The first step in analyzing data sets is to find some summary descriptive statistics for key variables. Since the assignment problems will
| This assignment covers the material presented in weeks 1 and 2. | Six Questions | |||||||||
| Before starting this assignment, make sure the the assignment data from the Employee Salary Data Set file is copied over to this Assignment file. | ||||||||||
| You can do this either by a copy and paste of all the columns or by opening the data file, right clicking on the Data tab, selecting Move or Copy, and copying the entire sheet to this file | ||||||||||
| (Weekly Assignment Sheet or whatever you are calling your master assignment file). | ||||||||||
| It is highly recommended that you copy the data columns (with labels) and paste them to the right so that whatever you do will not disrupt the original data values and relationships. | ||||||||||
| To Ensure full credit for each question, you need to show how you got your results. For example, Question 1 asks for several data values. If you obtain them using descriptive statistics, | ||||||||||
| then the cells should have an “=XX” formula in them, where XX is the column and row number showing the value in the descriptive statistics table. If you choose to generate each | ||||||||||
| value using fxfunctions, then each function should be located in the cell and the location of the data values should be shown. | ||||||||||
| So, Cell D31 – as an example – shoud contain something like “=T6” or “=average(T2:T26)”. Having only a numerical value will not earn full credit. | ||||||||||
| The reason for this is to allow instructors to provide feedback on Excel tools if the answers are not correct – we need to see how the results were obtained. | ||||||||||
| In starting the analysis on a research question, we focus on overall descriptive statistics and seeing if differences exist. Probing into reasons and mitigating factors is a follow-up activity. | ||||||||||
| 1 | The first step in analyzing data sets is to find some summary descriptive statistics for key variables. Since the assignment problems will | |||||||||
| focus mostly on the compa-ratios, we need to find the mean, standard deviations, and range for our groups: Males, Females, and Overall. | ||||||||||
| Sorting the compa-ratios into male and females will require you copy and paste the Compa-ratio and Gender1 columns, and then sort on Gender1. | ||||||||||
| The values for age, performance rating, and service are provided for you for future use, and – if desired – to test your approach to the compa-ratio answers | ||||||||||
| (see if you can replicate the values). | ||||||||||
| You can use either the Data Analysis Descriptive Statistics tool or the Fx =average and =stdev functions. | ||||||||||
| The range can be found using the difference between the =max and =min functions with Fx functions or from Descriptive Statistics. | ||||||||||
| Suggestion: Copy and paste the compa-ratio data to the right (Column T) and gender data in column U. | ||||||||||
| If you use Descriptive statistics, Place the output table in row 1 of a column to the right. | ||||||||||
| If you did not use Descriptive Statistics, make sure your cells show the location of the data (Example: =average(T2:T51) | ||||||||||
| Compa-ratio | Age | Perf. Rat. | Service | |||||||
| Overall | Mean | 1.0616 | 35.7 | 85.9 | 9.0 | |||||
| Standard Deviation | 0.0820501986715500 | 8.2513 | 11.4147 | 5.7177 | Note – remember the data is a sample from the larger company population | |||||
| Range | 0.399 | 30 | 45 | 21 | ||||||
| Female | Mean | 1.1163 | 32.5 | 84.2 | 7.9 | |||||
| Standard Deviation | 0.083183472 | 6.9 | 13.6 | 4.9 | ||||||
| Range | 0.329 | 26.0 | 45.0 | 18.0 | ||||||
| Male | Mean | 1.0516 | 38.9 | 87.6 | 10.0 | |||||
| Standard Deviation | 0.081345907 | 8.4 | 8.7 | 6.4 | ||||||
| Range | 0.321 | 28.0 | 30.0 | 21.0 | ||||||
| A key issue in comparing data sets is to see if they are distributed/shaped the same. At this point we can do this | ||||||||||
| by looking at the probabilities that males and females are distributed in the same way for a grade levels. | ||||||||||
| 2 | Empirical Probability: What is the probability for a: | Probability | ||||||||
| a. Randomly selected person being in grade E or above? | 18/50 | |||||||||
| b. Randomly selected person being a male in grade E or above? | 13/50 | |||||||||
| c. Randomly selected male being in grade E or above? | 13/25 | |||||||||
| d. Why are the results different? | the sample size increased | |||||||||
| 3 | Normal Curve based probability: For each group (overall, females, males), what are the values for each question below?: | |||||||||
| Make sure your answer cells show the Excel function and cell location of the data used. | ||||||||||
| A | The probability of being in the top 1/3 of the compa-ratio distribution. | |||||||||
| Note, we can find the cutoff value for the top 1/3 using the fx Large function: =large(range, value). | ||||||||||
| Value is the number that identifies the x-largest value. For the top 1/3 value would be the value that starts the top 1/3 of the range, | ||||||||||
| For the overall group, this would be the 50/3 or 17th (rounded), for the gender groups, it would be the 25/3 = 8th (rounded) value. | ||||||||||
| Overall | Female | |||||||||
| i. | How nany salaries are in the top 1/3 (rounded to nearest whole number) for each group? | 61 | 61 | |||||||
| ii | What Compa-ratio value starts the top 1/3 of the range for each group? | 1.093 | 1.116 | |||||||
| iii | What is the z-score for this value? | 0.382205 | -0.0035 | |||||||
| iv. | What is the normal curve probability of exceeding this score? | 0.648845 | 0.4986 | |||||||
| B | How do you interpret the relationship between the data sets? What does this suggest about our equal pay for equal work question? | |||||||||
| The data is very scewed. There is a obvious difference in pay according to gender for equal work. | ||||||||||
| 4 | Based on our sample data set, can the male and female compa-ratios in the population be equal to each other? | |||||||||
| A | First, we need to determine if these two groups have equal variances, in order to decide which t-test to use. | |||||||||
| What is the data input ranged used for this question: | ||||||||||
| Step 1: | Ho: | male mean salary = female mean salary | ||||||||
| Ha: | male mean salary not equal to female mean salary | |||||||||
| Step 2: | Decision Rule: | Reject Ho if p -value is less than Alpha = 0.025 for one tail | ||||||||
| Step 3: | Statistical test: | F-test | ||||||||
| Why? | F-test is for variance. Since we are testing for equality, there are 2 sample test and rejection region is in both tails and makes each tail 0.25 | |||||||||
| Step 4: | Conduct the test – place cell B77 in the output location box. | |||||||||
| 0.913760913 | F-Test Two-Sample for Variances | |||||||||
| Descriptive Statistics | Ratio of variances Var[0.947]/Var[0.877] | F [larger/smaller] | ||||||||
| 0.947 | 0.877 | F | 1.17981 | F | 1.17981 | |||||
| Sample size | 24 | 24 | F Critical value (5%) | 2.01442 | F Critical value (5%) | 2.01442 | ||||
| Mean | 1.07683 | 1.05892 | F Critical value (5%) 2-tailed | 2.31164 | F Critical value (5%) 2-tailed | 2.31164 | ||||
| Variance | 0.00652 | 0.00552 | p-level 2-tailed (H1: F ≠ 1) | 0.69506 | H1 rejected | H0 F=1 (5%)? | accepted | |||
| Standard Deviation | 0.08073 | 0.07432 | p-level 1-tailed (H1: F > 1) | 0.34753 | H1 rejected | |||||
| Mean Standard Error | 0.01648 | 0.01517 | p-level 1-tailed (H1: F < 1) | 0.65247 | H1 rejected | |||||
| Step 5: | Conclusion and Interpretation | p-level 2-tailed (H1: F ≠ 1) | 0.69506 | H1 rejected | ||||||
| What is the p-value: | p-level 1-tailed (H1: F > 1) | 0.34753 | H1 rejected | |||||||
| p-level 1-tailed (H1: F < 1) | 0.65247 | H1 rejected | ||||||||
| Is the P-value < 0.05 (for a one tail test) or 0.025 (for a two tail test)? | ||||||||||
| neither | ||||||||||
| What is your decision: REJ or NOT reject the null? | rejected | |||||||||
| What does this result say about our question of variance equality? | ||||||||||
| male mean salary not equal to female mean salary | ||||||||||
| B | Are male and female average compa-ratios equal? | |||||||||
| (Regardless of the outcome of the above F-test, assume equal variances for this test.) | ||||||||||
| What is the data input ranged used for this question: | ||||||||||
| Step 1: | Ho: | male mean salary = female mean salary | ||||||||
| Ha: | male mean salary not equal to female mean salary | |||||||||
| Step 2: | Decision Rule: | Reject the null hypothesis is less than Alpha = 0.05 | ||||||||
| Step 3: | Statistical test: | 2 sample unequal variance T-Test | ||||||||
| Why? | p-value greater than rejection alpha | |||||||||
| Step 4: | Conduct the test – place cell B109 in the output location box. | |||||||||
| 0.394338636 | ||||||||||
| Step 5: | Conclusion and Interpretation | |||||||||
| What is the p-value: | ||||||||||
| Is the P-value < 0.05 (for a one tail test) or 0.025 (for a two tail test)? | ||||||||||
| What is your decision: REJ or NOT reject the null? | ||||||||||
| What does your decision on rejecting the null hypothesis mean? | ||||||||||
| If the null hypothesis was rejected, calculate the effect size value: | ||||||||||
| If the effect size was calculated, what doe the result mean in terms of why the null hypothesis was rejected? | ||||||||||
| What does the result of this test tell us about our question on salary equality? | ||||||||||
| 5 | Is the Female average compa-ratio equal to or less than the midpoint value of 1.00? | |||||||||
| This question is the same as: Does the company, pay its females – on average – at or below the grade midpoint (which is considered the market rate)? | ||||||||||
| Suggestion: Use the data column T to the right for your null hypothesis value. | ||||||||||
| What is the data input ranged used for this question: | ||||||||||
| Step 1: | Ho: | |||||||||
| Ha: | ||||||||||
| Step 2: | Decision Rule: | |||||||||
| Step 3: | Statistical test: | |||||||||
| Why? | ||||||||||
| Step 4: | Conduct the test – place cell B162 in the output location box. | |||||||||
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