ID Salary Compa Midpoint Age 4 5 6 7 8 9 10 11 12 13 14 15 19 20 21 22 26 27 30 31 32 43 44 49 50 1 2 3 16 17 18 23 24 25 28 29 33 34 75.4 72 46 26.9 59.8 65.4 24.2 22.7 77.2 63.2 51 62.9 23 22.6 23.4 63.7 39.4 24 23.3 36.1 63.5 26.5 61.4 45.2 69.6 74.2 23.4 24.5 26.8 24.9 66.5 23.6 61.7 34.3 57.9 22.9 60 34.7 1.125 67 67 48 31 57 57 23 23 67 57 48 57 23 23 23 57 40 23 23 31 57 31 57 40 57 67 23 23 31 23 57 23 57 31 57 23 57 31 44 52 45 25 35 45 24 32 42 45 36 37 22 23 27 39 35 29 22 27 41 26 38 25 34 49 41 32 52 32 27 32 52 44 42 32 34 30 1.075 0.958 0.867 1.049 1.147 1.053 0.989 1.152 1.108 1.062 1.104 0.998 0.982 1.017 1.117 0.985 1.045 1.014 1.164 1.114 0.856 1.078 1.130 1.221 1.107 1.018 1.064 0.866 1.084 1.166 1.025 1.082 1.107 1.016 0.997 1.053 1.120 Performance Service Gender Rating 95 95 90 95 90 95 90 100 95 90 95 95 95 90 75 75 80 60 95 90 95 80 80 80 90 100 100 85 80 80 55 90 95 70 100 90 85 75 9 5 18 4 9 11 2 8 20 16 8 5 2 4 3 20 7 4 2 6 21 2 12 5 11 10 19 1 7 8 3 9 22 16 16 12 8 5 1 0 0 0 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 0 0 0 0 1 0 1 0 0 1 1 1 0 1 0 1 0 1 Raise ID 4.4 5.4 4.3 5.6 5.5 4.5 6.3 5.7 5.5 5.2 5.2 5.5 6.2 5.3 4.3 3.9 3.9 3.9 6.2 5.5 6.6 4.9 4.6 4.3 5.3 4 4.8 4.6 3.9 4.9 3 5.8 4.5 4.8 5.5 6 5.7 3.6 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 35 36 37 38 39 40 41 42 45 46 47 48 42 46.6 36.3 76.7 57.3 22.5 54.7 24.5 48.5 74.3 22.6 41.3 1.050 1.166 1.170 1.145 1.193 0.979 1.140 1.067 1.010 1.109 0.984 1.033 40 40 31 67 48 23 48 23 48 67 23 40 32 44 31 43 48 36 30 41 36 36 30 30 100 90 80 95 65 65 75 70 90 70 80 100 8 4 11 13 6 6 9 4 16 12 7 2 1 0 1 0 1 1 1 0 0 0 1 1 5.7 5.7 5.6 6.3 3.8 3.3 3.8 4 5.7 4.5 4.7 4.7 12 33 38 44 46 47 49 50 6 9 21 29 Salary Compa Midpoint Age 23.6 22.6 23.4 22.9 24.9 22.5 23 24 22.6 23.4 23.3 22.7 34.7 36.3 34.3 36.1 42 41.3 57.3 54.7 51 66.5 69.6 75.4 77.2 24.5 24.5 24.2 26.8 26.9 26.5 46.6 39.4 45.2 48.5 46 60 57.9 1.025 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 0.984 1.018 0.997 1.084 0.979 0.998 1.045 0.982 1.017 1.014 0.989 1.120 1.170 1.107 1.164 1.050 1.033 1.193 1.140 1.062 1.166 1.221 1.125 1.152 1.064 1.067 1.053 0.866 0.867 0.856 1.166 0.985 1.130 1.010 0.958 1.053 1.016 Performa Service nce Rating 90 9 80 7 100 19 90 12 80 8 65 6 95 2 60 4 90 4 75 3 95 2 100 8 75 5 80 11 70 16 90 6 100 8 100 2 65 6 75 9 95 8 55 3 90 11 95 9 95 20 85 1 70 4 90 2 80 7 95 4 80 2 90 4 80 7 80 5 90 16 90 18 85 8 100 16 Gender Raise Degree 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 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 61.7 59.8 65.4 63.2 63.7 62.9 63.5 61.4 74.3 74.2 76.7 72 1.082 1.049 1.147 1.108 1.117 1.104 1.114 1.078 1.109 1.107 1.145 1.075 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 0 0 0 0 0 0 0 0 0 0 0 0 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 Gender1 Gr 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 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 1.053 0.087 1.073 0.076 M M M M M M M M M M M M E E E E E E E E F F F F 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 ID Degree Perf. Rat. Salary Gender Grade Compa Gender 1 Age Midpoint Service b. For each variable that you did not call ratio, why did you make that decision? For Nominal, the ID Gender and Gender 1 are labels with no actual order to them For Ordinal, the Degree and Grade are rank ordered labels without a fixed differeance between levels For Interval, the Pref Rat. is done once than a break in the activity before completed again. It has no fixed 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 45.02 1.0632 35.7 85.9 9.0 Standard Deviation 19.2083 0.0816 8.2513 11.4147 5.7177 Note - data is a sample from th Range 54.7000 0.3650 30 45 21 Female Mean 38.2120 1.0734 32.5 84.2 7.9 Standard Deviation 18.5321 0.0758 6.9 13.6 4.9 Range 54.7000 0.2420 26.0 45.0 18.0 Male Mean 51.8320 1.0530 38.9 87.6 10.0 Standard Deviation 17.6984 0.0873 8.4 8.7 6.4 Range 52.5000 0.3100 28.0 30.0 21.0 3 4 5. What is the probability for a: Probability a. Randomly selected person being a male in grade E? b. Randomly selected male being in grade E? Note part b is the same as given a male, what is probabilty of being in grade E? c. Why are the results different? 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 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? Conclusions from looking at salary results: Men are generally paid higher Conclusions from looking at compa results: The male and female compa results look closer to each other. Slightly higher for the female Do both salary measures show the same results? No they do not. Can we make any conclusions about equal pay for equal work yet? Females are still not making the same amount as their male counterparts. ID Salary 1 2 3 60 26.8 34.7 Compa Midpoint 1.053 0.866 1.120 57 31 31 Age Perform ance Rating Service Gender Raise 34 52 30 85 80 75 8 7 5 0 0 1 5.7 3.9 3.6 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 57.9 48.5 74.3 42 23.6 74.2 22.6 23.4 61.7 41.3 22.9 24.9 46.6 66.5 36.3 24.5 34.3 76.7 57.3 22.5 54.7 24.5 23 39.4 75.4 72 46 24 26.9 59.8 26.5 22.6 23.4 23.3 65.4 36.1 24.2 45.2 22.7 77.2 63.2 1.016 1.010 1.109 1.050 1.025 1.107 0.984 1.018 1.082 1.033 0.997 1.084 1.166 1.166 1.170 1.064 1.107 1.145 1.193 0.979 1.140 1.067 0.998 0.985 1.125 1.075 0.958 1.045 0.867 1.049 0.856 0.982 1.017 1.014 1.147 1.164 1.053 1.130 0.989 1.152 1.108 57 48 67 40 23 67 23 23 57 40 23 23 40 57 31 23 31 67 48 23 48 23 23 40 67 67 48 23 31 57 31 23 23 23 57 31 23 40 23 67 57 42 36 36 32 32 49 30 41 52 30 32 32 44 27 31 32 44 43 48 36 30 41 22 35 44 52 45 29 25 35 26 23 27 22 45 27 24 25 32 42 45 100 90 70 100 90 100 80 100 95 100 90 80 90 55 80 85 70 95 65 65 75 70 95 80 95 95 90 60 95 90 80 90 75 95 95 90 90 80 100 95 90 16 16 12 8 9 10 7 19 22 2 12 8 4 3 11 1 16 13 6 6 9 4 2 7 9 5 18 4 4 9 2 4 3 2 11 6 2 5 8 20 16 0 0 0 1 1 0 1 1 0 1 1 1 0 1 1 0 1 0 1 1 1 0 1 0 1 0 0 1 0 0 0 1 1 1 0 1 0 0 1 1 0 5.5 5.7 4.5 5.7 5.8 4 4.7 4.8 4.5 4.7 6 4.9 5.7 3 5.6 4.6 4.8 6.3 3.8 3.3 3.8 4 6.2 3.9 4.4 5.4 4.3 3.9 5.6 5.5 4.9 5.3 4.3 6.2 4.5 5.5 6.3 4.3 5.7 5.5 5.2 45 46 47 48 49 50 51 63.7 62.9 69.6 63.5 61.4 1.062 1.117 1.104 1.221 1.114 1.078 48 57 57 57 57 57 36 39 37 34 41 38 95 75 95 90 95 80 8 20 5 11 21 12 1 0 0 1 0 0 5.2 3.9 5.5 5.3 6.6 4.6 f data we are looking at, some descriptive measurse, and a variable is, as riptive statistics differeance between levels ompleted again. It has no fixed 0 point, but differences are the same between values. tistics for key variables. dard deviation, and range for 3 groups: overall sample, Females, and Males. verage and =stdev functions. functions with Fx) functions. he right as well. Note - data is a sample from the larger company population Probability ng in grade E? e same. We can do this by looking at some measures of where above and below a comparable value. Overall Female Male 59.1667 41.3000 62.9000 "=large" function 0.7364 0.1666 0.6254 Excel's standize function 0.9894 0.9800 0.9981 1-normsdist function 1.1077 0.5447 1.0000 1.1200 0.6145 1.0000 1.1070 0.6179 1.0000 mean about our equal pay for equal work question? uality? Are all of the results consistent? er for the female Degree Gender1 0 0 1 M M F Gr E B B The ongoing question that the weekly assignments will focus on is: Are males Note: to simplfy the analysis, we will assume that jobs within each grade comp 1 1 1 1 1 1 1 1 0 0 1 1 0 1 0 1 0 1 1 0 0 0 0 1 0 0 0 1 0 1 1 0 0 0 0 0 0 0 1 0 1 M M M F F M F F M F F F M F F M F M F F F M F M F M M F M M M F F F M F M M F F M E D F C A F A A E C A A C E B A B F D A D A A C F F D A B E B A A A E B A C A F E The column labels in the table mean: ID - Employee sample number Salary - Salary in thousands Age - Age in years Performance Rating - Appraisal rating (e Service - Years of service (rou Gender - 0 = male, 1 = female Midpoint - salary grade midpoin Raise - percent of last raise Grade - job/pay grade Degree (0= BS\\BA 1 = MS) Gender1 (Male or Female) Compa - salary divided by midpoint 1 1 1 1 0 0 F M M F M M D E E E E E nments will focus on is: Are males and females paid the same for equal work (under the Equal Pay Act)? e that jobs within each grade comprise equal work. Salary in thousands ance Rating - Appraisal rating (employee evaluation score) - 0 = male, 1 = female percent of last raise 0= BS\\BA 1 = MS) salary divided by midpoint Week 2 1 Testing means - T-tests In questions 2, 3, and 4 be sure to include the null and alternate hypotheses you will be testing. In the first 4 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 Note: These values are not the same as the data the assignment uses. The purpose is to analyze the re Based on these results, how do you interpret the results and what do these results suggest about the po Males Ho: Mean salary = Ha: Mean salary =/= 45.00 45.00 Females Ho: Mean salary = Ha: Mean salary =/= 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 Ho Mean 52 45 Mean Variance 316 0 Variance Observations 25 25 Observations Hypothesized Mean Di 0 Hypothesized Mean Di df 24 df t Stat 1.9689038266 t Stat P(T<=t) one-tail 0.0303078503 P(T<=t) one-tail t Critical one-tail 1.7108820799 t Critical one-tail P(T<=t) two-tail 0.0606157006 P(T<=t) two-tail t Critical two-tail 2.0638985616 Conclusion: Do not reject Ho; mean equals 45 t Critical two-tail Conclusion: Do not reject Ho; mean equa Note: the Female results are done for you, please complete the male results. Is this a 1 or 2 tail test? Is this a 1 or 2 tail test? 2 tail - why? - why? P-value is: Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? 0.0606157006 P-value is: No Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? Why do we not reject the null hypothesis? Pvalue is > 0.05 Why do we not reject the null hypothesis? Interpretation of test outcomes: 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: Male salary mean = Female salary mean Male salary mean =/= Female salary mean t-Test: Two-Sample Assuming Equal Variances t-Test: Two-Sample Assuming Equal Variances Variable 1 Mean Variable 2 51.832 Variance 38.212 313.233933333 343.436933 Observations Pooled Variance 25 25 328.335433333 Hypothesized Mean Dif 0 df 48 t Stat 2.6574997806 P(T<=t) one-tail 0.005329795 t Critical one-tail 1.6772241966 P(T<=t) two-tail 0.01065959 t Critical two-tail 2.0106347219 P-value is: 0.01065959 Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? yes Reject or do not reject Ho: Reject Ho If the null hypothesis was rejected, calculate the effect size value: 0.5314999561 If calculated, what is the meaning of effect size measure: large Interpretation: There is significant associationg between salary and gender b. Is the one or two sample t-test the proper/correct apporach to comparing salary equality? Why? Yes. The sample size is small and the sample is taken from normal population and population standard 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: Male compas mean = female compas mean Ha: male compas mean =/= female compas mean Statistical test to use: t-Test: Two-Sample Assuming Equal Variances Variable 1 Mean Variable 2 1.05304 Variance 1.0734 0.0076257067 0.00575142 Observations 25 Pooled Variance 25 0.0066885617 Hypothesized Mean Dif 0 df 48 t Stat -0.8801696972 P(T<=t) one-tail 0.1915765138 t Critical one-tail 1.6772241966 P(T<=t) two-tail 0.3831530276 t Critical two-tail 2.0106347219 What is the p-value: 0.383 Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? No 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 effect size measure: - Interpretation: There is no effect size 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: Males performance rating mean = femlae performance rating mean Ha: Males performance rating mean =/= femlae performance rating mean Test to ust-Test: Two-Sample Assuming Unequal Variances t-Test: Two-Sample Assuming Unequal Variances Variable 1 Mean Variance Observations Hypothesized Mean Dif df Variable 2 87.6 84.2 75.25 184.75 25 25 0 41 t Stat 1.054295244 P(T<=t) one-tail 0.1489606745 t Critical one-tail 1.6828780026 P(T<=t) two-tail 0.297921349 t Critical two-tail 2.0195409483 What is the p-value: 0.29792 Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? No Do we REJ or Not reject the null? Do not reject If the null hypothesis was rejected, calculate the effect size value: If calculated, what is the 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 fema 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? Salary becoause there is large association between salary and gender heses you will be testing. ejecting 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 - a constant.) The purpose is to analyze the results of t-tests rather than directly answer our equal pay question. hese results suggest about the population means for male and female average salaries? 45.00 45.00 wo-Sample Assuming Unequal Variances, o the one-sample t-test outcome - we are tricking Excel into doing a one sample test for us. Female Ho 38 45 334.66667 0 25 25 0 24 -1.913206 0.0338621 1.7108821 0.0677242 2.0638986 n: Do not reject Ho; mean equals 45 2 tail Ho contains = 0.0677242 No P-value greater than (>) rejection alpha opulation male and female average salaries could be equal to each other. he data sets have statistically equal variances.) and gender ng salary equality? Why? pulation and population standard deviation is not known population be equal to each other? (Another 2-sample t-test.) rmance Rating the same for both genders? mance rating mean ormance rating mean ent results about male and female salary equality, out salary equity? Why? 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? Use the data columns at the right to set up the paired data set for the analysis. Null Hypothesis: Salaries are < = to market rate dbar< =0 Alt. Hypothesis: Salaries are > market rate dbar>0 Statistical test to use: t-Test: Paired Two Sample for Means Salary Mean Midpoint 45.022 Variance 41.76 368.95726122 263.45143 Observations 50 50 Pearson Correlati 0.9889305701 Hypothesized Mea df 0 49 t Stat 5.8092273485 P(T<=t) one-tail 2.29930E-007 t Critical one-tail 1.6765508931 P(T<=t) two-tail 4.59860E-007 t Critical two-tail 2.0095751993 What is the p-value: 2.30E-07 Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? Yes What else needs to be checked on a 1-tail test in t stat > tcritical for one taileed test order to reject the null? Do we REJ orrejected, Not reject the is null? If the null hypothesis was what the REJ effect value: meaning of size effect size 0.821549 measure: Effect size is large. There is storng association between salaries and Interpretation of test results: The mean salaries are greater than 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: Ho: means equal for all grades Alt. Hypothesis: Ha: at least one mean is unequal Place B17 in Outcome range box. Anova: Single Factor SUMMARY Groups Count Sum Average Variance A 15 1265 84.3333333 153.0952381 B 7 570 81.4285714 72.61904762 C 5 450 90 100 D 5 415 83 157.5 E 12 1045 87.0833333 152.0833333 F 6 550 91.6666667 116.6666667 ANOVA Source of Variation SS Between Groups 519.20238095 Within Groups Total 5865.297619 6384.5 df MS F P-value 5 103.840476 0.778985356 0.570215477 44 133.302219 49 Interpretation of test results: What is the p-value: 0.57 Is P-value < 0.05? No Do we REJ or Not reject the null? Not Reject If the null hypothesis was rejected, what is the effect size value (eta squared): 0.08132232 Meaning of effect size measure: Effect size is small There is no association between What does that decision mean in terms of our equal pay question: The performance rating is sam f 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 mean salaries are same for all grades Alt. Hypothesis: Atleast one mean is different from others Place B51 in Outcome range box. Anova: Single Factor SUMMARY Groups Count Sum Average Variance A 15 352.1 23.4733333 0.616380952 B 7 221.6 31.6571429 21.72619048 C 5 214.5 42.9 8.65 D 5 257.5 51.5 20.845 E 12 755.6 62.9666667 10.16787879 F 6 449.8 74.9666667 3.602666667 50 ANOVA Source of Variation SS Between Groups 17692.079324 Within Groups Total 386.82647619 18078.9058 df MS F P-value 5 3538.41586 402.480977 1.5137E-035 44 8.79151082 49 Note: Sometimes we see a p-value in the format of 3.4E-5; this means move the decimal point left What is the p-value: 1.514E-035 Is P-value < 0.05? Yes Do we REJ or Not reject the null? REJ If the null hypothesis was rejected, calculate the effect size value (eta squared): 0.97860344 If calculated, what is the meaning of effect size measure: Very large. The salaries are depe Interpretation: The mean salary of atleat one gr 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 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 degre 1.052 1.134 Ha: Average compas are not equal for each d 1.175 1.149 Ho: Interaction is not significant Female 1.043 1.074 1.020 0.903 0.982 1.086 1.075 1.052 1.096 1.025 1.000 0.956 1.000 1.043 1.043 1.210 1.187 1.043 1.043 1.145 1.043 1.134 1.000 1.122 0.903 1.052 1.140 1.087 1.050 1.161 1.096 1.000 1.041 1.043 1.119 1.043 1.000 0.956 1.129 1.149 Ha: Interaction is significant Perform analysis: Anova: Two-Factor With Replication SUMMARY BA Male Count 12 Sum 12.349 Average 1.029083333 Variance 0.006686447 Female Count 12 Sum 12.791 Average 1.065916667 Variance 0.006102447 Total Count Sum 24 25.14 Average 1.0475 Variance 0.006470348 ANOVA Source of Variation SS Sample 0.002255021 Columns 0.006233521 Interaction 0.006417188 Within Total 0.25873675 0.273642479 Interpretation: For Ho: Average compas by gender are equal Ha: Average compas by gender are not equal What is the p-value: 0.53893895 Is P-value < 0.05? No Do you reject or not reject the null hypothesis:Do not Reject If the null hypothesis was rejected, what is the effect size value (eta squared): 0.00824076 Meaning of effect size measure: Very small For Ho: Average compas are equal for all degrees Ha: Average compas are not equal for all gra What is the p-value: 0.30882956 Is P-value < 0.05? No Do you reject or not reject the null hypothesis:Do not Reject If the null hypothesis was rejected, what is the effect size value (eta squared): 0.0227798 Meaning of effect size measure: Small For: Ho: Interaction is not significant Ha: Interaction is significant What is the p-value: 0.30189151 Is P-value < 0.05? No Do you reject or not reject the null hypothesis:Do not Reject If the null hypothesis was rejected, what is the effect size value (eta squared): 0.02345099 Meaning of effect size measure: Small What do these three decisions mean in terms of our equal pay question: The mean compas are equal across differetn grades and gender 5. Using the results up thru this week, what are your conclusions about gender equal pay for equal w t-Test: Two-Sample Assuming Equal Variances Female Mean Variance Male 38.212 51.832 343.43693333 313.23393 Observations 25 Null hypothesis: Mean salaries for males and fem Alternate hypothesis : Mean salaries for males a 25 Pooled Variance 328.33543333 pvalue < 0.05 Hypothesized Mea We reject Ho df t Stat P(T<=t) one-tail 0 48 -2.657499781 0.005329795 t Critical one-tail 1.6772241966 P(T<=t) two-tail 0.01065959 t Critical two-tail 2.0106347219 Conclusions Performance rating is independent of gender and grades Salaries differ for grades and gender potheses and use alpha = 0.05 for our decision " - the salary needed to hire a new employee. e the market rate? sociation between salaries and midpoint mance ratings. between males and females in the population. erformace 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 90 100 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 crit 2.427040114 If the ANVOA was done correctly, this is the p-value shown. t size is small here is no association between grades and performance rating he performance rating is sam for all the grades to test this assumption. F 70 100 95 95 95 95 If desired, place salaries per grade in these columns A B C D E 24.5 26.8 46.6 48.5 60 24.5 26.9 39.4 46 57.9 24.2 26.5 45.2 57.3 61.7 23.6 34.7 42 54.7 59.8 22.6 36.3 41.3 51 65.4 23.4 34.3 63.2 22.9 36.1 63.7 24.9 62.9 22.5 63.5 23 61.4 24 66.5 22.6 69.6 23.4 23.3 22.7 F 74.3 74.2 76.7 72 75.4 77.2 F crit 2.427040114 ns move the decimal point left 5 places. In this example, the p-value is 0.000034 ery large. The salaries are dependent on grades he mean salary of atleat one grade is different from others lication. Please interpret the results. . The purpose of this question is to analyze the result of a 2-way ANOVA test rather than directly answer our equal pay question. ompas by gender are equal ompas by gender are not equal ompas are equal for each degree ompas are not equal for each degree is not significant is significant tor With Replication MA Total 12 24 12.9 25.249 1.075 1.0520417 0.006519818 0.006866 12 24 12.787 25.578 1.065583333 1.06575 0.004212811 0.0049334 24 25.687 1.070291667 0.005156129 df MS P-value F crit (This is the row variable or gender.) 1 0.0062335 1.060054 0.3088296 4.0617065 (This is the column variable or Degree.) 1 0.002255 0.3834821 0.538939 4.0617065 1 0.0064172 1.0912878 0.3018915 4.0617065 44 0.0058804 47 ompas by gender are not equal o not Reject F ompas are not equal for all grades o not Reject o not Reject Place data values in these columns t gender equal pay for equal work at this point? Mean salaries for males and females are equal esis : Mean salaries for males and females are not equal Female 23.6 22.6 23.4 22.9 24.9 22.5 23 24 22.6 23.4 23.3 22.7 34.7 36.3 34.3 36.1 42 41.3 57.3 54.7 Male 24.5 24.5 24.2 26.8 26.9 26.5 46.6 39.4 45.2 48.5 46 60 57.9 61.7 59.8 65.4 63.2 63.7 62.9 63.5 Dif 51 66.5 69.6 75.4 77.2 61.4 74.3 74.2 76.7 72 Salary Midpoint diff 23.6 23 0.6 22.6 23 -0.4 23.4 23 0.4 22.9 23 -0.1 24.9 23 1.9 22.5 23 -0.5 23 23 0 24 23 1 22.6 23 -0.4 23.4 23 0.4 23.3 23 0.3 22.7 23 -0.3 34.7 31 3.7 36.3 31 5.3 34.3 31 3.3 36.1 31 5.1 42 40 2 41.3 40 1.3 57.3 48 9.3 54.7 48 6.7 51 48 3 66.5 57 9.5 69.6 57 12.6 75.4 67 8.4 77.2 67 10.2 24.5 23 1.5 24.5 23 1.5 24.2 23 1.2 26.8 31 -4.2 26.9 31 -4.1 26.5 31 -4.5 46.6 40 6.6 39.4 40 -0.6 45.2 40 5.2 48.5 48 0.5 46 48 -2 60 57 3 57.9 57 0.9 61.7 57 4.7 59.8 57 2.8 65.4 57 8.4 63.2 57 6.2 63.7 57 6.7 62.9 57 5.9 63.5 57 6.5 61.4 57 4.4 74.3 67 7.3 74.2 67 7.2 76.7 67 9.7 72 67 5 Mean 3.262 Stdev 3.9705492 swer our equal pay question. 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 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 2.513961 2.063899 46.81 Females 38 2.587148 2.063899 32.66 for the mean salary difference between the genders in the population. High 16.4517164 s not lie in the interval Results are the same - means are not equal. a better choice than using 2 one-sample techniques when comparing two samples? ation do not impact compa rates. ss the grades and genders. es by grade? enough to perform this test, ignore this limitation for this exercise.) egrees by gender n by gender If desired, you can do manual calculations per cell here. A B C D E F M Grad 3.6 1.68 1.2 1.2 2.88 1.44 Fem Grad 3.9 1.82 1.3 1.3 3.12 1.56 Male Und 3.9 1.82 1.3 1.3 3.12 1.56 12 13 13 Female Und 3.6 15 1.68 1.2 1.2 7 5 5 Sum = 17.69231 pvalue 0.279187 For this exercise - ignore the requirement for a correction for expected values less than 5. 2.88 12 1.44 6 12 50 have the same distribution of degrees by gender emales are distributed across grades in a similar pattern s gender in a similar pattern cross gender in a similar pattern F 4 2 6 25 25 50 M F Do manual calculations per cell here (if desired) A B C D E F 2.7 0.071429 0.1 0.1 2.666667 0.333333 2.7 1.633333 4.033333 2.7 4.033333 4.033333 Sum = 25.10476 3 7.5 70% variation in distribution of gender d females are not distributed across gender in a similar pattern equal work? 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