ID Salary Compa Midpoint Age 1 2 3 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 57.7 27.8 34 59.2 49.5 75.7 41.7 23.4 80.8 23.6 23.6 66.9 41.6 21.5 24.4 39 68.8 34.9 23.2 36 75.3 56.7 22.6 51.5 25.5 22.9 43.5 74.4 73.5 45.7 23.7 26.9 55.1 28 21.9 23.7 23.2 57.6 1.012 57 31 31 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 34 52 30 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 0.897 1.096 1.038 1.031 1.130 1.043 1.018 1.206 1.027 1.024 1.174 1.041 0.936 1.059 0.975 1.207 1.126 1.008 1.160 1.124 1.182 0.984 1.072 1.109 0.994 1.088 1.111 1.097 0.952 1.031 0.867 0.967 0.904 0.953 1.032 1.010 1.010 Performance Service Gender Rating 85 80 75 100 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 8 7 5 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 0 0 1 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 Raise Degree 5.7 3.9 3.6 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 0 0 1 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 39 40 41 42 43 44 45 46 47 48 49 50 34.3 24.4 40.5 23.3 77.2 56.9 57.7 65.4 56.8 59.7 62.4 56.5 1.108 1.062 1.012 1.012 1.152 0.999 1.202 1.148 0.997 1.048 1.095 0.992 31 23 40 23 67 57 48 57 57 57 57 57 27 24 25 32 42 45 36 39 37 34 41 38 90 90 80 100 95 90 95 75 95 90 95 80 6 2 5 8 20 16 8 20 5 11 21 12 1 0 0 1 1 0 1 0 0 1 0 0 SS MS F Significance F 5.5 6.3 4.3 5.7 5.5 5.2 5.2 3.9 5.5 5.3 6.6 4.6 0 0 0 1 0 1 1 1 1 1 0 0 Regression Statistics Multiple R 0.705017948 R Square 0.497050308 Adjusted R Sq0.413225359 Standard Erro 0.056125269 Observations 50 ANOVA df Regression 7 0.13075 0.0186785825 5.929623 7.8E-005 Residual 42 0.1323 0.0031500458 Total 49 0.26305 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Intercept 0.948623877 0.08172 11.608680312 1E-014 0.783713 1.11353 0.78371 Mid 0.003499503 0.00065 5.3900133356 3E-006 0.002189 0.00481 0.00219 Age 0.000552774 0.00145 0.3822925256 0.704172 -0.00237 0.00347 -0.0024 EES -0.00184626 0.00103 -1.800846137 0.078911 -0.00392 0.00022 -0.0039 SR -0.00041823 0.00183 -0.228814135 0.820124 -0.00411 0.00327 -0.0041 G 0.064664996 0.01834 3.5259629624 0.001035 0.027654 0.10168 0.02765 Raise 0.014654956 0.01391 1.0536389608 0.298072 -0.01341 0.04272 -0.0134 Deg 0.001467599 0.01611 0.0910996125 0.927847 -0.03104 0.03398 t-Test: Two-Sample Assuming Equal Variances Variable 1 Variable 2 Mean Variance Observations 1.06684 1.04836 0.00430164 0.00648 25 Pooled Varian 0.005391315 Hypothesized df t Stat 0 48 0.889835278 P(T<=t) one-ta0.188996287 t Critical one-t 1.677224196 P(T<=t) two-tai0.377992574 t Critical two-ta2.010634758 25 -0.031 Gender 1 Gr M M F 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 E B B 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 Students: Copy the Student Data file data values into this sheet to assist in doing your weekly assignments. The ongoing question that the weekly assignments will focus on is: Are males and females paid Note: to simplfy the analysis, we will assume that jobs within each grade comprise equal work. The column labels in the table mean: ID - Employee sample number Salary - Salary in thousands Age - Age in years Performance Rating - Appraisal rating (employee ev Service - Years of service (rounded) Gender - 0 = male, 1 = female Midpoint - salary grade midpoint Raise - percent of last raise Grade - job/pay grade Degree (0= BS\\BA 1 = MS) Gender1 (Male or Female) Compa - salary divided by midpoint F M M F F M F M M F M M B A C A F E D E E E E E Regression Statistics Multiple R 0.993129 R Square 0.986305 Adjusted R 0.984022 Standard E 2.435282 Observatio 50 ANOVA df Regression SS F Significance F 7 17938.42 2562.632087409 432.1034 5.30E-037 Residual 42 249.0852 Total 49 18187.51 Upper 95.0% MS Coefficients Standard Error 1.11353 Intercept -4.871454 3.545701 0.00481 Mid 1.228416 0.028171 0.00347 Age 0.036828 0.00022 5.9305997175 t Stat -1.3739045839 P-value Lower 95%Upper 95% 0.17676 -12.02697 2.284059 43.6051641629 1.32E-036 1.171563 1.285268 0.06274 0.5869957178 0.560349 -0.089786 0.163442 EES -0.082158 0.044484 -1.8469014451 0.071815 -0.171931 0.007615 0.00327 SR -0.077848 0.079309 -0.9815852701 0.331925 0.10168 G 2.914508 0.795761 3.6625445343 0.000694 1.308599 4.520418 0.04272 Raise 0.676329 0.603509 1.1206622295 0.268799 -0.541601 1.894259 -0.2379 0.082203 0.03398 Deg 0.034504 0.699007 0.0493620731 0.960865 -1.376149 1.445158 assist in doing your on is: Are males and females paid the same for equal work (under the Equal Pay Act)? each grade comprise equal work. thousands g - Appraisal rating (employee evaluation score) , 1 = female vided by midpoint Lower 95.0% Upper 95.0% -12.02697 2.284059 1.171563 1.285268 -0.089786 0.163442 -0.171931 0.007615 -0.2379 0.082203 1.308599 4.520418 -0.541601 1.894259 -1.376149 1.445158 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 b. For each variable that you did not call ratio, why did you make that decision? 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 35.7 85.9 9.0 Standard Deviation 8.2513 11.4147 5.7177 Note - data is a sample from Range 30 45 21 Female Mean 32.5 84.2 7.9 Standard Deviation 6.9 13.6 4.9 Range 26.0 45.0 18.0 Male Mean 38.9 87.6 10.0 Standard Deviation 8.4 8.7 6.4 Range 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: Conclusions from looking at compa results: Do both salary measures show the same results? Can we make any conclusions about equal pay for equal work yet? ata we are looking at, some descriptive measurse, and a tive statistics 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 ame. We can do this by looking at some measures of where ove and below a comparable value. Overall Female Male "=large" function Excel's standize function 1-normsdist function an about our equal pay for equal work question? ty? Are all of the results consistent? 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 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 Variance 316 0 Observations 25 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 Conclusion: Do not reject Ho; mean equals 45 Note: the Female results are done for you, please complete the male results. Is this a 1 or 2 tail test? 2 tail test - why? Ha: 45 P-value is: 2.0638985616 Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? P-value <0.05 Why do we not reject the null because P-value > 0.05 hypothesis? Is this a 1 or 2 tail test? Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? Why do we not reject the null hypothesis? Interpretation of test outcomes: The mean salary of the population is 45,000. The mean pay of guys and that of t 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: Mean Variance Observations Pooled Variance Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Male salary mean = Female salary mean Male salary mean =/= Female salary mean t-Test: Two-Sample Assuming Equal Variances Male Female 52 38 316 334.6666666667 25 25 325.3333333333 0 48 2.7442189608 0.0042530089 1.6772241966 0.0085060177 2.0106347219 P-value is: P(T<=t) two-tail Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? Reject or do not reject Ho: <0.05 One tail test reject Ho If the null hypothesis was rejected, calculate the effect size value: 0.776182335 If calculated, what is the meaning of effect size measure: Since the impact size is huge, it shows that there is a critical distinction between Interpretation: There is sufficient proof to say that the mean compensation of all the male verse b. Is the one or two sample t-test the proper/correct apporach to comparing salary equality? Why? The two-tail test is the right approach when looking at pay uniformity. The one-tail test is not fitting s 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: the mean compa of male workers = the mean compa of female Ha: the mean compa of male workers the mean compa of female Statistical test to use: two-samples t-test with a supposition of equal variances Male Mean 1.05624 Variance 0.0070206067 Observations 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: Female 1.06872 0.0049483767 25 0.571087836 Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? P-value is 0.025 Reject or do not reject Ho: do not reject Ho If the null hypothesis was rejected, calculate the effect size value: If calculated, what is the meaning of effect size measure: does not apply does not apply Interpretation: There is sufficient proof to say that the mean compa of all the male verse the me 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: the mean execution rating of male workers = the mean execution rating of fema Ha: the mean execution rating of male workers the mean execution rating of fema Test to ust-Test: Two-Sample Assuming Unequal Variances Male Mean Variance Observations Pooled Variance Hypothesized Mean Di df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Female 87.6 75.25 25 130 0 48 1.054295244 0.1485129687 1.6772241966 0.2970259374 2.0106347219 84.2 184.75 25 What is the p-value: 0.29703 Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? P-value is 0.025 Do we REJ or Not reject the null? do not reject Ho If the null hypothesis was rejected, calculate the effect size value: If calculated, what is the meaning of effect size measure: does not apply does not apply Interpretation: There is sufficient proof to say that the mean execution rating 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? The salary mean test outcomes permit us to establish that the distinction amongst male and female me ternate hypotheses you will be testing. decisions on rejecting or not rejecting the null hypothesis. ale average salaries to the overall sample mean. selecting the 2-sample unequal variance t-test and making the second variable = Ho value - a constant.) gnment uses. The purpose is to analyze the results of t-tests rather than directly answer our equal pay question. nd what do these results suggest about the population means for male and female average salaries? Females Ho: Mean salary = Ha: Mean salary =/= 45.00 45.00 cel's t-Test: Two-Sample Assuming Unequal Variances, ions default to the one-sample t-test outcome - we are tricking Excel into doing a one sample test for us. Female Ho Mean 38 45 Variance 334.66667 0 Observations 25 25 Hypothesized Mean Di 0 df 24 t Stat -1.913206 P(T<=t) one-tail 0.0338621 t Critical one-tail 1.7108821 P(T<=t) two-tail 0.0677242 t Critical two-tail 2.0638986 Conclusion: Do not reject Ho; mean equals 45 Is this a 1 or 2 tail test? 2 tail - why? Ho contains = P-value is: 0.0677242 Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? No Why do we not reject the null hypothesis? P-value greater than (>) rejection alpha is 45,000. The mean pay of guys and that of the females are fundamentally not quite the same as the mean of the whole populatio o see if the population male and female average salaries could be equal to each other. lity, assume the data sets have statistically equal variances.) ows that there is a critical distinction between the mean compensations of male and female representatives. the mean compensation of all the male verse the mean salary of all the female representatives is essentially unique. ch to comparing salary equality? Why? ay uniformity. The one-tail test is not fitting since we have to decide centrality and it doesn't permit us. ompas in the population be equal to each other? (Another 2-sample t-test.) of male workers = the mean compa of female workers of male workers the mean compa of female workers st with a supposition of equal variances the mean compa of all the male verse the mean compensation of all the female representatives is fundamentally unique. verage Performance Rating the same for both genders? in this situation. workers = the mean execution rating of female representatives workers the mean execution rating of female representatives nt proof to say that the mean execution rating of all the male verse the mean pay of all the female workers is essentially unique. provide different results about male and female salary equality, e question about salary equity? Why? t the distinction amongst male and female mean pay rates is factually huge. The compa mean test outcomes permit us to decide th value - a constant.) er our equal pay question. erage salaries? e sample test for us. 2 tail test Ha: 45 0 P-value <0.05 because P-value > 0.05 the same as the mean of the whole population. emale representatives. sentatives is essentially unique. doesn't permit us. esentatives is fundamentally unique. l the female workers is essentially unique. pa mean test outcomes permit us to decide the contrast amongst male and female compa, which is not critical. The compa mean t ical. The compa mean test outcomes appear to be more suitable in light of the fact that by utilizing this estimation it neglects any ng this estimation it neglects any effect of various review, which lessens inclination. By and large, I infer that guys and females a er that guys and females appear to be paid the same for equivalent work. 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: Alt. Hypothesis: Statistical test to use: What is the p-value: Is P-value < 0.05 (one tail test) or 0.025 (two tail test)? What else needs to be checked on a 1-tail test in order to reject the null? Do we REJ orrejected, Not reject the is null? If the null hypothesis was what the effect size value: meaning of effect size measure: Interpretation of test results: 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. SUMMARY Groups Count Sum Average Variance A 15 353 23.5333333 0.695238095 B 7 222 31.7142857 14.9047619 C 5 213 42.6 7.3 D 5 258 51.6 17.8 E 12 F 6 751 62.5833333 14.81060606 453 75.5 3.5 ANOVA Source of Variation SS Between Groups 17686.021429 Within Groups Total 379.97857143 18066 df MS 5 3537.20429 F 409.59412 P-value 1.04E-35 44 8.63587662 49 Interpretation of test results: What is the p-value: 0.57 Is P-value < 0.05? Yes Do we REJ or Not reject the null? Not reject the null hypotheses If the null hypothesis was rejected, what is the effect size value Effect size not needed because t (eta squared): accepted measures the size of an effect as Meaning of effect size measure: population. What does that decision mean in terms of our equal pay question: From the ANOVA Table we see 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: Alt. Hypothesis: Place B51 in Outcome range box. 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: 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 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: 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 not equal for all gra 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 three 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 potheses and use alpha = 0.05 for our decision " - the salary needed to hire a new employee. e the market rate? Salary Midpoint Diff 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 70 100 95 95 95 95 F crit 2.42704012 If the ANVOA was done correctly, this is the p-value shown. ot reject the null hypotheses fect size not needed because the null was cepted easures the size of an effect as it exists in the opulation. om the ANOVA Table we see that correponding to the F =409.594119969171 the p value is 1.04E-35 which is greater than 0.05 to test this assumption. If desired, place salaries per grade in these columns A B C D E F ns move the decimal point left 5 places. In this example, the p-value is 0.000034 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 F 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 ompas are not equal for all grades t gender equal pay for equal work at this point? Place data values in these columns Dif 5 which is greater than 0.05 at 5% level of significance which means we accept the null hypothesis so the average performance r swer our equal pay question. average performance rating is the same for all grades. 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 Females for the mean salary difference between the genders in the population. High 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.) If desired, you can do manual calculations per cell here. A B C D E F M Grad Fem Grad Male Und Female Und Sum = For this exercise - ignore the requirement for a correction for expected values less than 5. emales are distributed across grades in a similar pattern Do manual calculations per cell here (if desired) A B C D E F M F Sum = F 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