Statics Problem Set - Week 3
Data
| ID | Salary | Compa-ratio | Midpoint | Age | Performance Rating | Service | Gender | Raise | Degree | Gender1 | Grade | Do not manipuilate Data set on this page, copy to another page to make changes | ||||
| 1 | 58 | 1.018 | 57 | 34 | 85 | 8 | 0 | 5.7 | 0 | M | E | The ongoing question that the weekly assignments will focus on is: Are males and females paid the same for equal work (under the Equal Pay Act)? | ||||
| 2 | 28 | 0.902 | 31 | 52 | 80 | 7 | 0 | 3.9 | 0 | M | B | Note: to simplfy the analysis, we will assume that jobs within each grade comprise equal work. | ||||
| 3 | 33.8 | 1.089 | 31 | 30 | 75 | 5 | 1 | 3.6 | 1 | F | B | |||||
| 4 | 64.8 | 1.137 | 57 | 42 | 100 | 16 | 0 | 5.5 | 1 | M | E | The column labels in the table mean: | ||||
| 5 | 48.7 | 1.014 | 48 | 36 | 90 | 16 | 0 | 5.7 | 1 | M | D | ID – Employee sample number | Salary – Salary in thousands | |||
| 6 | 75.4 | 1.126 | 67 | 36 | 70 | 12 | 0 | 4.5 | 1 | M | F | Age – Age in years | Performance Rating - Appraisal rating (employee evaluation score) | |||
| 7 | 41.6 | 1.039 | 40 | 32 | 100 | 8 | 1 | 5.7 | 1 | F | C | Service – Years of service (rounded) | Gender – 0 = male, 1 = female | |||
| 8 | 23.6 | 1.028 | 23 | 32 | 90 | 9 | 1 | 5.8 | 1 | F | A | Midpoint – salary grade midpoint | Raise – percent of last raise | |||
| 9 | 73.8 | 1.102 | 67 | 49 | 100 | 10 | 0 | 4 | 1 | M | F | Grade – job/pay grade | Degree (0= BS\BA 1 = MS) | |||
| 10 | 22.5 | 0.980 | 23 | 30 | 80 | 7 | 1 | 4.7 | 1 | F | A | Gender1 (Male or Female) | Compa-ratio - salary divided by midpoint | |||
| 11 | 24.1 | 1.046 | 23 | 41 | 100 | 19 | 1 | 4.8 | 1 | F | A | |||||
| 12 | 58.3 | 1.023 | 57 | 52 | 95 | 22 | 0 | 4.5 | 0 | M | E | |||||
| 13 | 41.4 | 1.035 | 40 | 30 | 100 | 2 | 1 | 4.7 | 0 | F | C | |||||
| 14 | 23.2 | 1.008 | 23 | 32 | 90 | 12 | 1 | 6 | 1 | F | A | |||||
| 15 | 22.5 | 0.978 | 23 | 32 | 80 | 8 | 1 | 4.9 | 1 | F | A | |||||
| 16 | 44.7 | 1.117 | 40 | 44 | 90 | 4 | 0 | 5.7 | 0 | M | C | |||||
| 17 | 63.8 | 1.120 | 57 | 27 | 55 | 3 | 1 | 3 | 1 | F | E | |||||
| 18 | 34.9 | 1.127 | 31 | 31 | 80 | 11 | 1 | 5.6 | 0 | F | B | |||||
| 19 | 24.9 | 1.083 | 23 | 32 | 85 | 1 | 0 | 4.6 | 1 | M | A | |||||
| 20 | 33.4 | 1.076 | 31 | 44 | 70 | 16 | 1 | 4.8 | 0 | F | B | |||||
| 21 | 73.5 | 1.097 | 67 | 43 | 95 | 13 | 0 | 6.3 | 1 | M | F | |||||
| 22 | 51.4 | 1.071 | 48 | 48 | 65 | 6 | 1 | 3.8 | 1 | F | D | |||||
| 23 | 25.1 | 1.090 | 23 | 36 | 65 | 6 | 1 | 3.3 | 0 | F | A | |||||
| 24 | 48.6 | 1.013 | 48 | 30 | 75 | 9 | 1 | 3.8 | 0 | F | D | |||||
| 25 | 24.4 | 1.062 | 23 | 41 | 70 | 4 | 0 | 4 | 0 | M | A | |||||
| 26 | 24.3 | 1.058 | 23 | 22 | 95 | 2 | 1 | 6.2 | 0 | F | A | |||||
| 27 | 39.2 | 0.980 | 40 | 35 | 80 | 7 | 0 | 3.9 | 1 | M | C | |||||
| 28 | 75.8 | 1.131 | 67 | 44 | 95 | 9 | 1 | 4.4 | 0 | F | F | |||||
| 29 | 76.3 | 1.139 | 67 | 52 | 95 | 5 | 0 | 5.4 | 0 | M | F | |||||
| 30 | 48.6 | 1.012 | 48 | 45 | 90 | 18 | 0 | 4.3 | 0 | M | D | |||||
| 31 | 23.4 | 1.018 | 23 | 29 | 60 | 4 | 1 | 3.9 | 1 | F | A | |||||
| 32 | 27.5 | 0.886 | 31 | 25 | 95 | 4 | 0 | 5.6 | 0 | M | B | |||||
| 33 | 58 | 1.018 | 57 | 35 | 90 | 9 | 0 | 5.5 | 1 | M | E | |||||
| 34 | 27 | 0.871 | 31 | 26 | 80 | 2 | 0 | 4.9 | 1 | M | B | |||||
| 35 | 23 | 0.998 | 23 | 23 | 90 | 4 | 1 | 5.3 | 0 | F | A | |||||
| 36 | 23 | 1.001 | 23 | 27 | 75 | 3 | 1 | 4.3 | 0 | F | A | |||||
| 37 | 23.6 | 1.026 | 23 | 22 | 95 | 2 | 1 | 6.2 | 0 | F | A | |||||
| 38 | 60.8 | 1.066 | 57 | 45 | 95 | 11 | 0 | 4.5 | 0 | M | E | |||||
| 39 | 35 | 1.129 | 31 | 27 | 90 | 6 | 1 | 5.5 | 0 | F | B | |||||
| 40 | 24.3 | 1.056 | 23 | 24 | 90 | 2 | 0 | 6.3 | 0 | M | A | |||||
| 41 | 40.5 | 1.013 | 40 | 25 | 80 | 5 | 0 | 4.3 | 0 | M | C | |||||
| 42 | 22.6 | 0.982 | 23 | 32 | 100 | 8 | 1 | 5.7 | 1 | F | A | |||||
| 43 | 76 | 1.134 | 67 | 42 | 95 | 20 | 1 | 5.5 | 0 | F | F | |||||
| 44 | 63.1 | 1.107 | 57 | 45 | 90 | 16 | 0 | 5.2 | 1 | M | E | |||||
| 45 | 52.9 | 1.103 | 48 | 36 | 95 | 8 | 1 | 5.2 | 1 | F | D | |||||
| 46 | 59.1 | 1.036 | 57 | 39 | 75 | 20 | 0 | 3.9 | 1 | M | E | |||||
| 47 | 62.6 | 1.099 | 57 | 37 | 95 | 5 | 0 | 5.5 | 1 | M | E | |||||
| 48 | 67.4 | 1.183 | 57 | 34 | 90 | 11 | 1 | 5.3 | 1 | F | E | |||||
| 49 | 58.4 | 1.025 | 57 | 41 | 95 | 21 | 0 | 6.6 | 0 | M | E | |||||
| 50 | 61.4 | 1.077 | 57 | 38 | 80 | 12 | 0 | 4.6 | 0 | M | E |
Week 1
| Week 1: Descriptive Statistics, including Probability | Salary | Gender1 | ||||||||||||||||||
| While the lectures will examine our equal pay question from the compa-ratio viewpoint, our weekly assignments will focus on | 33.8 | F | ||||||||||||||||||
| examining the issue using the salary measure. | 41.6 | F | ||||||||||||||||||
| 23.6 | F | |||||||||||||||||||
| The purpose of this assignmnent is two fold: | 22.5 | F | ||||||||||||||||||
| 1. Demonstrate mastery with Excel tools. | 24.1 | F | ||||||||||||||||||
| 2. Develop descriptive statistics to help examine the question. | 41.4 | F | ||||||||||||||||||
| 3. Interpret descriptive outcomes | 23.2 | F | ||||||||||||||||||
| 22.5 | F | |||||||||||||||||||
| The first issue in examining salary data to determine if we - as a company - are paying males and females equally for doing equal work is to develop some | 63.8 | F | ||||||||||||||||||
| descriptive statistics to give us something to make a preliminary decision on whether we have an issue or not. | 34.9 | F | ||||||||||||||||||
| 33.4 | F | |||||||||||||||||||
| 1 | Descriptive Statistics: Develop basic descriptive statistics for Salary | 51.4 | F | |||||||||||||||||
| The first step in analyzing data sets is to find some summary descriptive statistics for key variables. | 25.1 | F | ||||||||||||||||||
| Suggestion: Copy the gender1 and salary columns from the Data tab to columns T and U at the right. | 48.6 | F | ||||||||||||||||||
| Then use Data Sort (by gender1) to get all the male and female salary values grouped together. | 24.3 | F | ||||||||||||||||||
| 75.8 | F | |||||||||||||||||||
| a. | Use the Descriptive Statistics function in the Data Analysis tab | Place Excel outcome in Cell K19 | 23.4 | F | ||||||||||||||||
| to develop the descriptive statistics summary for the overall | Column1 | 23 | F | |||||||||||||||||
| group's overall salary. (Place K19 in output range.) | 23 | F | ||||||||||||||||||
| Highlight the mean, sample standard deviation, and range. | Mean | 44.364 | 23.6 | F | ||||||||||||||||
| Standard Error | 2.6490133088 | 35 | F | |||||||||||||||||
| Median | 41.5 | 22.6 | F | |||||||||||||||||
| Mode | 23.6 | 76 | F | |||||||||||||||||
| b. | Using Fx (or formula) functions find the following (be sure to show the formula | Standard Deviation | 18.7313527411 | 52.9 | F | |||||||||||||||
| and not just the value in each cell) asked for salary statistics for each gender: | Sample Variance | 350.8635755102 | 67.4 | F | ||||||||||||||||
| Male | Female | Male | Female | Kurtosis | -1.3765917226 | 58 | M | |||||||||||||
| Mean: | 51.252 | 37.476 | 24.300 | 76.300 | =AVERAGE(T27:T51) | =AVERAGE(T2:T26) | Skewness | 0.2827691781 | 28 | M | ||||||||||
| Sample Standard Deviation: | 17.4001465511 | 17.7408727707 | 22.500 | 76 | =STDEV.S(T27:T51) | =STDEV.S(T2:T26) | Range | 53.8 | 64.8 | M | ||||||||||
| Range: | 52.000 | 53.500 | =H28-G28 | =H29-G29 | Minimum | 22.5 | 48.7 | M | ||||||||||||
| Maximum | 76.3 | 75.4 | M | |||||||||||||||||
| Sum | 2218.2 | 73.8 | M | |||||||||||||||||
| Count | 50 | 58.3 | M | |||||||||||||||||
| 44.7 | M | |||||||||||||||||||
| 2 | Develop a 5-number summary for the overall, male, and female SALARY variable. | 24.9 | M | |||||||||||||||||
| For full credit, use the excel formulas in each cell rather than simply the numerical answer. | 73.5 | M | ||||||||||||||||||
| Overall | Males | Females | Overall | Males | Females | 24.4 | M | |||||||||||||
| Max | 76.300 | 76.300 | 76.000 | =QUARTILE(T$2:T$51,4) | =QUARTILE($T$27:$T$51,4) | =QUARTILE($T$2:$T$26,4) | 39.2 | M | ||||||||||||
| 3rd Q | 60.375 | 62.600 | 48.600 | =QUARTILE(T$2:T$51,3) | =QUARTILE($T$27:$T$51,3) | =QUARTILE($T$2:$T$26,3) | 76.3 | M | ||||||||||||
| Midpoint | 41.500 | 58.000 | 33.400 | =QUARTILE(T$2:T$51,2) | =QUARTILE($T$27:$T$51,2) | =QUARTILE($T$2:$T$26,2) | 48.6 | M | ||||||||||||
| 1st Q | 24.525 | 39.200 | 23.400 | =QUARTILE(T$2:T$51,1) | =QUARTILE($T$27:$T$51,1) | =QUARTILE($T$2:$T$26,1) | 27.5 | M | ||||||||||||
| Min | 22.500 | 24.300 | 22.500 | =QUARTILE(T$2:T$51,0) | =QUARTILE($T$27:$T$51,0) | =QUARTILE($T$2:$T$26,0) | 58 | M | ||||||||||||
| 27 | M | |||||||||||||||||||
| 3 | Location Measures: comparing Male and Female midpoints to the overall Salary data range. | 60.8 | M | |||||||||||||||||
| For full credit, show the excel formulas in each cell rather than simply the numerical answer. | 24.3 | M | ||||||||||||||||||
| Using the entire Salary range and the M and F midpoints found in Q2 | Male | Female | Male | Female | 40.5 | M | ||||||||||||||
| a. What would each midpoint's percentile rank be in the overall range? | 0.645 | 0.42 | Use Excel's =PERCENTRANK.EXC function | =PERCENTRANK.EXC(E38:E42,F40) | =PERCENTRANK.EXC(E38:E42,G40) | 63.1 | M | |||||||||||||
| b. What is the normal curve z value for each midpoint within overall range? | 0.7280 | -0.5853 | Use Excel's =STANDARDIZE function | =STANDARDIZE(F40,L21,L25) | =STANDARDIZE(G40,L21,L25) | 59.1 | M | |||||||||||||
| 62.6 | M | |||||||||||||||||||
| 4 | Probability Measures: comparing Male and Female midpoints to the overall Salary data range | 58.4 | M | |||||||||||||||||
| For full credit, show the excel formulas in each cell rather than simply the numerical answer. | 61.4 | M | ||||||||||||||||||
| Using the entire Salary range and the M and F midpoints found in Q2, find | Male | Female | Male | Female | ||||||||||||||||
| a. The Empirical Probability of equaling or exceeding (=>) that value for | 0.36 | 0.64 | Show the calculation formula = value/50 or =countif(range,">="&cell)/50 | =COUNTIF(T2:T51,">="&F40)/50 | =COUNTIF(T2:T51,">="&G40)/50 | |||||||||||||||
| b. The Normal curve Prob of => that value for each group | 0.233313734430809 | 0.720836669337808 | Use "=1-NORM.S.DIST" function | =1-NORM.S.DIST(I48, TRUE) | =1-NORM.S.DIST(J48, TRUE) | |||||||||||||||
| Note: be sure to use the ENTIRE salary range for part a when finding the probability. | ||||||||||||||||||||
| 5 | Conclusions: What do you make of these results? | Be sure to include findings from this week's lectures as well. | ||||||||||||||||||
| In comparing the overall, male, and female outcomes, what relationship(s) see, to exist between the data sets? | ||||||||||||||||||||
| Your findings: | Although with a lower mean, females end up making more in the long run, with a higher empiical probability as well. | |||||||||||||||||||
| The lecture's related findings: From the lecture, the found the same, females tend to have a higher average of pay than males. | ||||||||||||||||||||
| Overall conclusion: Males tend to start off with a higher salary, but females end up with a higher salary in the peak of their careers. | ||||||||||||||||||||
| What does this suggest about our equal pay for equal work question? It suggests, that equal pay is given for both genders. | ||||||||||||||||||||
| Looking at the mean, standard deviation, and the range the numbers are close, we can say that the pay is pretty much equal. | ||||||||||||||||||||
Week 2
| Week 2: Identifying Significant Differences - part 1 | Salary | Gender | male salaries | female salaries | Male salaries | female salaries | Salary | Degree | Salaries with advanced degree | salary without advanced degree | Salaries with advanced degree | Salaries without advanced degrees | |||||||||||||||||
| 58 | 0 | 58 | 58 | 33.8 | 58 | 0 | 58 | 33.8 | 58 | ||||||||||||||||||||
| To Ensure full credit for each question, you need to show how you got your results. This involves either showing where the data you used is located | 28 | 0 | 28 | 28 | 41.6 | 28 | 0 | 28 | 64.8 | 28 | |||||||||||||||||||
| or showing the excel formula in each cell. | Be sure to copy the appropriate data columns from the data tab to the right for your use this week. | 33.8 | 1 | 33.8 | 64.8 | 23.6 | 33.8 | 1 | 33.8 | 48.7 | 58.3 | ||||||||||||||||||
| 64.8 | 0 | 64.8 | 48.7 | 22.5 | 64.8 | 1 | 64.8 | 75.4 | 41.4 | ||||||||||||||||||||
| As with our examination of compa-ratio in the lecture, the first question we have about salary between the genders involves equality - are they the same or different? | 48.7 | 0 | 48.7 | 75.4 | 24.1 | 48.7 | 1 | 48.7 | 41.6 | 44.7 | |||||||||||||||||||
| What we do, depends upon our findings. | 75.4 | 0 | 75.4 | 73.8 | 41.4 | 75.4 | 1 | 75.4 | 23.6 | 34.9 | |||||||||||||||||||
| 41.6 | 1 | 41.6 | 58.3 | 23.2 | 41.6 | 1 | 41.6 | 73.8 | 33.4 | ||||||||||||||||||||
| 1 | As with the compa-ratio lecture example, we want to examine salary variation within the groups - are they equal? | . | Use Cell K10 for the Excel test outcome location. | 23.6 | 1 | 23.6 | 44.7 | 22.5 | 23.6 | 1 | 23.6 | 22.5 | 25.1 | ||||||||||||||||
| a | What is the data input ranged used for this question: | F-Test Two-Sample for Variances | 73.8 | 0 | 73.8 | 24.9 | 63.8 | 73.8 | 1 | 73.8 | 24.1 | 48.6 | |||||||||||||||||
| V2:W26 on this sheet | 22.5 | 1 | 22.5 | 73.5 | 34.9 | 22.5 | 1 | 22.5 | 23.2 | 24.4 | |||||||||||||||||||
| b | Which is needed for this question: a one- or two-tail hypothesis statement and test ? | Variable 1 | Variable 2 | 24.1 | 1 | 24.1 | 24.4 | 33.4 | 24.1 | 1 | 24.1 | 22.5 | 24.3 | ||||||||||||||||
| Answer: | One-tailed test | Mean | 51.252 | 37.476 | 58.3 | 0 | 58.3 | 39.2 | 51.4 | 58.3 | 0 | 58.3 | 63.8 | 75.8 | |||||||||||||||
| Why: | It uses F-test which is a one-tailed test. | Variance | 302.7651 | 314.7385666667 | 41.4 | 1 | 41.4 | 76.3 | 25.1 | 41.4 | 0 | 41.4 | 24.9 | 76.3 | |||||||||||||||
| Observations | 25 | 25 | 23.2 | 1 | 23.2 | 48.6 | 48.6 | 23.2 | 1 | 23.2 | 73.5 | 48.6 | |||||||||||||||||
| c. Step 1: | Ho: | male salary variance=female salary variance | df | 24 | 24 | 22.5 | 1 | 22.5 | 27.5 | 24.3 | 22.5 | 1 | 22.5 | 51.4 | 27.5 | ||||||||||||||
| Ha: | male salary variance≠female salary variance | F | 0.9619574214 | 44.7 | 0 | 44.7 | 58 | 75.8 | 44.7 | 0 | 44.7 | 39.2 | 23 | ||||||||||||||||
| Step 2: | Significance (Alpha): | 0.05 | P(F<=f) one-tail | 0.4625480001 | 63.8 | 1 | 63.8 | 27 | 23.4 | 63.8 | 1 | 63.8 | 23.4 | 23 | |||||||||||||||
| Step 3: | Test Statistic and test: | Test statistic is the F and test is the F-test | F Critical one-tail | 0.5040933467 | 34.9 | 1 | 34.9 | 60.8 | 23 | 34.9 | 0 | 34.9 | 58 | 23.6 | |||||||||||||||
| Why this test? | Because it tests for the equality of variances | In this case, we test if male population salary variance is equal to female population salary variance | 24.9 | 0 | 24.9 | 24.3 | 23 | 24.9 | 1 | 24.9 | 27 | 60.8 | |||||||||||||||||
| Step 4: | Decision rule: | Reject the null hypothesis if pvalue is less than 0.05 | 33.4 | 1 | 33.4 | 40.5 | 23.6 | 33.4 | 0 | 33.4 | 22.6 | 35 | |||||||||||||||||
| Step 5: | Conduct the test - place test function in cell k10 | 73.5 | 0 | 73.5 | 63.1 | 35 | 73.5 | 1 | 73.5 | 63.1 | 24.3 | ||||||||||||||||||
| 51.4 | 1 | 51.4 | 59.1 | 22.6 | 51.4 | 1 | 51.4 | 52.9 | 40.5 | ||||||||||||||||||||
| Step 6: | Conclusion and Interpretation | 25.1 | 1 | 25.1 | 62.6 | 76 | 25.1 | 0 | 25.1 | 59.1 | 76 | ||||||||||||||||||
| What is the p-value: | 0.4625 | 48.6 | 1 | 48.6 | 58.4 | 52.9 | 48.6 | 0 | 48.6 | 62.6 | 58.4 | ||||||||||||||||||
| What is your decision: REJ or NOT reject the null? | NOT reject the null | 24.4 | 0 | 24.4 | 61.4 | 67.4 | 24.4 | 0 | 24.4 | 67.4 | 61.4 | ||||||||||||||||||
| Why? | p-value>0.05 | 24.3 | 1 | 24.3 | 24.3 | 0 | 24.3 | ||||||||||||||||||||||
| What is your conclusion about the variance in the population for male and female salaries? | 39.2 | 0 | 39.2 | 39.2 | 1 | 39.2 | |||||||||||||||||||||||
| the variance in population and female variances are equal | 75.8 | 1 | 75.8 | 75.8 | 0 | 75.8 | |||||||||||||||||||||||
| 76.3 | 0 | 76.3 | 76.3 | 0 | 76.3 | ||||||||||||||||||||||||
| 2 | Once we know about variance quality, we can move on to means: Are male and female average salaries equal? | Use Cell K35 for the Excel test outcome location. | 48.6 | 0 | 48.6 | 48.6 | 0 | 48.6 | |||||||||||||||||||||
| (Regardless of the outcome of the above F-test, assume equal variances for this test.) | 23.4 | 1 | 23.4 | 23.4 | 1 | 23.4 | |||||||||||||||||||||||
| 27.5 | 0 | 27.5 | 27.5 | 0 | 27.5 | ||||||||||||||||||||||||
| a | What is the data input ranged used for this question: | 58 | 0 | 58 | 58 | 1 | 58 | ||||||||||||||||||||||
| V2:W26 on this sheet | t-Test: Two-Sample Assuming Equal Variances | 27 | 0 | 27 | 27 | 1 | 27 | ||||||||||||||||||||||
| b | Does this question need a one or two-tail hypothesis statement and test? | Two-tailed test | 23 | 1 | 23.0 | 23 | 0 | 23 | |||||||||||||||||||||
| Why: | we are testing for the possibility of the relationship in both directions; either less than or greater than | Variable 1 | Variable 2 | 23 | 1 | 23.0 | 23 | 0 | 23 | ||||||||||||||||||||
| c. Step 1: | Ho: | population mean male salary=population mean female salary | Mean | 51.252 | 37.476 | 23.6 | 1 | 23.6 | 23.6 | 0 | 23.6 | ||||||||||||||||||
| Ha: | population mean male salary≠population mean female salary | Variance | 302.7651 | 314.7385666667 | 60.8 | 0 | 60.8 | 60.8 | 0 | 60.8 | |||||||||||||||||||
| Step 2: | Significance (Alpha): | 0.05 | Observations | 25 | 25 | 35 | 1 | 35.0 | 35 | 0 | 35 | ||||||||||||||||||
| Step 3: | Test Statistic and test: | t-statistic and t-test | Pooled Variance | 308.7518333333 | 24.3 | 0 | 24.3 | 24.3 | 0 | 24.3 | |||||||||||||||||||
| Why this test? | We are testing for the differences between group means | Hypothesized Mean Difference | 0 | 40.5 | 0 | 40.5 | 40.5 | 0 | 40.5 | ||||||||||||||||||||
| Step 4: | Decision rule: | Reject the null hypothesis if p-value is less than 0.05 | df | 48 | 22.6 | 1 | 22.6 | 22.6 | 1 | 22.6 | |||||||||||||||||||
| Step 5: | Conduct the test - place test function in cell K35 | t Stat | 2.7718732548 | 76 | 1 | 76.0 | 76 | 0 | 76 | ||||||||||||||||||||
| P(T<=t) one-tail | 0.003954294 | 63.1 | 0 | 63.1 | 63.1 | 1 | 63.1 | ||||||||||||||||||||||
| Step 6: | Conclusion and Interpretation | t Critical one-tail | 1.6772241961 | 52.9 | 1 | 52.9 | 52.9 | 1 | 52.9 | ||||||||||||||||||||
| What is the p-value: | 0.0079 | P(T<=t) two-tail | 0.0079085881 | 59.1 | 0 | 59.1 | 59.1 | 1 | 59.1 | ||||||||||||||||||||
| What is your decision: REJ or NOT reject the null? | Reject the null hypothesis | t Critical two-tail | 2.0106347576 | 62.6 | 0 | 62.6 | 62.6 | 1 | 62.6 | ||||||||||||||||||||
| Why? | p-value<0.05 | 67.4 | 1 | 67.4 | 67.4 | 1 | 67.4 | ||||||||||||||||||||||
| What is your conclusion about the means in the population for male and female salaries? | 58.4 | 0 | 58.4 | 58.4 | 0 | 58.4 | |||||||||||||||||||||||
| 61.4 | 0 | 61.4 | 61.4 | 0 | 61.4 | ||||||||||||||||||||||||
| The population means for male and female salaries are not equal | |||||||||||||||||||||||||||||
| 3 | Education is often a factor in pay differences. | ||||||||||||||||||||||||||||
| Do employees with an advanced degree (degree = 1) have higher average salaries? | Use Cell K60 for the Excel test outcome location. | ||||||||||||||||||||||||||||
| Note: assume equal variance for the salaries in each degree for this question. | |||||||||||||||||||||||||||||
| a | What is the data input ranged used for this question: | ||||||||||||||||||||||||||||
| AC2:AD26 in this sheet | |||||||||||||||||||||||||||||
| b | Does this question need a one or two-tail hypothesis statement and test? | one-tailed | |||||||||||||||||||||||||||
| Why: | We are only tesing for the possibility in one direction (ie higher average salaries) | t-Test: Two-Sample Assuming Equal Variances | |||||||||||||||||||||||||||
| c. Step 1: | Ho: | Salaries for those with advanced degrees is equal to salaries for those without advanced degrees | |||||||||||||||||||||||||||
| Ha: | Salaries for those with advanced degrees is greater than salaries for those without advanced degrees | Variable 1 | Variable 2 | ||||||||||||||||||||||||||
| Step 2: | Significance (Alpha): | 0.05 | Mean | 45.716 | 43.012 | ||||||||||||||||||||||||
| Step 3: | Test Statistic and test: | t-statistic and t-test | Variance | 382.1664 | 330.3719333333 | ||||||||||||||||||||||||
| Why this test? | We are testing for the equality of means | Observations | 25 | 25 | |||||||||||||||||||||||||
| Step 4: | Decision rule: | Reject the null hypothesis if p-value<0.05 | Pooled Variance | 356.2691666667 | |||||||||||||||||||||||||
| Step 5: | Conduct the test - place test function in cell K60 | Hypothesized Mean Difference | 0 | ||||||||||||||||||||||||||
| df | 48 | ||||||||||||||||||||||||||||
| Step 6: | Conclusion and Interpretation | t Stat | 0.5064919822 | ||||||||||||||||||||||||||
| What is the p-value: | 0.3074 | P(T<=t) one-tail | 0.3074150127 | ||||||||||||||||||||||||||
| Is the t value in the t-distribution tail indicated by the arrow in the Ha claim? | t Critical one-tail | 1.6772241961 | |||||||||||||||||||||||||||
| No | P(T<=t) two-tail | 0.6148300254 | |||||||||||||||||||||||||||
| What is your decision: REJ or NOT reject the null? | NOT reject the null | t Critical two-tail | 2.0106347576 | ||||||||||||||||||||||||||
| Why? | p-value>0.05 | ||||||||||||||||||||||||||||
| What is your conclusion about the impact of education on average salaries? | |||||||||||||||||||||||||||||
| Salaries for those with advanced degrees is equal to salaries for those without advanced degrees | |||||||||||||||||||||||||||||
| Therefore, education does not have an impact on salaries | |||||||||||||||||||||||||||||
| 4 | Considering both the compa-ratio information from the lectures and your salary information, what conclusions can you reach about equal pay for equal work? | ||||||||||||||||||||||||||||
| Your findings: | We can see that there is a significant difference between male and female salaries | ||||||||||||||||||||||||||||
| The lecture's related findings: | It does not appear that the equal pay act is not being violated. It seems that male and femals are being paid around the same. | ||||||||||||||||||||||||||||
| Overall conclusion: | Equal work for equal pay does not hold | ||||||||||||||||||||||||||||
| Why - what statistical results support this conclusion? | |||||||||||||||||||||||||||||
| the t-test for equality of means in number 2 | |||||||||||||||||||||||||||||
Week 3
| Week 3: Identifying Significant Differences - part 2 | Data Input Table: | Salary Range Groups | |||||||||||||||||||||||||
| Group name: | A | B | C | D | E | F | |||||||||||||||||||||
| To Ensure full credit for each question, you need to show how you got your results. This involves either showing where the data you used is located | List salaries within each grade | ||||||||||||||||||||||||||
| or showing the excel formula in each cell. | Be sure to copy the appropriate data columns from the data tab to the right for your use this week. | ||||||||||||||||||||||||||
| 1 | A good pay program will have different average salaries by grade. Is this the case for our company? | ||||||||||||||||||||||||||
| a | What is the data input ranged used for this question: | Use Cell K08 for the Excel test outcome location. | |||||||||||||||||||||||||
| Note: assume equal variances for each grade, even though this may not be accurate, for purposes of this question. | |||||||||||||||||||||||||||
| b. Step 1: | Ho: | ||||||||||||||||||||||||||
| Ha: | |||||||||||||||||||||||||||
| Step 2: | Significance (Alpha): | ||||||||||||||||||||||||||
| Step 3: | Test Statistic and test: | ||||||||||||||||||||||||||
| Why this test? | |||||||||||||||||||||||||||
| Step 4: | Decision rule: | ||||||||||||||||||||||||||
| Step 5: | Conduct the test - place test function in cell K08 | ||||||||||||||||||||||||||
| Step 6: | Conclusion and Interpretation | ||||||||||||||||||||||||||
| What is the p-value: | |||||||||||||||||||||||||||
| What is your decision: REJ or NOT reject the null? | |||||||||||||||||||||||||||
| Why? | |||||||||||||||||||||||||||
| What is your conclusion about the means in the population for grade salaries? | |||||||||||||||||||||||||||
| 2 | If the null hypothesis in question 1 was rejected, which pairs of means differ? | ||||||||||||||||||||||||||
| (Use the values from the ANOVA table to complete the follow table.) | |||||||||||||||||||||||||||
| Groups Compared | Mean Diff. | T value used | +/- Term | Low | to | High | Difference Significant? | Why? | |||||||||||||||||||
| A-B | |||||||||||||||||||||||||||
| A-C | |||||||||||||||||||||||||||
| A-D | |||||||||||||||||||||||||||
| A-E | |||||||||||||||||||||||||||
| A-F | |||||||||||||||||||||||||||
| B-C | |||||||||||||||||||||||||||
| B-D | |||||||||||||||||||||||||||
| B-E | |||||||||||||||||||||||||||
| B-E | |||||||||||||||||||||||||||
| C-D | |||||||||||||||||||||||||||
| C-E | |||||||||||||||||||||||||||
| C-F | |||||||||||||||||||||||||||
| D-E | |||||||||||||||||||||||||||
| D-F | |||||||||||||||||||||||||||
| E-F | |||||||||||||||||||||||||||
| 3 | One issue in salary is the grade an employee is in - higher grades have higher salaries. | ||||||||||||||||||||||||||
| This suggests that one question to ask is if males and females are distributed in a similar pattern across the salary grades? | |||||||||||||||||||||||||||
| a | What is the data input ranged used for this question: | Use Cell K54 for the Excel test outcome location. | |||||||||||||||||||||||||
| b. Step 1: | Ho: | ||||||||||||||||||||||||||
| Ha: | |||||||||||||||||||||||||||
| Step 2: | Significance (Alpha): | ||||||||||||||||||||||||||
| Step 3: | Test Statistic and test: | Place the actual distribution in the table below. | |||||||||||||||||||||||||
| Why this test? | A | B | C | D | E | F | Sum | ||||||||||||||||||||
| Step 4: | Decision rule: | Male | 0 | ||||||||||||||||||||||||
| Step 5: | Conduct the test - place test function in cell K54 | Female | 0 | ||||||||||||||||||||||||
| Sum: | 0 | 0 | 0 | 0 | 0 | 0 | 0 | ||||||||||||||||||||
| Step 6: | Conclusion and Interpretation | Place the expected distribution in the table below. | |||||||||||||||||||||||||
| What is the p-value: | A | B | C | D | E | F | |||||||||||||||||||||
| What is your decision: REJ or NOT reject the null? | Male | 0 | |||||||||||||||||||||||||
| Why? | Female | 0 | |||||||||||||||||||||||||
| What is your conclusion about the means in the population for male and female salaries? | Sum: | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||||||||||||||||||
| 4 | What implications do this week's analysis have for our equal pay question? | ||||||||||||||||||||||||||
| Your findings: | |||||||||||||||||||||||||||
| The lecture's related findings: | |||||||||||||||||||||||||||
| Overall conclusion: | |||||||||||||||||||||||||||
| Why - what statistical results support this conclusion? | |||||||||||||||||||||||||||
Week 4
| Week 4: Identifying relationships - correlations and regression | ||||||||||
| To Ensure full credit for each question, you need to show how you got your results. This involves either showing where the data you used is located | ||||||||||
| or showing the excel formula in each cell. | Be sure to copy the appropriate data columns from the data tab to the right for your use this week. | |||||||||
| 1 | What is the correlation between and among the interval/ratio level variables with salary? (Do not include compa-ratio in this question.) | |||||||||
| a. Create the correlation table. | Use Cell K08 for the Excel test outcome location. | |||||||||
| i. | What is the data input ranged used for this question: | |||||||||
| ii. | Create a correlation table in cell K08. | |||||||||
| b. Technically, we should perform a hypothesis testing on each correlation to determine | ||||||||||
| if it is significant or not. However, we can be faithful to the process and save some | ||||||||||
| time by finding the minimum correlation that would result in a two tail rejection of the null. | ||||||||||
| We can then compare each correlation to this value, and those exceeding it (in either a | ||||||||||
| positive or negative direction) can be considered statistically significant. | ||||||||||
| i. What is the t-value we would use to cut off the two tails? | T = | |||||||||
| ii. What is the associated correlation value related to this t-value? r = | ||||||||||
| c. What variable(s) is(are) significantly correlated to salary? | ||||||||||
| d. Are there any surprises - correlations you though would be significant and are not, or non significant correlations you thought would be? | ||||||||||
| e. Why does or does not this information help answer our equal pay question? | ||||||||||
| 2 | Perform a regression analysis using salary as the dependent variable and all of the variables used in Q1. Add the | |||||||||
| two dummy variables - gender and education - to your list of independent variables. Show the result, and interpret your findings by answering the following questions. | ||||||||||
| Suggestion: Add the dummy variables values to the right of the last data columns used for Q1. | ||||||||||
| What is the multiple regression equation predicting/explaining salary using all of our possible variables except compa-ratio? | ||||||||||
| a. | What is the data input ranged used for this question: | |||||||||
| b. | Step 1: State the appropriate hypothesis statements: | Use Cell M34 for the Excel test outcome location. | ||||||||
| Ho: | ||||||||||
| Ha: | ||||||||||
| Step 2: | Significance (Alpha): | |||||||||
| Step 3: | Test Statistic and test: | |||||||||
| Why this test? | ||||||||||
| Step 4: | Decision rule: | |||||||||
| Step 5: | Conduct the test - place test function in cell M34 | |||||||||
| Step 6: | Conclusion and Interpretation | |||||||||
| What is the p-value: | ||||||||||
| What is your decision: REJ or NOT reject the null? | ||||||||||
| Why? | ||||||||||
| What is your conclusion about the factors influencing the population salary values? | ||||||||||
| c. | If we rejected the null hypothesis, we need to test the significance of each of the variable coefficients. | |||||||||
| Step 1: State the appropriate coefficient hypothesis statements: | (Write a single pair, we will use it for each variable separately.) | |||||||||
| Ho: | ||||||||||
| Ha: | ||||||||||
| Step 2: | Significance (Alpha): | |||||||||
| Step 3: | Test Statistic and test: | |||||||||
| Why this test? | ||||||||||
| Step 4: | Decision rule: | |||||||||
| Step 5: | Conduct the test | |||||||||
| Note, in this case the test has been performed and is part of the Regression output above. | ||||||||||
| Step 6: | Conclusion and Interpretation | |||||||||
| Place the t and p-values in the following table | ||||||||||
| Identify your decision on rejecting the null for each variable. If you reject the null, place the coefficient in the table. | ||||||||||
| Midpoint | Age | Perf. Rat. | Seniority | Raise | Gender | Degree | ||||
| t-value: | ||||||||||
| P-value: | ||||||||||
| Rejection Decision: | ||||||||||
| If Null is rejected, what is the variable's coefficient value? | ||||||||||
| Using the intercept coefficient and only the significant variables, what is the equation? | ||||||||||
| Salary = | ||||||||||
| d. | Is gender a significant factor in salary? | |||||||||
| e. | Regardless of statistical significance, who gets paid more with all other things being equal? | |||||||||
| f. | How do we know? | |||||||||
| 3 | After considering the compa-ratio based results in the lectures and your salary based results, what else would you like to know | |||||||||
| before answering our question on equal pay? Why? | ||||||||||
| 4 | Between the lecture results and your results, what is your answer to the question | |||||||||
| of equal pay for equal work for males and females? Why? | ||||||||||
| Your findings: | ||||||||||
| The lecture's related findings: | ||||||||||
| Overall conclusion: | ||||||||||
| 5 | What does regression analysis show us about analyzing complex measures? | |||||||||