Fundamental Statistics for the Behavior Sciences - Assignment Example

Wk3 Assignment Wk3 Assignment Question The following table was generated in order to perform Analysis of Variance using Analysis Tool Pack of MS Excel 2007. A B C D E F 23 27 41 47 58 76 22 34 42 57 66 77 23 36 47 50 60 76 24 34 40 49 69 75 24 28 43 55 64 72 24 28     56 77 23 35     60   24       65   24       62   24       65   24       60   23       66   22           25           24           The ANOVA test employed in this case is single factor because there is only one variable that has been used to group data. Moreover the ANOVA test has been employed assuming equal variances. Therefore, the output summary is given below. Anova: Single Factor SUMMARY Groups Count Sum Average Variance A 15 353 23.53333 0.695238 B 7 222 31.71429 14.90476 C 5 213 42.6 7.3 D 5 258 51.6 17.8 E 12 751 62.58333 14.81061 F 6 453 75.5 3.5 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 17686.02 5 3537.204 409.5941 1.04E-35 2.42704 Within Groups 379.9786 44 8.635877 Total 18066 49         Hypothesis Testing Ho: µA=µB=µC= µD=µE=µF H1: µA≠µB≠µC≠ µD≠µE≠µF Where µA shows the average salaries for Grade A, µB shows the average salaries for Grade B, µC shows the average salaries for Grade C, µD shows the average salaries for Grade D, µE shows the average salaries for Grade E and µF shows the average salaries of Grade F employees. Level of significance The level of significance chosen for hypothesis testing is 0.05. Decision Rule We reject Ho, if the P-value is found to be less than 0.05 significance level. Test Statistics The test statistics employed to test the sample means is F statistics while the decision making criteria involves the use of P-value. Decision Since the P-value, as mentioned in the table, is found to be less than 0.05, therefore, we cannot accept null hypothesis which states that the average salaries of employees of each grade is equal. It can thus be concluded that significant difference exists in the average salaries of employees at different grades. Question #02 Anova: Two-Factor With Replication SUMMARY A B C D E F Total M               Count 2 2 2 2 2 2 12 Sum 49 55 87 96 122 153 562 Average 24.5 27.5 43.5 48 61 76.5 46.8333 Variance 0.5 0.5 24.5 2 50 0.5 364.515 F               Count 2 2 2 2 2 2 12 Sum 46 70 83 107 134 152 592 Average 23 35 41.5 53.5 67 76 49.3333 Variance 2 2 0.5 24.5 8 2 367.333 Total             Count 4 4 4 4 4 4 Sum 95 125 170 203 256 305 Average 23.75 31.25 42.5 50.75 64 76.25 Variance 1.58333 19.5833 9.66667 18.9167 31.3333 0.91667 ANOVA Source of Variation SS df MS F P-value F crit Sample 37.5 1 37.5 3.84615 0.07348 4.74723 Columns 7841.83 5 1568.37 160.858 1.5E-10 3.10588 Interaction 91.5 5 18.3 1.87692 0.17231 3.10588 Within 117 12 9.75 Total 8087.83 23         Interpretation of output The first null hypothesis that states that average salaries for all grades are equal is accepted as the associated P value is found to be equal to 0.07 which is greater than 0.05. The second null hypothesis that states that average salaries for both male and female employees are equal is, however, rejected as the associated P value is found to be equal to 1.5 x 10-10 which is quite less than the significance level 0.05. The third null hypothesis that tests whether the interaction is statistically significant is found to be statistically significant as the associated P value is found to be 0.17 which is significantly greater than the significance level of 0.05. It means that the grades and gender do have an interaction that impacts the equality of pays amongst employees on average. Moreover it shows that the mean salaries are equal amongst different grades while they are found to be unequal amongst male and female employees without considering their grades. Moreover in order to avoid biasness in the results the observations have been chosen at random so that the sample results can be employed for the interpretation of population parameters. Therefore mean value in this case can be used to interpret the results of mean values of population. Question #3 In order to test that the compa values in the population are equal by grade and/or gender, and are independent of each factor the ANOVA has been done using Data Analysis Tool pack of MS Excel 2007. The values are chosen on convenience basis for each category. The summary of outputs thus obtained is mentioned below. The two factor ANOVA has been performed with replication. Anova: Two-Factor With Replication SUMMARY A B C D E F Total Male               Count 2 2 2 2 2 2 12 Sum 2.086 1.773 2.175 1.999 2.209 2.283 12.525 Average 1.043 0.8865 1.0875 0.9995 1.1045 1.1415 1.04375 Variance 0 0.000545 0.015313 0.000841 0.005512 0.000113 0.009638 Female               Count 2 2 2 2 2 2 12 Sum 1.956 2.257 2.075 2.228 2.35 2.329 13.195 Average 0.978 1.1285 1.0375 1.114 1.175 1.1645 1.099583 Variance 0.000968 0.002112 0.000313 0.010658 0.00245 0.004141 0.007255 Total       Count 4 4 4 4 4 4 Sum 4.042 4.03 4.25 4.227 4.559 4.612 Average 1.0105 1.0075 1.0625 1.05675 1.13975 1.153 Variance 0.001731 0.020407 0.006042 0.008203 0.004311 0.001594 ANOVA Source of Variation SS df MS F P-value F crit Sample 0.018704 1 0.018704 5.224141 0.041256 4.747225 Columns 0.077663 5 0.015533 4.338302 0.01736 3.105875 Interaction 0.065194 5 0.013039 3.641802 0.030919 3.105875 Within 0.042964 12 0.00358 Total 0.204525 23         The hypotheses in this case will be Ho: Average Compaq values are equal for all grades Ha: Average Compaq values are not equal for all grades Ho: Average Compaq values by gender are equal Ha: Average Compaq values by gender are not equal Ho: Interaction is not significant Ha: Interaction is significant Interpretation of outputs All three null hypotheses are rejected as the associated P values are found to be less than the significance level of 0.05. It means that average Compaq values are not equal by gender as well as by grade. Moreover the interaction is also found to be insignificant. It can thus be concluded that on average the Compaq values are found to be significantly different amongst employees on the basis of gender as well as on the basis of grades. Question #04 Using the Pivot Table option the following table was generated that contains the average ages of employees grouped by gender and grade. The reason of choosing “Age” as variable for analysis is that, I wanted to test whether male and female employees are hired on each level at same ages or any distinction exists on the basis of gender. Average of Age Column Labels Row Labels A B C D E F Grand Total F 30 33 31 38 31 43 33 M 32 34 35 41 41 45 39 Grand Total 30 34 33 39 39 44 36 Now this data table is utilized to perform two-way ANOVA without replication using Data Analysis Tool pack of MS Excel 2007. This test requires that there is no interaction between the groups of sample and so the concerned variable is tested on the basis of each group independent of the other. The summary of output is mentioned below. Anova: Two-Factor Without Replication SUMMARY Count Sum Average Variance F 6 205.3333333 34.22222222 27.26296296 M 6 227.6333333 37.93888889 23.94685185 A 2 62.16666667 31.08333333 3.125 B 2 67.33333333 33.66666667 0.888888889 C 2 65.66666667 32.83333333 6.722222222 D 2 78.5 39.25 3.125 E 2 71.3 35.65 53.045 F 2 88 44 2 ANOVA Source of Variation SS df MS F P-value F crit Rows 41.44083333 1 41.44083333 7.544222503 0.04047 6.607891 Columns 228.5837963 5 45.71675926 8.322646439 0.018169 5.050329 Error 27.46527778 5 5.493055556 Total 297.4899074 11         Hypotheses In order to perform Analysis of Variance the following hypotheses are drawn. Ho: The mean age of employees is equal by gender Ha: The mean age of employees is not equal by gender Ho: The mean age of employees is equal for all grades Ha: The mean age of employees is not equal for all grades Interpretation The first null hypothesis is rejected at 0.05 significance level as the associated P-value is found to be less than 0.05. It shows that there is a significant difference amongst mean ages of male and female employees. The second null hypothesis is also rejected at 0.05 significance level as the associated P-value is found to be less than 0.05. Therefore, it can be said that mean age varies amongst employees at various grades. Question#05 From the analysis performed it seems that there exists a considerable variation amongst the salaries on the basis of gender and the concept of equal pay for equal work does not seem to have been implemented in its true sense. However, these conclusions are based on the statistical analysis performed which may be biased due to the selection of values for each variable or biasness may also have resulted during the collection of data (Howell, 2011). References Howell, D. C. (2011). Fundamental Statistics for the Behavior Sciences, 7th ed. Belmont, CA: Wadsworth, Cengage Learning. Read More