because in part "A" we are taking a sample from the entire group of 50 individuals whereas, in parts "B" and "C" we narrow our field to 25 individuals in each ratio.
4 For each group (overall, females, and males) find: Overall Female Male
a. The value that cuts off the top 1/3 salary in each group. 58 41 62
b. The z score for each value: 0.67703386 0.163989116 0.56254395
c. The normal curve probability of exceeding this score: 24.92% 43.49% 28.69%
d. What is the empirical probability of being at or exceeding this salary value? 25.08% 6.51% 21.31%
e. The value that cuts off the top 1/3 compa in each group. 1.096 1.096 1.087
f. The z score for each value: 0.4363159 0.387804631 0.36711236
g. The normal curve probability of exceeding this score: 33.13% 34.91% 35.68%
h. What is the empirical probability of being at or exceeding this compa value? 16.87% 15.09% 14.32%
i. How do you interpret the relationship between the data sets? What do they mean about our equal pay for equal work question?
I interperate the relationship between the data sets as such; the female group has a lower salary mean that that of the male group and the salary mean of the overall group is 7 points larger that the female group and 7 points smaller than the male group.
5.      What conclusions can you make about the issue of male and female pay equality? Are all of the results consistent?
The conclusions I can make about the issue of male and female pay equality is that there is more consistancy of males earning more money in the top 1/3 percentile than there are women in the same catagory. Also, the data has a skewed distribution.
What is the difference between the sal and compa measures of pay?
The difference between the salary and compa measures of pay are such that the overall salary mean of 45, the mean for females is 38 and the mean for males is 52 are the midpoint within the industry as where the standard deviation of 19.201 for the overall group, 18.294 for the female group and 17.776 for the male group indicate how much deviation there is from the mean of their respective groups.
Conclusions from looking at salary results:
Conclusions from looking at salary results are that the range of all three groups are relatively close to one another. There are no major disparaties with one group over the next which indicates a normal distorbution.
Conclusions from looking at compa results:
The conclusion from looking at the compa results are that with the overall, female and male groups the mean indicates the groups are being paid above the midpoint within the industry.
Do both salary measures show the same results?
Althought the numbers are not identical I believe that the numbers do show the same results but in a different format.
Can we make any conclusions about equal pay for equal work yet?
Yes, we can make conclusions about equal pay for equal work.

Week 2

Week 2 Testing means Q3
In questions 2 and 3, be sure to include the null and alternate hypotheses you will be testing. Ho Female Male Female
In the first 3 questions use alpha = 0.05 in making your decisions on rejecting or not rejecting the null hypothesis. 45 34 1.017 1.096
45 41 0.870 1.025
1 Below are 2 one-sample t-tests comparing male and female average salaries to the overall sample mean. 45 23 1.157 1.000
(Note: a one-sample t-test in Excel can be performed by selecting the 2-sample unequal variance t-test and making the second variable = Ho value -- see column S) 45 22 0.979 0.956
Based on our sample, how do you interpret the results and what do these results suggest about the population means for male and female average salaries? 45 23 1.134 1.000
Males Females 45 42 1.149 1.050
Ho: Mean salary = 45 Ho: Mean salary = 45 45 24 1.052 1.043
Ha: Mean salary =/= 45 Ha: Mean salary =/= 45 45 24 1.175 1.043
Based on the sample, I interpret the results as being homogeneous and the results suggest that male and female salaries are equal to one another.
45 69 1.043 1.210
Note: While the results both below are actually from Excel's t-Test: Two-Sample Assuming Unequal Variances, 45 36 1.134 1.161
having no variance in the Ho variable makes the calculations default to the one-sample t-test outcome - we are tricking Excel into doing a one sample test for us. 45 34 1.043 1.096
Male Ho Female Ho 45 57 1.000 1.187
Mean 52 45 Mean 38 45 45 23 1.074 1.000
Variance 316 0 Variance 334.6666666667 0 45 50 1.020 1.041
Observations 25 25 Observations 25 25 45 24 0.903 1.043
Hypothesized Mean Difference 0 Hypothesized Mean Difference 0 45 75 1.122 1.119
df 24 df 24 45 24 0.903 1.043
t Stat 1.9689038266 t Stat -1.9132063573 45 24 0.982 1.043
P(T<=t) one-tail 0.0303078503 P(T<=t) one-tail 0.0338621184 45 23 1.086 1.000
t Critical one-tail 1.7108820799 t Critical one-tail 1.7108820799 45 22 1.075 0.956
P(T<=t) two-tail 0.0606157006 P(T<=t) two-tail 0.0677242369 45 35 1.052 1.129
t Critical two-tail 2.0638985616 t Critical two-tail 2.0638985616 45 24 1.140 1.043
Conclusion: Do not reject Ho; mean equals 45 Conclusion: Do not reject Ho; mean equals 45 45 77 1.087 1.149
Is this a 1 or 2 tail test? A 1 Tail Test Is this a 1 or 2 tail test?
- why? Because the t Stat is 1.968903827 - why?
P-value is: P-value is: 45 55 1.052 1.145
Is P-value > 0.05? Is P-value > 0.05? 45 65 1.157 1.140
Why do we not reject Ho? Why do we not reject Ho?
Interpretation:
2 Based on our sample data set, perform a 2-sample t-test to see if the population male and female average salaries could be equal to each other.
(Since we have not yet covered testing for variance equality, assume the data sets have statistically equal variances.)
Ho:
Ha:
Test to use:
Place B43 in Outcome range box.
P-value is:
Is P-value < 0.05?
Reject or do not reject Ho:
If the null hypothesis was rejected, what is the effect size value:
Meaning of effect size measure:
Interpretation:
b. Since the one and two tail t-test results provided different outcomes, which is the proper/correct apporach to comparing salary equality? Why?
3 Based on our sample data set, can the male and female compas in the population be equal to each other? (Another 2-sample t-test.)
Ho:
Ha:
Statistical test to use:
Place B75 in Outcome range box.
What is the p-value:
Is P-value < 0.05?
Reject or do not reject Ho:
If the null hypothesis was rejected, what is the effect size value:
Meaning of effect size measure:
Interpretation:
4 Since performance is often a factor in pay levels, is the average Performance Rating the same for both genders?
Ho:
Ha:
Test to use:
Place B106 in Outcome range box.
What is the p-value:
Is P-value < 0.05?
Do we REJ or Not reject the null?
If the null hypothesis was rejected, what is the effect size value:
Meaning of effect size measure:
Interpretation:
5 If the salary and compa mean tests in questions 2 and 3 provide different results about male and female salary equality,
which would be more appropriate to use in answering the question about salary equity? Why?
What are your conclusions about equal pay at this point?

Week 3

Week 3
At this point we know the following about male and female salaries.
a. Male and female overall average salaries are not equal in the population.
b. Male and female overall average compas are equal in the population, but males are a bit more spread out.
c. The male and female salary range are almost the same, as is their age and service.
d. Average performance ratings per gender are equal.
Let's look at some other factors that might influence pay - education(degree) and performance ratings.
1 Last week, we found that average performance ratings do not differ between males and females in the population.
Now we need to see if they differ among the grades. Is the average performace rating the same for all grades?
(Assume variances are equal across the grades for this ANOVA.) A B C D E F
Null Hypothesis:
Alt. Hypothesis:
Place B17 in Outcome range box.
Interpretation:
What is the p-value:
Is P-value < 0.05?
Do we REJ or Not reject the null?
If the null hypothesis was rejected, what is the effect size value (eta squared):
Meaning of effect size measure:
What does that decision mean in terms of our equal pay question:
2 While it appears that average salaries per each grade differ, we need to test this assumption.
Is the average salary the same for each of the grade levels? (Assume equal variance, and use the analysis toolpak function ANOVA.)
Use the input table to the right to list salaries under each grade level.
Null Hypothesis:
Alt. Hypothesis: A B C D E F
Place B55 in Outcome range box.
What is the p-value:
Is P-value < 0.05?
Do you reject or not reject the null hypothesis:
If the null hypothesis was rejected, what is the effect size value (eta squared):
Meaning of effect size measure:
Interpretation:
3 The table and analysis below demonstrate a 2-way ANOVA with replication. Please interpret the results.
BA MA Ho: Average compas by gender are equal
Male 1.017 1.157 Ha: Average compas by gender are not equal
0.870 0.979 Ho: Average compas are equal for each degree
1.052 1.134 Ho: Average compas are not equal for each degree
1.175 1.149 Ho: Interaction is not significant
1.043 1.043 Ha: Interaction is significant
1.074 1.134
1.020 1.000 Perform analysis:
0.903 1.122
0.982 0.903 Anova: Two-Factor With Replication
1.086 1.052
1.075 1.140 SUMMARY BA MA Total
1.052 1.087 Male
Female 1.096 1.050 Count 12 12 24
1.025 1.161 Sum 12.349 12.9 25.249
1.000 1.096 Average 1.0290833333 1.075 1.0520416667
0.956 1.000 Variance 0.006686447 0.0065198182 0.0068660417
1.000 1.041
1.043 1.043 Female
1.043 1.119 Count 12 12 24
1.210 1.043 Sum 12.791 12.787 25.578
1.187 1.000 Average 1.0659166667 1.0655833333 1.06575
1.043 0.956 Variance 0.006102447 0.0042128106 0.004933413
1.043 1.129
1.145 1.149 Total
Count 24 24
Sum 25.14 25.687
Average 1.0475 1.0702916667
Variance 0.0064703478 0.0051561286
ANOVA
Source of Variation SS df MS F P-value F crit
Sample 0.0022550208 1 0.0022550208 0.3834821171 0.5389389507 4.0617064601 (This is the row variable or gender.)
Columns 0.0062335208 1 0.0062335208 1.0600539609 0.3088295633 4.0617064601 (This is the column variable or Degree.)
Interaction 0.0064171875 1 0.0064171875 1.0912877664 0.3018915062 4.0617064601
Within 0.25873675 44 0.0058803807
Total 0.2736424792 47
Interpretation:
For Ho: Average compas by gender are equal Ha: Average compas by gender are not equal
What is the p-value:
Is P-value < 0.05?
Do you reject or not reject the null hypothesis:
If the null hypothesis was rejected, what is the effect size value (eta squared):
Meaning of effect size measure:
For Ho: Average salaries are equal for all grades Ha: Average salaries are not equal for all grades
What is the p-value:
Is P-value < 0.05?
Do you reject or not reject the null hypothesis:
If the null hypothesis was rejected, what is the effect size value (eta squared):
Meaning of effect size measure:
For: Ho: Interaction is not significant Ha: Interaction is significant
What is the p-value:
Do you reject or not reject the null hypothesis:
If the null hypothesis was rejected, what is the effect size value (eta squared):
Meaning of effect size measure:
What do these decisions mean in terms of our equal pay question:
4 Many companies consider the grade midpoint to be the "market rate" - what is needed to hire a new employee. Midpoint Salary
Does the company, on average, pay its existing employees at or above the market rate?
Null Hypothesis:
Alt. Hypothesis:
Statistical test to use:
Place the cursor in B160 for correl.
What is the p-value:
Is P-value < 0.05?
Do we REJ or Not reject the null?
If the null hypothesis was rejected, what is the effect size value: Since the effect size was not discussed in this chapter, we do not have a formula for it - it differs from the non-paired t.
Meaning of effect size measure: NA
Interpretation:
5.   Using the results up thru this week, what are your conclusions about gender equal pay for equal work at this point?

Week 4

Week 4 Confidence Intervals and Chi Square (Chs 11 - 12)
For questions 3 and 4 below, be sure to list the null and alternate hypothesis statements. Use .05 for your significance level in making your decisions.
For full credit, you need to also show the statistical outcomes - either the Excel test result or the calculations you performed.
1 Using our sample data, construct a 95% confidence interval for the population's mean salary for each gender.
Interpret the results. How do they compare with the findings in the week 2 one sample t-test outcomes (Question 1)?
Mean St error t value Low to High
Males
Females
<Reminder: standard error is the sample standard deviation divided by the square root of the sample size.>
Interpretation:
2 Using our sample data, construct a 95% confidence interval for the mean salary difference between the genders in the population.
How does this compare to the findings in week 2, question 2?
Difference St Err. T value Low to High
Yes/No
Can the means be equal? Why?
How does this compare to the week 2, question 2 result (2 sampe t-test)?
a. Why is using a two sample tool (t-test, confidence interval) a better choice than using 2 one-sample techniques when comparing two samples?
3 We found last week that the degrees compa values within the population.
do not impact compa rates. This does not mean that degrees are distributed evenly across the grades and genders.
Do males and females have athe same distribution of degrees by grade?
(Note: while technically the sample size might not be large enough to perform this test, ignore this limitation for this exercise.)
What are the hypothesis statements:
Ho:
Ha:
Note: You can either use the Excel Chi-related functions or do the calculations manually.
Data input tables - graduate degrees by gender and grade level
OBSERVED A B C D E F Total Do manual calculations per cell here (if desired)
M Grad A B C D E F
Fem Grad M Grad
Male Und Fem Grad
Female Und Male Und
Female Und
Sum =
EXPECTED
M Grad For this exercise - ignore the requirement for a correction
Fem Grad for expected values less than 5.
Male Und
Female Und
Interpretation:
What is the value of the chi square statistic:
What is the p-value associated with this value:
Is the p-value <0.05?
Do you reject or not reject the null hypothesis:
If you rejected the null, what is the Cramer's V correlation:
What does this correlation mean?
What does this decision mean for our equal pay question:
4 Based on our sample data, can we conclude that males and females are distributed across grades in a similar pattern
within the population?
What are the hypothesis statements:
Ho:
Ha:
Do manual calculations per cell here (if desired)
A B C D E F A B C D E F
OBS COUNT - m M
OBS COUNT - f F
Sum =
EXPECTED
What is the value of the chi square statistic:
What is the p-value associated with this value:
Is the p-value <0.05?
Do you reject or not reject the null hypothesis:
If you rejected the null, what is the Phi correlation:
What does this correlation mean?
What does this decision mean for our equal pay question:
5.      How do you interpret these results in light of our question about equal pay for equal work?

Week 5

Week 5 Correlation and Regression
1.     Create a correlation table for the variables in our data set. (Use analysis ToolPak or StatPlus:mac LE function Correlation.)
a. Reviewing the data levels from week 1, what variables can be used in a Pearson's Correlation table (which is what Excel produces)?
b. Place table here (C8 in Output range box):
c. Using r = approximately .28 as the signicant r value (at p = 0.05) for a correlation between 50 values, what variables are
significantly related to Salary?
To compa?
d. Looking at the above correlations - both significant or not - are there any surprises -by that I
mean any relationships you expected to be meaningful and are not and vice-versa?
e. Does this help us answer our equal pay for equal work question?
2 Below is a regression analysis for salary being predicted/explained by the other variables in our sample (Midpoint,
age, performance rating, service, gender, and degree variables. (Note: since salary and compa are different ways of
expressing an employee’s salary, we do not want to have both used in the same regression.)
Plase interpret the findings.
Ho: The regression equation is not significant.
Ha: The regression equation is significant.
Ho: The regression coefficient for each variable is not significant Note: technically we have one for each input variable.
Ha: The regression coefficient for each variable is significant Listing it this way to save space.
Sal
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.9915590747
R Square 0.9831893985
Adjusted R Square 0.9808437332
Standard Error 2.6575925726
Observations 50
ANOVA
df SS MS F Significance F
Regression 6 17762.2996738743 2960.383278979 419.1516111294 1.8121523852609E-36
Residual 43 303.7003261257 7.062798282
Total 49 18066
Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept -1.7496212123 3.6183676583 -0.4835388157 0.6311664899 -9.0467550427 5.547512618 -9.0467550427 5.547512618
Midpoint 1.2167010505 0.0319023509 38.1382881163 8.66416336978111E-35 1.1523638283 1.2810382727 1.1523638283 1.2810382727
Age -0.0046280102 0.065197212 -0.0709847876 0.9437389875 -0.1361107191 0.1268546987 -0.1361107191 0.1268546987
Performace Rating -0.0565964405 0.0344950678 -1.6407110971 0.1081531819 -0.1261623747 0.0129694936 -0.1261623747 0.0129694936
Service -0.0425003573 0.0843369821 -0.5039350033 0.6168793519 -0.2125820912 0.1275813765 -0.2125820912 0.1275813765
Gender 2.420337212 0.8608443176 2.8115852804 0.0073966188 0.684279192 4.156395232 0.684279192 4.156395232
Degree 0.2755334143 0.7998023048 0.3445019009 0.732148119 -1.3374216547 1.8884884833 -1.3374216547 1.8884884833
Note: since Gender and Degree are expressed as 0 and 1, they are considered dummy variables and can be used in a multiple regression equation.
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?
Do you reject or not reject the null hypothesis:
What does this decision mean for our equal pay question:
For each of the coefficients: Intercept Midpoint Age Perf. Rat. Service Gender Degree
What 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 only the significant variables, what is the equation? Salary =
Is gender a significant factor in salary:
If so, who gets paid more with all other things being equal?
How do we know?
3 Perform a regression analysis using compa as the dependent variable and the same independent
variables as used in question 2. Show the result, and interpret your findings by answering the same questions.
Note: be sure to include the appropriate hypothesis statements.
Regression hypotheses
Ho:
Ha:
Coefficient hypotheses (one to stand for all the separate variables)
Ho:
Ha:
Put C94 in output range 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?
Do you reject or not reject the null hypothesis:
What does this decision mean for our equal pay question:
For each of the coefficients: Intercept Midpoint Age Perf. Rat. Service Gender Degree
What 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 only the significant variables, what is the equation? Compa =
Is gender a significant factor in compa:
If so, who gets paid more with all other things being equal?
How do we know?
4 Based on all of your results to date, do we have an answer to the question of are males and females paid equally for equal work?
If so, which gender gets paid more?
How do we know?
Which is the best variable to use in analyzing pay practices - salary or compa? Why?
What is most interesting or surprising about the results we got doing the analysis during the last 5 weeks?
5 Why did the single factor tests and analysis (such as t and single factor ANOVA tests on salary equality) not provide a complete answer to our salary equality question?
What outcomes in your life or work might benefit from a multiple regression examination rather than a simpler one variable test?

Sheet1

dollar sales population mean = 305
235 sample mean = 340.1
340 standard deviation = 73.196
228 n = 10
430 squar root of n = 3.1622776602
378 standard error of the mean = 23.1466075614
394 t = 1.5164209229
285 critical value of t = 2.262
312
374
425
Variable #1 (Var1)
Count 10 Skewness -0.3496
Mean 340.1 Skewness Standard Error 0.61451
Mean LCL 274.79328 Kurtosis 1.80154
Mean UCL 405.40672 Kurtosis Standard Error 0.92244
Variance 5,357.65556 Alternative Skewness (Fisher's) -0.41458
Standard Deviation 73.19601 Alternative Kurtosis (Fisher's) -1.15443
Mean Standard Error 23.14661 Coefficient of Variation 0.21522
Minimum 228 Mean Deviation 60.1
Maximum 430 Second Moment 4,821.89
Range 202 Third Moment -117,058.068
Sum 3,401 Fourth Moment 41,886,833.4137
Sum Standard Error 231.4661 Median 357
Total Sum Squares 1,204,899 Median Error 9.17376
Adjusted Sum Squares 48,218.9 Percentile 25% (Q1) 298.5
Geometric Mean 332.445 Percentile 75% (Q2) 409.5
Harmonic Mean 324.33568 IQR 111
Mode #N/A MAD (Median Absolute Deviation) 56.5
Coefficient of Dispersion (COD) 0.16835