ID Salary Compa Midpoint Age Performance Rating Service 8 10 11 14 15 23 26 31 35 36 37 42 3 18 20 39 7 13 22 24 45 17 48 28 43 19 25 40 2 32 34 16 27 41 5 30 1 4 23 24 24 23 24 23 24 23 24 25 24 22 35 36 34 36 42 41 52 52 53 64 71 75 77 24 25 24 26 28 29 44 41 45 48 47 67 56 0.990 23 23 23 23 23 23 23 23 23 23 23 23 31 31 31 31 40 40 48 48 48 57 57 67 67 23 23 23 31 31 31 40 40 40 48 48 57 57 32 30 41 32 32 36 22 29 23 27 22 32 30 31 44 27 32 30 48 30 36 27 34 44 42 32 41 24 52 25 26 44 35 25 36 45 34 42 90 80 100 90 80 65 95 60 90 75 95 100 75 80 70 90 100 100 65 75 95 55 90 95 95 85 70 90 80 95 80 90 80 80 90 90 85 100 9 7 19 12 8 6 2 4 4 3 2 8 5 11 16 6 8 2 6 9 8 3 11 9 20 1 4 2 7 4 2 4 7 5 16 18 8 16 1.046 1.026 1.001 1.024 0.996 1.057 0.986 1.047 1.069 1.027 0.975 1.140 1.158 1.084 1.164 1.037 1.035 1.085 1.078 1.101 1.118 1.242 1.122 1.145 1.045 1.080 1.059 0.851 0.886 0.918 1.102 1.035 1.117 0.990 0.987 1.168 0.983 12 33 38 44 46 47 49 50 6 9 21 29 63 59 63 66 63 55 62 57 76 74 75 73 1.113 1.040 1.100 1.163 1.096 0.967 1.083 1.006 1.127 1.104 1.117 1.086 2237 52.976 149 3.5317333333 57 57 57 57 57 57 57 57 67 67 67 67 52 35 45 45 39 37 41 38 36 49 43 52 95 90 95 90 75 95 95 80 70 100 95 95 22 9 11 16 20 5 21 12 12 10 13 5 2088 1786 4295 448 139.2 119.06666667 286.3333333333 29.866666667 Gender 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 Raise Degree Gender 1 5.8 4.7 4.8 6 4.9 3.3 6.2 3.9 5.3 4.3 6.2 5.7 3.6 5.6 4.8 5.5 5.7 4.7 3.8 3.8 5.2 3 5.3 4.4 5.5 4.6 4 6.3 3.9 5.6 4.9 5.7 3.9 4.3 5.7 4.3 5.7 5.5 1 1 1 1 1 0 0 1 0 0 0 1 1 0 0 0 1 0 1 0 1 1 1 0 0 1 0 0 0 0 1 0 1 0 1 0 0 1 F F F F F F F F F F F F F F F F F F F F F F F F F M M M M M M M M M M M M M Gr A A A A A A A A A A A A B B B B C C D D D E E F F A A A B B B C C C D D E E Students: Copy the Student Data file data values into this sheet to assis weekly assignments. Midpoint - salary grade midpoint Gender1 (Male or Female) Raise - percent of last ra Compa - salary divided b Service - Years of service (rounded) Gender - 0 = male, 1 = f Note: to simplfy the analysis, we will assume that jobs within each g ID - Employee sample number Salary - Salary in thousa The ongoing question that the weekly assignments will focus on is: The column labels in the table mean: 0 0 0 0 0 0 0 0 0 0 0 0 25 1.66667 4.5 5.5 4.5 5.2 3.9 5.5 6.6 4.6 4.5 4 6.3 5.4 0 1 0 1 1 1 0 0 1 1 1 0 246.9 25 16.46 1.6667 M M M M M M M M M M M M E E E E E E E E F F F F 0 0 Age - Age in years Grade - job/pay grade Performance Rating - Ap Degree (0= BS\\BA 1 = M ues into this sheet to assist in doing your nments. Raise - percent of last raise Compa - salary divided by midpoint Gender - 0 = male, 1 = female sume that jobs within each grade comprise equal work. Salary - Salary in thousands assignments will focus on is: Are males and females paid the same for equal work (under the Equal Pay Act)? Performance Rating - Appraisal rating (employee evaluation score) Degree (0= BS\\BA 1 = MS) Week 1. Measurement and Description - chapters 1 and 2 1 The goal this week is to gain an understanding of our data set - what kind of data we are looking at, some d look at how the data is distributed (shape). 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 Performance RaSalary Degree I.D. Raise Gender 1 Service Age b. For each variable that you did not call ratio, why did you make that decision? The variables Compa and Midpoint which I did not call ratio does not fall under either data set description 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 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. Some of the values are completed for you - please finish the table. Salary Compa Age Perf. Rat. Service Overall Mean 43.53 43.5267 35.7 85.9 9.0 Standard Deviation 2.8543 2.8543 8.2513 11.4147 5.7177 Note - data is a sample from Range 45 45.000 30 45 21 Female Mean 41.55 124.6400 32.5 84.2 7.9 Standard Deviation 4.6 4.5530 6.9 13.6 4.9 Range 45.0 45.000 26.0 45.0 18.0 Male Mean 45.51 136.5200 38.9 87.6 10.0 Standard Deviation 1.3 1.2628 8.4 8.7 6.4 Range 30.0 30.000 28.0 30.0 21.0 3 What is the probability for a: Probability a. Randomly selected person being a male in grade E? 25/50 b. Randomly selected male being in grade E? 1 out of 12 Note part b is the same as given a male, what is probabilty of being i The probability of a male b c. Why are the results different? A total of 25 persons being a male in grade E is greater than male being in grade E. 4 A key issue in comparing data sets is to see if they are distributed/shaped the same. We can do this by look some selected values are within each data set - that is how many values are above and below a comparable For each group (overall, females, and males) find: A The value that cuts off the top 1/3 salary value in each group i The z score for this value within each group? ii The normal curve probability of exceeding this score: iii What is the empirical probability of being at or exceeding this salary value? B The value that cuts off the top 1/3 compa value in each group. i The z score for this value within each group? ii The normal curve probability of exceeding this score: iii What is the empirical probability of being at or exceeding this compa value? C How do you interpret the relationship between the data sets? What do they mean about our equal pay for e The relationship between the data sets show the salary for both male and female is different and mean male 5. What conclusions can you make about the issue of male and female pay equality? Are all of the results con What is the difference between the sal and compa measures of pay? When it comes to mean, males make more than female. In terms of standard deviation and range, female sa Sal is how much pay in shorter term than compa. Compa seem to have a lower digit number. Conclusions from looking at salary results: Male salary is more than female by looking at salary results. Conclusions from looking at compa results: Male compa is more than female looking at compa results. Do both salary measures show the same results? No, male and female salary measures show similar results. Can we make any conclusions about equal pay for equal Not enough information to conclude equal pay fo ata we are looking at, some descriptive measurse, and a tive statistics er either data set description. ics for key variables. d deviation, and range for 3 groups: overall sample, Females, and Males. age and =stdev functions. nctions with Fx) functions. right as well. Note - data is a sample from the larger company population Probability 1 out of 12 The probability of a male being in grade E is 25 out of 50. ame. We can do this by looking at some measures of where ove and below a comparable value. Overall Female Male "=large" function 0.9 -0.5 11.9 Excel's standize function 1-normsdist function -0.5 17.4 84.3 an about our equal pay for equal work question? e is different and mean male have higher salary than female because of the work. It is not equal pay for equal pay based on data. ty? Are all of the results consistent? eviation and range, female salary is more than male. digit number. on to conclude equal pay for equal work. pay for equal pay based on data. Week 2 1 Testing means - T-tests In questions 2 and 3, be sure to include the null and alternate hypotheses you will be testing. In the first 3 questions use alpha = 0.05 in making your decisions on rejecting or not rejecting the nul Below are 2 one-sample t-tests comparing male and female average salaries to the overall sample mea (Note: a one-sample t-test in Excel can be performed by selecting the 2-sample unequal variance t-tes Based on our sample, how do you interpret the results and what do these results suggest about the pop Males Females Ho: Mean salary = 45 Ho: Mean salary = 45 Ha: Mean salary =/= 45 Ha: Mean salary =/= 45 Note: While the results both below are actually from Excel's t-Test: Two-Sample Assuming Unequal V having no variance in the Ho variable makes the calculations default to the one-sample t-test outcome Male Mean 52 Variance 316 Observations 25 Hypothesized Mean Di 0 df 24 t Stat 1.9689038266 P(T<=t) one-tail 0.0303078503 t Critical one-tail 1.7108820799 P(T<=t) two-tail 0.0606157006 t Critical two-tail 2.0638985616 Ho 45 0 25 Mean Variance Observations Hypothesized Mean D df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Conclusion: Do not reject Ho; mean equals 45 Conclusion: Do not reject Ho; mean equa Is this a 1 or 2 tail test? 2 tail test Is this a 1 or 2 tail test? - why? The Ho can be (+) or (-) than 0.05 - why? P-value is: 0.0660060288 P-value is: 0.05 Is P-value < 0.05 (one tail test) or Is P-value < 0.05 (one tail 0.025 (two tail test)? test) or 0.025 (two tail test)? P value is greater Why do we not reject the null than 0.05 Why do we not reject the null hypothesis? Interpretation of test outcomes: hypothesis? In both cases, ther is insufficient evidence to support a claim that the mean salary for the group of ma from the mean for the combined group (=45) 2 Based on our sample data set, perform a 2-sample t-test to see if the population male and female avera (Since we have not yet covered testing for variance equality, assume the data sets have statistically eq Ho: Ha: Test to use: 1=2 1=2 2-sample t-test assuming equal variances t-Test: Two-Sample Assuming Equal Variances Male Mean Female 1.06684 Variance 1.04836 I got the chart to the left by using the f 0.00430164 0.006481 Observations Pooled Variance Hypothesized Mean Dif df 25 25 0.005391315 0 48 t Stat 0.8898352784 P(T<=t) one-tail 0.188996287 t Critical one-tail 1.6772241961 P(T<=t) two-tail 0.3779925741 t Critical two-tail 2.0106347576 P-value is: 0.1055857773 0.025 Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? Reject or do not reject Ho: Reject Ho 53.9425679859 If the null hypothesis was rejected, calculate the effect size value: If calculated, what is the meaning of There is a large variation in the mean salary between male and female effect size measure: Interpretation: mean salary of men is different than the mean salary of females b. Is the one or two sample t-test the proper/correct apporach to comparing salary equality? Why? Two Sample t-test - it provides significant results 3 Based on our sample data set, can the male and female compas in the population be equal to each othe Again, please assume equal variances for these groups. Ho: compa of male employess = compa of female employees Ha: compa of male employess = compa of female employees Statistical test to use: Two- Sample Assuming Equal Variances Female Mean Variance Observations Pooled Variance Hypothesized Mean Dif df Male 37.196 1.0574 307.7745666667 0.0084 25 25 153.8914835417 0 48 t Stat 10.2995686681 4.77571541989E-014 P(T<=t) one-tail 4.7757154E-014 9.55143083978E-014 t Critical one-tail 1.6772241961 2.0106347576 P(T<=t) two-tail 9.5514308E-014 1.6772241961 t Critical two-tail 2.0106347576 What is the p-value: 0.0573992348 0.025 Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? Reject or do not reject Ho: Do not Reject If the null hypothesis was rejected, calculate the effect size value: If calculated, what is the meaning of There isn't a large variation in the compa between male and female employees effect size measure: Interpretation: Compa is comparable between male employees and female employees 4 Since performance is often a factor in pay levels, is the average Performance Rating the same for both NOTE: do NOT assume variances are equal in this situation. Ho: Average performance rating is equal for both genders Ha: Average performance rating is unequal for both genders Test to use: t-Test: Two-Sample Assuming Equal Variances Female Mean Variance Observations Pooled Variance Male 84.2 87.6 184.75 75.25 25 25 130 Hypothesized Mean Dif 0 0.1485129687 48 0.2970259373 t Stat -1.054295244 2.0106347576 P(T<=t) one-tail 0.1485129687 1.6772241961 t Critical one-tail 1.6772241961 P(T<=t) two-tail 0.2970259373 t Critical two-tail 2.0106347576 df What is the p-value: 0.093936 Is P-value < 0.05 (one tail test) or yes 0.025 (two tail test)? Do we REJ or Not reject the null? Reject If the null hypothesis was rejected, -0.02196 calculate the effect size value: If calculated, what is the meaning of there is a significant difference in performance rating between m effect size measure: Interpretation: Male employees have a higher performance rating than female e 5 If the salary and compa mean tests in questions 2 and 3 provide different results about male and fema which would be more appropriate to use in answering the question about salary equity? Why? I would use the test in question two because it shows a larger variation that demands more hypothesis What are your conclusions about equal pay at this point? At this point, men employees are paid more than female employees eses you will be testing. rejecting or not rejecting the null hypothesis. alaries to the overall sample mean. 2-sample unequal variance t-test and making the second variable = Ho value -- see column S) ese results suggest about the population means for male and female average salaries? n salary = 45 n salary =/= 45 wo-Sample Assuming Unequal Variances, to the one-sample t-test outcome - we are tricking Excel into doing a one sample test for us. Female 38 334.6667 25 0 24 -1.91321 0.033862 1.710882 0.067724 2.063899 Ho 45 0 25 on: Do not reject Ho; mean equals 45 2 tail test The Ho can be (+) or (-) than 0.05 0.066806 0.05 value < 0.05 (one tail 0.025 (two tail test)? P value is greater than 0.05 Male - work shown (see formula in answer) P(T<=t) one-tail 0.0303078503 P(T<=t) two-tail 0.0606157006 t Critical two-tail 2.0638985616 t Critical one-tail 1.7108820799 Female - work shown (see formula in answer) P(T<=t) one-tail 0.0338621184 P(T<=t) two-tail 0.0677242369 t Critical two-tail t Critical one-tail 2.0638985616 1.7108820799 mean salary for the group of males or females is different population male and female average salaries could be equal to each other. the data sets have statistically equal variances.) chart to the left by using the following function: ween male and female alary of females 0.188996287 0.3779925741 2.0106347576 1.6772241961 ing salary equality? Why? population be equal to each other? (Another 2-sample t-test.) ale employees ale employees en male and female employees and female employees rmance Rating the same for both genders? in performance rating between male and female employees performance rating than female employees rent results about male and female salary equality, bout salary equity? Why? n that demands more hypothesis testing to show a more normalized results Q3 Ho Female Male Female 45 45 45 45 45 45 45 45 45 45 45 34 41 23 22 23 42 24 24 69 36 34 1.017 0.870 1.157 0.979 1.134 1.149 1.052 1.175 1.043 1.134 1.043 1.096 1.025 1.000 0.956 1.000 1.050 1.043 1.043 1.210 1.161 1.096 45 45 45 45 45 45 45 45 45 45 45 57 23 50 24 75 24 24 23 22 35 24 1.000 1.074 1.020 0.903 1.122 0.903 0.982 1.086 1.075 1.052 1.140 1.187 1.000 1.041 1.043 1.119 1.043 1.043 1.000 0.956 1.129 1.043 45 45 45 77 55 65 1.087 1.052 1.157 1.149 1.145 1.140 Week 3 Paired T-test and ANOVA For this week's work, again be sure to state the null and alternate hypotheses and use alpha = 0.05 value in the reject or do not reject decision on the null hypothesis. 1 Many companies consider the grade midpoint to be the "market rate" - the salary needed to hire a n Does the company, on average, pay its existing employees at or above the market rate?: Yes, since Use the data columns at the right to set up the paired data set for the analysis. Null Hypothesis: on average the pay is equal to the market rate Alt. Hypothesis: on average the pay is at or above the market rate What is the p-value: 1.84E-02 Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? P-value < 0.05 (one tail test) Yes What else needs to be checked on a 1-tail test in order to reject the null? T sample statistic value Do we REJ or Not reject the null? If the null hypothesis was rejected, is the effect size value: 0.177 If calculated, what iswhat the meaning of effect size measure: There is a positive correlation Interpretation of test results: We conclude that on average the pay is at or above the market rate Let's look at some other factors that might influence pay - education(degree) and performance ratings. 2 Last week, we found that average performance ratings do not differ between males and females in Now we need to see if they differ among the grades. Is the average performace rating the same for (Assume variances are equal across the grades for this ANOVA.) The rating values sorted by grade have been placed in columns I - N for you. Null Hypothesis: The average performance ratings is the same for all grades. Alt. Hypothesis: The average performance ratings is different for all grades. Place B17 in Outcome range box. SUMMARY Groups Count Sum Average Variance A 15 1265 84.333333 153.095238 B 7 570 81.428571 72.6190476 C 5 450 90 100 D 5 415 83 157.5 E 12 1045 87.083333 152.083333 F 6 550 91.666667 116.666667 ANOVA Source of Variation SS df Between Groups 519.2023809524 Within Groups 5865.297619048 Total MS F P-value 5 103.84048 0.77898536 0.570215477 44 133.30222 6384.5 49 Interpretation of test results: 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: 3 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? Use the input table to the right to list salaries under each grade level. (Assume equal variance, and use the analysis toolpak function ANOVA.) Null Hypothesis: The average salary is the same for each of the grade levels Alt. Hypothesis: The average salary is not the same for each grade levels Place B55 in Outcome range box. Summary Groups A B C D E F ANOVA Source of variat Between Group Within Group Count 15 7 5 5 12 6 Sum 354.9 223.3 213.1 251.5 745.4 449 Average Variance 23.66 0.56542857 31.9 18.21 42.62 2.687 50.3 6.925 62.116667 21.4178788 74.833333 1.77866667 SS 15517.17 288.7119 DF 5 38 MS F 3103.4349 408.4713 7.5976817 Total 15805.8819 43 Is P-value < 0.05? Do we REJ or Not reject the null? If the null hypothesis was rejected, calculate the effect size value (eta squared): If calculated, what is the meaning of effect size measure: Interpretation: 4 The table and analysis below demonstrate a 2-way ANOVA with replication. Please interpret the r Note: These values are not the same as the data the assignment uses. The purpose of this question BA MA Ho: Average compas by ge Male 1.017 1.157 Ha: Average compas by ge 0.870 0.979 Ho: Average compas are eq 1.052 1.134 Ha: Average compas are no 1.175 1.149 Ho: Interaction is not signi 1.043 1.043 Ha: Interaction is significa 1.074 1.134 1.020 1.000 Perform analysis: 0.903 1.122 0.982 0.903 Anova: Two-Factor With Rep 1.086 1.052 1.075 1.140 SUMMARY 1.052 1.087 Male 1.096 1.050 Female Count 1.025 1.161 Sum 1.000 1.096 Average 0.956 1.000 Variance 1.000 1.041 1.043 1.043 Female 1.043 1.119 Count 1.210 1.043 Sum 1.187 1.000 Average 1.043 0.956 Variance 1.043 1.129 1.145 1.149 Total Count Sum Average Variance ANOVA Source of Variation Sample Columns Interaction Within Total Interpretation: For Ho: Average compas by gender are equal Ha: Average compas by ge 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 compas are equal for all degrees Ha: Average compas are no 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: 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: What do these decisions mean in terms of our equal pay question: 5. Using the results up thru this week, what are your conclusions about gender equal pay for equal w Salary Midpoint Diff ses and use alpha = 0.05 for our decision e salary needed to hire a new employee. market rate?: Yes, since it is above the mean for salary 43.53 een males and females in the population. mace rating the same for all grades? Here are the data values sorted by grade level. A 90 80 100 90 80 85 65 70 95 60 90 75 95 B 80 75 80 70 95 80 90 C 100 100 90 80 80 D 90 65 75 90 95 E 85 100 95 55 90 95 90 75 95 90 95 80 F 70 100 95 95 95 95 90 100 F crit 2.42704012 0.57 If the ANVOA was done correctly, this is the p-value shown. No Do Not reject the null N/A N/A st this assumption. P-Value 1.04E-35 F Crit 2.4862548 If desired, place salaries per grade in these columns A B C D E 23 26.4 41.5 47.5 66.6 24 35.3 41.4 52.1 56 24 35.9 44.1 51.7 63.4 23 33.6 41.4 47.4 63.7 24 27.5 44.7 52.8 59.3 24 28.5 62.7 23 36.1 66.3 25 62.5 24 55.1 23 70.8 24 61.7 25 57.3 24 24 22 F 75.5 74 74.8 75.2 72.8 76.7 Yes Reject 0.98 It means that the effect size is large, meaning that the change in the dependant variable can be accounted by the indep The p-value is less that 5 ho had to be rejected. on. Please interpret the results. e purpose of this question is to analyze the result of a 2-way ANOVA test rather than directly answer our equal pay question. o: Average compas by gender are equal a: Average compas by gender are not equal o: Average compas are equal for each degree a: Average compas are not equal for each degree o: Interaction is not significant a: Interaction is significant erform analysis: nova: Two-Factor With Replication BA MA Total 12 12 24 12.349 12.9 25.249 1.029083333 1.075 1.0520417 0.006686447 0.006519818 0.006866 12 12 24 12.791 12.787 25.578 1.065916667 1.065583333 1.06575 0.006102447 0.004212811 0.0049334 24 24 25.14 25.687 1.0475 1.070291667 0.006470348 0.005156129 SS df MS F P-value F crit 0.006233521 (This is the row variable or gender.) 1 0.0062335 1.060054 0.3088296 4.0617065 (This is the column variable or Degree.) 0.006417188 1 0.0064172 1.0912878 0.3018915 4.0617065 0.002255021 0.25873675 0.273642479 1 0.002255 0.3834821 0.538939 4.0617065 44 0.0058804 47 a: Average compas by gender are not equal 0.538939 No Do Not Reject a: Average compas are not equal for all grades 0.30883 No Do Not Reject n is significant 0.301892 No Do Not Reject our equal pay question: The data shows that the pay is different between male and female, but is comparable overall in regards to the degree status. Place data values in these columns der equal pay for equal work at this point? Dif be accounted by the independent variable r equal pay question. ble or gender.) ariable or Degree.) arable overall in regards ese columns 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 For full credit, you need to also show the statistical outcomes - either the Excel test result or the calculations you perf 1 Interpretation: 2 Using our sample data, construct a 95% confidence interval for the population's mean salary f Interpret the results. Mean St error t value Low to Males 52 3.492846 2.063899 44.34 Females 38 3.609949 2.063899 30.49 es is within this interval (44.34, 58.76) les male is within this interval (30.49, 45.39) mean salary difference between the genders in the population. High 23.7076536 0 is not within the interval Results are the same - means are not equal. r choice than using 2 one-sample techniques when comparing two samples? e than one test o not impact compa rates. grades and genders. to perform this test, ignore this limitation for this exercise.) If desired, you can do manual calculations per cell here. A B C D E F M Grad 1.8777778 0.2752381 0.0333333 0.0333333 1.5605556 1.69 Fem Grad 0.3102564 0.7650549 0.0692308 0.0692308 1.4405128 0.1241026 Male Und 0.925641 0.0178022 0.3769231 0.0692308 1.1328205 0.2010256 Female Salary 23 24 24 23 24 23 24 23 24 25 24 22 35 36 34 36 42 41 52 52 53 64 71 75 77 Female Und 3.2111111 0.2752381 0.0333333 0.5333333 1.2272222 Sum = 1.44 17.69231 For this exercise - ignore the requirement for a correction for expected values less than 5. rees by grade are distributed across grades in a similar pattern s grades in a similar pattern cross grades in a similar pattern F 2 4 6 Do manual calculations per cell here (if desired) A B C D E 25 25 50 M F 2.7 0.0714286 0.1 0.1 2.6666667 0.3333333333 2.7 0.0714286 0.1 0.1 2.6666667 0.3333333333 Sum = 11.94286 3 3 F a medium association between Gender and grade similar pattern ders are also not evenly distributed across grades. Thus the salary difference can be caused sn't any evidence of an unequal pay. Male Salary 24 25 24 26 28 29 44 41 45 48 47 67 56 63 59 63 66 63 55 62 57 76 74 75 73 Female Salary Mean Standard Error Male Salary 37.94 Mean 3.609949 Standard Error Median 33.6 Median Mode 23.6 Mode 51.548 3.492846 56 #N/A Standard Deviation 18.04975 Standard Deviat 17.46423 Sample Variance Kurtosis Skewness 325.7933 Sample Varianc 304.9993 -0.163187 Kurtosis 1.037207 Skewness -1.166612 -0.373883 Range 54.3 Range Minimum 22.4 Minimum 24 Maximum 76.7 Maximum 75.5 Sum Count 948.5 Sum 25 Count 51.5 1288.7 25 Week 5 Correlation and Regression 1. Create a correlation table for the variables in our data set. (Use analysis ToolPak or StatPlus:mac LE fu a. Reviewing the data levels from week 1, what variables can be used in a Pearson's Correlation tab b. Place table here (C8): 2 c. Using r = approximately .28 as the signicant r value (at p = 0.05) for a correlation between 50 val 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? Below is a regression analysis for salary being predicted/explained by the other variables in our s age, performance rating, service, gender, and degree variables. (Note: since salary and compa ar expressing an employee's salary, we do not want to have both used in the same regression.) Plase interpret the findings. Note: These values are not the same as the data the assignment uses. The purpose is to analyze the res 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 hav Ha: The regression coefficient for each variable is significant Listing it this way to sav Sal SUMMARY OUTPUT Regression Statistics Multiple R 0.99155907 R Square 0.9831894 Adjusted R Square 0.98084373 Standard Error 2.65759257 Observations 50 ANOVA df Regression Residual Total SS MS F Significance F 6 17762.3 2960.383 419.15161 1.812E-036 43 303.70033 7.062798 49 18066 Standard Coefficients Error t Stat P-value Lower 95% Upper 95% Intercept -1.74962121 3.6183677 -0.483539 0.6311665 -9.046755 5.54751262 Midpoint 1.21670105 0.0319024 38.13829 8.66E-035 1.15236383 1.28103827 Age -0.00462801 0.0651972 -0.070985 0.943739 -0.1361107 0.1268547 Performace Rating -0.05659644 0.0344951 -1.640711 0.1081532 -0.1261624 0.01296949 Service -0.04250036 0.084337 -0.503935 0.6168794 -0.2125821 0.12758138 Gender 2.420337212 0.8608443 2.811585 0.0073966 0.68427919 4.15639523 Degree 0.27553341 0.7998023 0.344502 0.7321481 -1.3374217 1.88848848 Note: since Gender and Degree are expressed as 0 and 1, they are considered dummy variables an 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 What is the coefficient's p-value for each of the variables: NA Is the p-value < 0.05? NA Do you reject or not reject each null hypothesis: NA What are the coefficients for the significant variables? Using the intercept coefficient and 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 Age 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 sam 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 c94 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 What is the coefficient's p-value for each of the variables: NA Is the p-value < 0.05? NA Do you reject or not reject each null hypothesis: NA What are the coefficients for the significant variables? Midpoint Age Using the intercept coefficient and only the significant variables, what is the equation? Compa = Is genderofa statistical significantsignificance, factor in compa: Regardless 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? Does the company pay employees equally for for equal work? 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 5 Why did the single factor tests and analysis (such as t and single factor ANOVA tests on salary eq What outcomes in your life or work might benefit from a multiple regression examination rather t k or StatPlus:mac LE function Correlation.) earson's Correlation table (which is what Excel produces)? rrelation between 50 values, what variables are surprises -by that I other variables in our sample (Midpoint, nce salary and compa are different ways of e same regression.) ose is to analyze the result of a regression test rather than directly answer our equal pay question. Note: technically we have one for each input variable. Listing it this way to save space. Lower 95.0% Upper 95.0% -9.0467550427 5.547512618 1.1523638283 1.2810382727 -0.1361107191 0.1268546987 -0.1261623747 0.0129694936 -0.2125820912 0.1275813765 0.684279192 4.156395232 -1.3374216547 1.8884884833 ered dummy variables and can be used in a multiple regression equation. Perf. Rat. Service the same independent ngs by answering the same questions. Gender Degree Perf. Rat. Service Gender Degree ally for equal work? mpa? Why? analysis during the last 5 weeks? NOVA tests on salary equality) not provide a complete answer to our salary equality question? sion examination rather than a simpler one variable test? Note: These values are not the same as in the data the assignment uses. The purpose is to analyze the result of a 2-wa nalyze the result of a 2-way ANOVA test rather than directly answer our equal pay