Stats Homework - Correlation and Regression

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stats_week_5_problems_v1_0402.xlsx
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Data Sheet

See comments at the right of the data set.
ID Salary Compa Midpoint Age Performance Rating Service Gender Raise Degree Gender1 Grade
8 23 1.000 23 32 90 9 1 5.8 0 F A 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 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
14 24 1.043 23 32 90 12 1 6 0 F A The column labels in the table mean:
15 24 1.043 23 32 80 8 1 4.9 0 F A ID – Employee sample number Salary – Salary in thousands
23 23 1.000 23 36 65 6 1 3.3 1 F A 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 Service – Years of service (rounded) Gender: 0 = male, 1 = female
31 24 1.043 23 29 60 4 1 3.9 0 F A Midpoint – salary grade midpoint Raise – percent of last raise
35 24 1.043 23 23 90 4 1 5.3 1 F A Grade – job/pay grade Degree (0= BS\BA 1 = MS)
36 23 1.000 23 27 75 3 1 4.3 1 F A Gender1 (Male or Female) Compa - salary divided by midpoint
37 22 0.956 23 22 95 2 1 6.2 1 F A
42 24 1.043 23 32 100 8 1 5.7 0 F A
3 34 1.096 31 30 75 5 1 3.6 0 F B
18 36 1.161 31 31 80 11 1 5.6 1 F B
20 34 1.096 31 44 70 16 1 4.8 1 F B
39 35 1.129 31 27 90 6 1 5.5 1 F B
7 41 1.025 40 32 100 8 1 5.7 0 F C
13 42 1.050 40 30 100 2 1 4.7 1 F C
22 57 1.187 48 48 65 6 1 3.8 0 F D
24 50 1.041 48 30 75 9 1 3.8 1 F D
45 55 1.145 48 36 95 8 1 5.2 0 F D
17 69 1.210 57 27 55 3 1 3 0 F E
48 65 1.140 57 34 90 11 1 5.3 1 F E
28 75 1.119 67 44 95 9 1 4.4 1 F F
43 77 1.149 67 42 95 20 1 5.5 1 F F
19 24 1.043 23 32 85 1 0 4.6 1 M A
25 24 1.043 23 41 70 4 0 4 0 M A
40 25 1.086 23 24 90 2 0 6.3 0 M A
2 27 0.870 31 52 80 7 0 3.9 0 M B
32 28 0.903 31 25 95 4 0 5.6 0 M B
34 28 0.903 31 26 80 2 0 4.9 1 M B
16 47 1.175 40 44 90 4 0 5.7 0 M C
27 40 1.000 40 35 80 7 0 3.9 1 M C
41 43 1.075 40 25 80 5 0 4.3 0 M C
5 47 0.979 48 36 90 16 0 5.7 1 M D
30 49 1.020 48 45 90 18 0 4.3 0 M D
1 58 1.017 57 34 85 8 0 5.7 0 M E
4 66 1.157 57 42 100 16 0 5.5 1 M E
12 60 1.052 57 52 95 22 0 4.5 0 M E
33 64 1.122 57 35 90 9 0 5.5 1 M E
38 56 0.982 57 45 95 11 0 4.5 0 M E
44 60 1.052 57 45 90 16 0 5.2 1 M E
46 65 1.140 57 39 75 20 0 3.9 1 M E
47 62 1.087 57 37 95 5 0 5.5 1 M E
49 60 1.052 57 41 95 21 0 6.6 0 M E
50 66 1.157 57 38 80 12 0 4.6 0 M E
6 76 1.134 67 36 70 12 0 4.5 1 M F
9 77 1.149 67 49 100 10 0 4 1 M F
21 76 1.134 67 43 95 13 0 6.3 1 M F
29 72 1.074 67 52 95 5 0 5.4 0 M F

Week 1

Score: Week 1. Measurement and Description - chapters 1 and 2
<1 point> 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 Degree Compa Salary
Gender1 Grade Midpoint Age
Performance rating Service
Raise
b. For each variable that you did not call ratio, why did you make that decision?
Performance rating is a scaled measure and there is no absolute zero there and hence it is an interval scale variable.
Compa is interval because it is scaled by dividing it with a midpint of an interval variable
Midpoint is an interval variable because it is the midpoint of an interval and hence the difference between any two values is not absolute
Degree is ordinal, because 0 represents degree and 1 represents MS
Gender is a nomial variable because it represents just naming of males and females. There is no mathematical relation between 0 and 1
Gender1 is a nomial variable because it represents just naming of males and females. There is no mathematical relation between F andM
<1 point> 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. Compa
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 45.0 1.06248 35.7 85.9 9.0
Standard Deviation 19.2014 0.0768 8.2513 11.4147 5.7177
Range 55 0.3400 30 45 21
Female Mean 38.0 1.06872 32.5 84.2 7.9
Standard Deviation 18.294 0.0703 6.881 13.592 4.907
Range 55 0.254 26 45 18
Male Mean 52.0 1.05624 38.9 87.6 10.0
Standard Deviation 17.776 0.0837890605 8.386 8.675 6.357
Range 53 0.305 28 30 21
<1 point> 3 What is the probability for a: Probability
a.       Randomly selected person being a male in grade E? 0.2 10 50
b.      Randomly selected male being in grade E? 0.4 10 25
Note part b is the same as given a male, what is probabilty of being in grade E?
c. Why are the results different?
In the first case we considered the probability to get a male with E grade among the entire population. But in case b) we considered the probability of getting E among the males. Hence, the results are different
<1 point> 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. 57.6666666667 41 62 Hint: can use these Fx functions
b. The z score for each value: 0.6596740176 0.1639891163 0.5625439505 Excel's standize function
c. The normal curve probability of exceeding this score: 0.2547315219 0.4348698559 0.2868727339 1-normsdist function
d. What is the empirical probability of being at or exceeding this salary value? 0.3333333333 0.3333333333 0.3333333333
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: 0.436315904 0.3878046309 0.3671123629
g. The normal curve probability of exceeding this score: 0.3313037722 0.3490803091 0.356767603
h. What is the empirical probability of being at or exceeding this compa value? 0.3333333333 0.3333333333 0.3333333333
i. How do you interpret the relationship between the data sets? What do they mean about our equal pay for equal work question?
Table shows that the value that cuts off top (1/3) salary for females is 41 and that of males is 62. To compare the two values we calculte the Z scores of these two values.
When we calculate Z score, we can see that the Z score for the females is 0.164 and that of males is 0.563. Therefore , we can conclude that males are getting better salery than females
So, we conclude that there is a pay difference among males and females
<2 points> 5.      What conclusions can you make about the issue of male and female pay equality? Are all of the results consistent?
What is the difference between the sal and compa measures of pay?
The above calculations shows that the Z score for males salary is hgher than females. But the Z score for the compa salary is higher females than males
Conclusions from looking at salary results:
The salary shows that mean salary for males is higher than females. The z score for the top (1/3) is higher for males than females
Conclusions from looking at compa results:
But the compa salary shows that mean salary for females is higher than females. The z score for the top (1/3) is also higher for females than males
Do both salary measures show the same results?
No. The two groups shows different results as far as salary and compa are concerned.
Can we make any conclusions about equal pay for equal work yet?
Since the compa is a scaled variable, we can use the average salary to take a conlusions. It shows that the average salary is higher for males and the probability of a male in the top (1/3) group is higher than females
So there is an evidence of unequal pay for equal work

Week 2

Week 2 Testing means Q3 Salary Compa Performance Rating
In questions 2 and 3, be sure to include the null and alternate hypotheses you will be testing. Ho Female Male Female Female Male Female Male Female Male
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 23 24 1.000 1.043 90 85
45 41 0.870 1.025 22 24 0.956 1.043 80 70
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 23 25 1.000 1.086 100 90
(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 24 27 1.043 0.870 90 80
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 24 28 1.043 0.903 80 95
Males Females 45 42 1.149 1.050 23 28 1.000 0.903 65 80
Ho: Mean salary = 45 Ho: Mean salary = 45 45 24 1.052 1.043 24 47 1.043 1.175 95 90
Ha: Mean salary =/= 45 Ha: Mean salary =/= 45 45 24 1.175 1.043 24 40 1.043 1.000 60 80
45 69 1.043 1.210 24 43 1.043 1.075 90 80
Note: While the results both below are actually from Excel's t-Test: Two-Sample Assuming Unequal Variances, 45 36 1.134 1.161 23 47 1.000 0.979 75 90
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 22 49 0.956 1.020 95 90
Male Ho Female Ho 45 57 1.000 1.187 24 58 1.043 1.017 100 85
Mean 52 45 Mean 38 45 45 23 1.074 1.000 34 66 1.096 1.157 75 100
Variance 316 0 Variance 334.6666666667 0 45 50 1.020 1.041 36 60 1.161 1.052 80 95
Observations 25 25 Observations 25 25 45 24 0.903 1.043 34 64 1.096 1.122 70 90
Hypothesized Mean Difference 0 Hypothesized Mean Difference 0 45 75 1.122 1.119 35 56 1.129 0.982 90 95
df 24 df 24 45 24 0.903 1.043 41 60 1.025 1.052 100 90
t Stat 1.9689038266 t Stat -1.9132063573 45 24 0.982 1.043 42 65 1.050 1.140 100 75
P(T<=t) one-tail 0.0303078503 P(T<=t) one-tail 0.0338621184 45 23 1.086 1.000 57 62 1.187 1.087 65 95
t Critical one-tail 1.7108820799 t Critical one-tail 1.7108820799 45 22 1.075 0.956 50 60 1.041 1.052 75 95
P(T<=t) two-tail 0.0606157006 P(T<=t) two-tail 0.0677242369 45 35 1.052 1.129 55 66 1.145 1.157 95 80
t Critical two-tail 2.0638985616 t Critical two-tail 2.0638985616 45 24 1.140 1.043 69 76 1.210 1.134 55 70
Conclusion: Do not reject Ho; mean equals 45 Conclusion: Do not reject Ho; mean equals 45 45 77 1.087 1.149 65 77 1.140 1.149 90 100
Is this a 1 or 2 tail test? 2 tail test Is this a 1 or 2 tail test? 2 tail test 75 76 1.119 1.134 95 95
- why? Ha: µ ≠ 45 - why? Ha: µ ≠ 45 77 72 1.149 1.074 95 95
P-value is: 0.0606157006 P-value is: 0.0677242369 45 55 1.052 1.145
Is P-value > 0.05? Yes Is P-value > 0.05? Yes 45 65 1.157 1.140
Why do we not reject Ho? P-value > 0.05 Why do we not reject Ho? P-value > 0.05
As the P-value is greater than the significance level, α = 0.05, we do not reject the Ho. The P-value gives the probability of rejecting a true null hypothesis.
Interpretation: There is no evidence to suggest, at 95% level of confidence, that the mean salary of the male employees is significantly different from the mean salary of the population, which is 45 thousands.
There is no evidence to suggest, at 95% level of confidence, that the mean salary of the female employees is significantly different from the mean salary of the population, which is 45 thousands.
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: Mean salary of male employees = Mean salary of female employees
Ha: Mean salary of male employees ≠ Mean salary of female employees
Statistical test to use: Two-Sample t-Test Assuming Equal Variances
t-Test: Two-Sample Assuming Equal Variances
Male Female
Mean 52 38
Variance 316 334.6666666667
Observations 25 25
Pooled Variance 325.3333333333
Hypothesized Mean Difference 0
df 48
t Stat 2.7442189608
P(T<=t) one-tail 0.0042530089
t Critical one-tail 1.6772241966
P(T<=t) two-tail 0.0085060177
t Critical two-tail 2.0106347219
P-value is: 0.0085060177
Is P-value < 0.05? Yes
Reject or do not reject Ho: Reject Ho
If the null hypothesis was rejected, what is the effect size value: 0.7761823345
Meaning of effect size measure: The effect size is large. The large effect size indicates that there is considerable difference between the mean salaries of male and female employees.
Interpretation: There is sufficient evidence to suggest, at 0.05 significance level, that the mean salary of the male employees is significantly different from the mean salary of the female employees.
b. Since the one and two tail t-test results provided different outcomes, which is the proper/correct apporach to comparing salary equality? Why?
The two sample t-test is the proper/correct approach to comparing salary equality among male and female employees.
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: Mean compa of male employees = Mean compa of female employees
Ha: Mean compa of male employees ≠ Mean compa of female employees
Statistical test to use: Two-Sample t-Test Assuming Equal Variances
t-Test: Two-Sample Assuming Equal Variances
Male Female
Mean 1.05624 1.06872
Variance 0.0070206067 0.0049483767
Observations 25 25
Pooled Variance 0.0059844917
Hypothesized Mean Difference 0
df 48
t Stat -0.5703690595
P(T<=t) one-tail 0.2855439182
t Critical one-tail 1.6772241966
P(T<=t) two-tail 0.5710878365
t Critical two-tail 2.0106347219
What is the p-value: 0.5710878365
Is P-value < 0.05? No
Reject or do not reject Ho: Do Not Reject Ho
If the null hypothesis was rejected, what is the effect size value: Not Applicable
Meaning of effect size measure: Not Applicable
Interpretation: There is no evidence to suggest, at 0.05 significance level, that the mean compa of the male employees is significantly different from the mean compa of the female employees.
4 Since performance is often a factor in pay levels, is the average Performance Rating the same for both genders?
Ho: Mean performance rating of male employees = Mean performance rating of female employees
Ha: Mean performance rating of male employees ≠ Mean performance rating of female employees
Statistical test to use: Two-Sample t-Test Assuming Equal Variances
t-Test: Two-Sample Assuming Equal Variances
Male Female
Mean 87.6 84.2
Variance 75.25 184.75
Observations 25 25
Pooled Variance 130
Hypothesized Mean Difference 0
df 48
t Stat 1.054295244
P(T<=t) one-tail 0.1485129687
t Critical one-tail 1.6772241966
P(T<=t) two-tail 0.2970259374
t Critical two-tail 2.0106347219
What is the p-value: 0.2970259374
Is P-value < 0.05? No
Do we REJ or Not reject the null? Do Not Reject Ho
If the null hypothesis was rejected, what is the effect size value: Not Applicable
Meaning of effect size measure: Not Applicable
Interpretation: There is no evidence to suggest, at 0.05 significance level, that the mean performance rating of the male employees is significantly different from the mean performance rating of the female employees.
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?
The salary mean test output implies that there is a statistically significant difference between the male and female mean salaries. The compa mean test output implies that there is no significant difference between the male and female mean compas. Compa is the more appropriate variable as it is a measure of salary which removes the impact of grade and thus reduces chances of bias due to grade.
The overall conclusion is that, males and females paid the same for equal work.

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
90 75 100 65 55
Null Hypothesis: All the grades have the same mean performance rating. 80 80 100 75 90
Alt. Hypothesis: At least one of the grades has a different mean performance rating than the rest. 100 70 90 95 85
90 90 80 90 100
Anova: Single Factor 80 80 80 90 95
65 95 90
SUMMARY 95 80 95
Groups Count Sum Average Variance 60 90
A 15 1265 84.3333333333 153.0952380952 90 75
B 7 570 81.4285714286 72.619047619 75 95
C 5 450 90 100 95 95
D 5 415 83 157.5 100 80
E 12 1045 87.0833333333 152.0833333333 85
F 6 550 91.6666666667 116.6666666667 70
90
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 519.2023809524 5 103.8404761905 0.7789853558 0.5702154774 2.4270401139
Within Groups 5865.2976190476 44 133.3022186147
Total 6384.5 49
Interpretation:
What is the p-value: 0.5702154774
Is P-value < 0.05? No
Do we REJ or Not reject the null? Do Not Reject Ho
If the null hypothesis was rejected, what is the effect size value (eta squared): Not Applicable
Meaning of effect size measure: Not Applicable
What does that decision mean in terms of our equal pay question: There is no sufficient evidence to suggest, at 0.05 level of significance, that the average performance ratings are not the same for all grades.
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: All the grades have the same mean salary.
Alt. Hypothesis: At least one of the grades has a different mean salary than the rest. A B C D E
23 34 41 57 69
22 36 42 50 65
23 34 47 55 58
Anova: Single Factor 24 35 40 47 66
24 27 43 49 60
SUMMARY 23 28 64
Groups Count Sum Average Variance 24 28 56
A 15 353 23.5333333333 0.6952380952 24 60
B 7 222 31.7142857143 14.9047619048 24 65
C 5 213 42.6 7.3 23 62
D 5 258 51.6 17.8 22 60
E 12 751 62.5833333333 14.8106060606 24 66
24
24
ANOVA 25
Source of Variation SS df MS F P-value F crit
Between Groups 11343.4077922078 4 2835.8519480519 305.1165908598 0.00000 2.6123056118
Within Groups 362.4785714286 39 9.2943223443
Total 11705.8863636364 43
What is the p-value: 0.00000
Is P-value < 0.05? Yes
Do you reject or not reject the null hypothesis: Reject Ho
If the null hypothesis was rejected, what is the effect size value (eta squared): 0.9690345045
Meaning of effect size measure: The effect size is large. The large effect size indicates that there is considerable difference in the mean salaries for the different grades.
Interpretation: There is sufficient evidence to suggest, at 0.05 level of significance, that the average salaries are not the same for all grades.
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: 0.5389389507
Is P-value < 0.05? No
Do you reject or not reject the null hypothesis: Do Not Reject Ho
If the null hypothesis was rejected, what is the effect size value (eta squared): Not Applicable
Meaning of effect size measure: Not Applicable
For Ho: Average compas are equal for all degrees Ha: Average compas are not equal for all degrees
What is the p-value: 0.3088295633
Is P-value < 0.05? No
Do you reject or not reject the null hypothesis: Do Not Reject Ho
If the null hypothesis was rejected, what is the effect size value (eta squared): Not Applicable
Meaning of effect size measure: Not Applicable
For: Ho: Interaction is not significant Ha: Interaction is significant
What is the p-value: 0.3018915062
Do you reject or not reject the null hypothesis: Do Not Reject Ho
If the null hypothesis was rejected, what is the effect size value (eta squared): Not Applicable
Meaning of effect size measure: Not Applicable
What do these decisions mean in terms of our equal pay question: There is no sufficient evidence to suggest, at 0.05 level of significance, that the average compas are not equal across gender.
There is no sufficient evidence to suggest, at 0.05 level of significance, that the average compas are not equal across degrees.
There is no sufficient evidence to suggest, at 0.05 level of significance, that there is a significant interaction between gender and degree.
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? 23 23
23 22
23 23
Null Hypothesis: The mean difference between the salary and midpoint of employees = 0 23 24
Alt. Hypothesis: The mean difference between the salary and midpoint of employees > 0 23 24
23 23
Statistical test to use: Paired Two Sample t-Test for Means 23 24
23 24
t-Test: Paired Two Sample for Means 23 24
23 23
Salary Midpoint 23 22
Mean 45 41.76 23 24
Variance 368.693877551 263.4514285714 31 34
Observations 50 50 31 36
Pearson Correlation 0.9889717827 31 34
Hypothesized Mean Difference 0 31 35
df 49 40 41
t Stat 5.7827044981 40 42
P(T<=t) one-tail 0.0000002525 48 57
t Critical one-tail 1.6765508931 48 50
P(T<=t) two-tail 0.000000505 48 55
t Critical two-tail 2.0095751993 57 69
57 65
23 24
What is the p-value: 1.00000 23 24
Is P-value < 0.05? No 23 25
Do we REJ or Not reject the null? Do Not Reject Ho 31 27
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. 31 28
Meaning of effect size measure: NA 31 28
40 47
Interpretation: There is no sufficient evidence, at 0.05 significance level, that the company on average pay its existing employees above the market rate. 40 40
40 43
48 47
5.   Using the results up thru this week, what are your conclusions about gender equal pay for equal work at this point? 48 49
57 58
There is sufficient evidence to suggest, at 0.05 level of significance, that the average salaries are not equal for male and female employees. 57 66
Thus the provision of equal pay for equal work is not followed and the male employees are paid significantly greater than female employees for equal work. 57 60
57 64
57 56
57 60
57 65
57 62
57 60
57 66
67 76
67 77
67 76
67 72

Week 4

Score: 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 point> 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 Results are mean +/-2.064*standard error
Males 52 3.65878 2.064 44.4483 59.5517 2.064 is t value for 95% interval
Females 38 3.62275 2.064 30.5226 45.4774 <Reminder: standard error is the sample standard deviation divided by the square root of the sample size.>
Interpretation:
The two intervals overlap each other. It indicates that there is no significant difference in the mean salaries of males and females. This is the same that is found in the week 2 one sample t-test outcomes.
<1 point> 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?
Results are mean +/-2.064*standard error
Mean St error t value Low to High 2.064 is t value for 95% interval
Males 10 1.2715 2.064 7.3757 12.6243 <Reminder: standard error is the sample standard deviation divided by the square root of the sample size.>
Females 7.92 0.9814 2.064 5.8945 9.9455
Interpretation:
Yes/No
Can the means be equal? no Why? (no the means of male and females are not equal because sum is different for both data.
How does this compare to the week 2, question 2 result (2 sampe t-test)?
I can reject my null hypothesis because the results are significant at .05 significance level.
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?
ans two tail t test is the proper/correct apporach to compare one sample t-test because the test is is to be used whether there is any significant difference b/w the mean salaries of men and women.
<1 point> 3 We found last week that the degree 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: The populaton correlation between grade and degree is 0.
Ha: The population correlation between grade and degree is > 0
Perform analysis:
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
COUNT - M or 0 7 5 3 2 5 3 25
COUNT - F or 1 8 2 2 3 7 3 25
total 15 7 5 5 12 6 50
EXPECTED
7.5 3.5 2.5 2.5 6 3 25
7.5 3.5 2.5 2.5 6 3 25
15 7 5 5 12 6 50
By using either the Excel Chi Square functions or calculating the results directly as the text shows, do we
reject or not reject the null hypothesis? What does your conclusion mean?
Interpretation:
Male A B C D E F Total
fo 7 5 3 2 5 3 25
fe 7.5 3.5 2.5 2.5 6 3 25
fo – fe -0.5 1.5 0.5 -0.5 -1 0
(fo – fe)2 0.25 2.25 0.25 0.25 1 0
(fo – fe)2/fe 0.0333333333 0.6428571429 0.1 0.1 0.1666666667 0
X2 1.0428571429 df=5 1.04286 <11.07 accept null hypothesis
<Highlighting each cell with show how the value
Female A B C D E F Total is found: row total times column total divided by
fo 8 2 2 3 7 3 25 grand total.>
fe 7.5 3.5 2.5 2.5 6 3 25
fo – fe 0.5 -1.5 -0.5 0.5 1 0
(fo – fe)2 0.25 2.25 0.25 0.25 1 0
(fo – fe)2/fe 0.0333333333 0.6428571429 0.1 0.1 0.1666666667 0
X2 1.0428571429 df=5 1.04286 <11.07 accept null hypothesis
Chi-Square 0.03333 0.64286 0.1 0.1 0.16667 0 Chi-Square Statistic 2.08572
0.03333 0.64286 0.1 0.1 0.16667 0 p-value 0.8371614452
Critical Value 11.0704976935
By using either the Excel Chi Square functions or calculating the results directly as the text shows, do we
reject or not reject the null hypothesis? What does your conclusion mean?
<1 point> 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: The populaton correlation between grade and degree is 0.
Ha: The population correlation between grade and degree is > 0
Perform analysis:
OBSERVED A B C D E F Total
COUNT - M (Gen1) 12 4 2 3 2 2 25
COUNT - F (Gen1) 3 3 3 2 10 4 25
Total 15 7 5 5 12 6 50
EXPECTED
7.5 3.5 2.5 2.5 6 3 25
7.5 3.5 2.5 2.5 6 3 25
15 7 5 5 12 6 50
Chi-Square 2.7 0.0714285714 0.1 0.1 2.6666666667 0.3333333333 Chi-Square Statistic 11.9428571429
2.7 0.0714285714 0.1 0.1 2.6666666667 0.3333333333 p-value 0.0355792145
Critical Value 11.0704976935
<2 points> 5.      How do you interpret these results in light of our question about equal pay for equal work?
Ans: Using the above results of t confidence intervals we have a conclusion about that there is no significant difference b/w the average salaries for males and females so it can be lead to a conclusion that both males and females have equal pay for equal work in this population.

Week 5 to complete

Score: Week 5 Correlation and Regression
<1 point> 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):
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?
<1 point> 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?
<1 point> 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 hyhpotheses (one to stand for all the separate variables)
Ho:
Ha:
Place D94 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?
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?
<1 point> 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?
<2 points> 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?