Project 3
Jacob Huebner
11/26/2019
Mean Differences
Intro
For this project, you should use the data file labeled ‘Project 3 Data’ available on the Google Classroom site. These data are taken from publicly available data regarding CPS employees https://cps.edu/About_CPS/Financial_information/Pages/EmployeePositionFiles.aspx
Objectives
Use the data contained in the files.
Answer 8 questions based on your R experimentation.
Outline
Independent Sample t-tests
Paired sample t-tests
One-way ANOVA
Independent Sample t-tests
Run the following independent sample t-tests:
DV = Annual Salary 19, IV = Principal versus Assistant Principals
DV = Annual Salary 19, IV = Special Education Classroom Assistant I versus Special Education Classroom Assistant II
DV = Annual Salary 19, IV = Lunchroom Attendant versus Custodial Worker
- Report the results of each of these three analyses and provide an interpretation of the results (including tests of the key assumptions). Your results should be written as if you were preparing them for inclusion in a research paper.
Part 1
DV = Annual Salary 19, IV = Principal versus Assistant Principals
# Code
# Create subset of data set that only includes data points with the job title Principal and Assistant Principal
indepentenddata1 <- subset(Proj3, Job.Title == "Principal" | Job.Title == "Assistant Principal")
# Look at annual salary by job title "Principal" and "Assistant principle"
ttest(Annual.Salary19 ~ Job.Title, data = indepentenddata1)##
## Compare Annual.Salary19 across Job.Title levels Principal and Assistant Principal
## --------------------------------------------------------------
##
##
## ------ Description ------
##
## Annual.Salary19 for Job.Title Principal: n.miss = 0, n = 377, mean = 147641.180, sd = 8947.850
## Annual.Salary19 for Job.Title Assistant Principal: n.miss = 0, n = 413, mean = 116026.995, sd = 8310.191
##
## Sample Mean Difference of Annual.Salary19: 31614.185
##
## Within-group Standard Deviation: 8620.340
##
##
## ------ Assumptions ------
##
## Note: These hypothesis tests can perform poorly, and the
## t-test is typically robust to violations of assumptions.
## Use as heuristic guides instead of interpreting literally.
##
## Null hypothesis, for each group, is a normal distribution of Annual.Salary19.
## Group Principal: Sample mean assumed normal because n>30, so no test needed.
## Group Assistant Principal: Sample mean assumed normal because n>30, so no test needed.
##
## Null hypothesis is equal variances of Annual.Salary19, i.e., homogeneous.
## Variance Ratio test: F = 80064019.308/69059267.408 = 1.159, df = 376;412, p-value = 0.142
## Levene's test, Brown-Forsythe: t = 2.522, df = 788, p-value = 0.012
##
##
## ------ Inference ------
##
## --- Assume equal population variances of Annual.Salary19 for each Job.Title
##
## t-cutoff: tcut = 1.963
## Standard Error of Mean Difference: SE = 614.034
##
## Hypothesis Test of 0 Mean Diff: t = 51.486, df = 788, p-value = 0.000
##
## Margin of Error for 95% Confidence Level: 1205.335
## 95% Confidence Interval for Mean Difference: 30408.850 to 32819.521
##
##
## --- Do not assume equal population variances of Annual.Salary19 for each Job.Title
##
## t-cutoff: tcut = 1.963
## Standard Error of Mean Difference: SE = 616.105
##
## Hypothesis Test of 0 Mean Diff: t = 51.313, df = 767.159, p-value = 0.000
##
## Margin of Error for 95% Confidence Level: 1209.451
## 95% Confidence Interval for Mean Difference: 30404.734 to 32823.637
##
##
## ------ Effect Size ------
##
## --- Assume equal population variances of Annual.Salary19 for each Job.Title
##
## Standardized Mean Difference of Annual.Salary19, Cohen's d: 3.667
##
##
## ------ Practical Importance ------
##
## Minimum Mean Difference of practical importance: mmd
## Minimum Standardized Mean Difference of practical importance: msmd
## Neither value specified, so no analysis
##
##
## ------ Graphics Smoothing Parameter ------
##
## Density bandwidth for Job.Title Principal: 3113.308
## Density bandwidth for Job.Title Assistant Principal: 1957.381
Results
The mean salary for the job title ‘Assistant Principal’ is 116026.995. The standard deviation for the job title ‘Assistant Principal’ is 8310.191. The mean salary for the job title ‘Principal’ is 147641.180. The standard deviation for the job title ‘Principal’ is 8947.850. The degrees of freedom is 788. The t-value is 51.486. The p-value is 0.000.
Background
The null hypothesis is that the two groups have the same mean. We want to disprove the null hypothesis using the results from the t-test. If the p is less than 0.05, then we reject the null hypothesis. If the p is greater than 0.05, then we don’t have enough infomration to be statistically different.
Interpretation
The p-value of 0.000 is less than 0.05, so we disprove the null hypothesis. This means that Assistant Principles are not paid the same as Principals on average. Or more accurately, Principals are paid more than Assistant Principles on average.
Assumptions
Null hypothesis, for each group, is a normal distribution of Annual.Salary19. Group Principal: Sample mean assumed normal because n>30, so no test needed. Group Assistant Principal: Sample mean assumed normal because n>30, so no test needed.
Null hypothesis is equal variances of Annual.Salary19, i.e., homogeneous. Variance Ratio test: F = 80064019.308/69059267.408 = 1.159, df = 376;412, p-value = 0.142 Levene’s test, Brown-Forsythe: t = 2.522, df = 788, p-value = 0.012
APA Style Answer
Assistant Principles did have significantly different Annual Salaries (M = 116,027.00, SD = 8310.19) than Principles (M = 147641.18, SD = 8947.85), t(788) = 51.49, ns (two-tailed).
Part 2
DV = Annual Salary 19, IV = Special Education Classroom Assistant I versus Special Education Classroom Assistant II
# Code
# Create subset of data set that only includes data points with the job title Special Ed Classroom Assist and Special Ed Classroom Assist II
indepentenddata2 <- subset(Proj3, Job.Title == "Special Ed Classroom Assist" | Job.Title == "Special Ed Classroom Assist II")
# Look at annual salary by job title "Special Ed Classroom Assist" and "Special Ed Classroom Assist II"
ttest(Annual.Salary19 ~ Job.Title, data = indepentenddata2)##
## Compare Annual.Salary19 across Job.Title levels Special Ed Classroom Assist II and Special Ed Classroom Assist
## --------------------------------------------------------------
##
##
## ------ Description ------
##
## Annual.Salary19 for Job.Title Special Ed Classroom Assist II: n.miss = 0, n = 1396, mean = 37284.754, sd = 3988.161
## Annual.Salary19 for Job.Title Special Ed Classroom Assist: n.miss = 0, n = 1260, mean = 37220.454, sd = 3784.195
##
## Sample Mean Difference of Annual.Salary19: 64.300
##
## Within-group Standard Deviation: 3892.737
##
##
## ------ Assumptions ------
##
## Note: These hypothesis tests can perform poorly, and the
## t-test is typically robust to violations of assumptions.
## Use as heuristic guides instead of interpreting literally.
##
## Null hypothesis, for each group, is a normal distribution of Annual.Salary19.
## Group Special Ed Classroom Assist II: Sample mean assumed normal because n>30, so no test needed.
## Group Special Ed Classroom Assist: Sample mean assumed normal because n>30, so no test needed.
##
## Null hypothesis is equal variances of Annual.Salary19, i.e., homogeneous.
## Variance Ratio test: F = 15905427.880/14320133.093 = 1.111, df = 1395;1259, p-value = 0.057
## Levene's test, Brown-Forsythe: t = -1.828, df = 2654, p-value = 0.068
##
##
## ------ Inference ------
##
## --- Assume equal population variances of Annual.Salary19 for each Job.Title
##
## t-cutoff: tcut = 1.961
## Standard Error of Mean Difference: SE = 151.266
##
## Hypothesis Test of 0 Mean Diff: t = 0.425, df = 2654, p-value = 0.671
##
## Margin of Error for 95% Confidence Level: 296.611
## 95% Confidence Interval for Mean Difference: -232.311 to 360.911
##
##
## --- Do not assume equal population variances of Annual.Salary19 for each Job.Title
##
## t-cutoff: tcut = 1.961
## Standard Error of Mean Difference: SE = 150.860
##
## Hypothesis Test of 0 Mean Diff: t = 0.426, df = 2647.365, p-value = 0.670
##
## Margin of Error for 95% Confidence Level: 295.816
## 95% Confidence Interval for Mean Difference: -231.515 to 360.116
##
##
## ------ Effect Size ------
##
## --- Assume equal population variances of Annual.Salary19 for each Job.Title
##
## Standardized Mean Difference of Annual.Salary19, Cohen's d: 0.017
##
##
## ------ Practical Importance ------
##
## Minimum Mean Difference of practical importance: mmd
## Minimum Standardized Mean Difference of practical importance: msmd
## Neither value specified, so no analysis
##
##
## ------ Graphics Smoothing Parameter ------
##
## Density bandwidth for Job.Title Special Ed Classroom Assist II: 1024.142
## Density bandwidth for Job.Title Special Ed Classroom Assist: 1034.600
Results
The mean salary for the job title ‘Special Ed Classroom Assist’ is 37,220.454 The standard deviation for the job title ‘Special Ed Classroom Assist’ is 3784.195. The mean salary for the job title ‘Special Ed Classroom Assist II’ is 37,284.754. The standard deviation for the job title ‘Special Ed Classroom Assist II’ is 3988.161. The degrees of freedom is 2654. The t-value is 0.425. The p-value is 0.671.
Background
The null hypothesis is that the two groups have the same mean. We want to disprove the null hypothesis using the results from the t-test. If the p is less than 0.05, then we reject the null hypothesis. If the p is greater than 0.05, then we don’t have enough infomration to be statistically different.
Interpretation
The p-value of 0.000 is greater than 0.05, so we can’t disprove the null hypothesis. This means that there is not enough information to prove that there is a significant difference between the salaries of Special Ed Classroom Assistants I and Special Ed Classroom Assistants II. More accurately, Special Ed Classroom Assistants I are paid the same as Special Ed Classroom Assistants II.
Assumptions
Null hypothesis, for each group, is a normal distribution of Annual.Salary19. Group Special Ed Classroom Assist II: Sample mean assumed normal because n>30, so no test needed. Group Special Ed Classroom Assist: Sample mean assumed normal because n>30, so no test needed.
Null hypothesis is equal variances of Annual.Salary19, i.e., homogeneous. Variance Ratio test: F = 15905427.880/14320133.093 = 1.111, df = 1395;1259, p-value = 0.057 Levene’s test, Brown-Forsythe: t = -1.828, df = 2654, p-value = 0.068
APA Style Answer
Special Ed Classroom Assistants I did not have significantly different Annual Salaries (M = 37,220.45, SD = 3784.20) than Special Ed Classroom Assistants II (M = 37,284.75, SD = 3988.16), t(2654) = 0.425, ns (two-tailed).
Part 3
DV = Annual Salary 19, IV = Lunchroom Attendant versus Custodial Worker
# Code
# Create subset of data set that only includes data points with the job title Lunchroom Attendant and Custodial Worker
indepentenddata3 <- subset(Proj3, Job.Title == "Lunchroom Attendant" | Job.Title == "Custodial Worker")
# Look at annual salary by job title "Lunchroom Attendant" and "Custodial Worker"
ttest(Annual.Salary19 ~ Job.Title, data = indepentenddata3)##
## Compare Annual.Salary19 across Job.Title levels Custodial Worker and Lunchroom Attendant
## --------------------------------------------------------------
##
##
## ------ Description ------
##
## Annual.Salary19 for Job.Title Custodial Worker: n.miss = 0, n = 547, mean = 35362.146, sd = 4336.202
## Annual.Salary19 for Job.Title Lunchroom Attendant: n.miss = 0, n = 690, mean = 18189.778, sd = 2977.944
##
## Sample Mean Difference of Annual.Salary19: 17172.368
##
## Within-group Standard Deviation: 3641.462
##
##
## ------ Assumptions ------
##
## Note: These hypothesis tests can perform poorly, and the
## t-test is typically robust to violations of assumptions.
## Use as heuristic guides instead of interpreting literally.
##
## Null hypothesis, for each group, is a normal distribution of Annual.Salary19.
## Group Custodial Worker: Sample mean assumed normal because n>30, so no test needed.
## Group Lunchroom Attendant: Sample mean assumed normal because n>30, so no test needed.
##
## Null hypothesis is equal variances of Annual.Salary19, i.e., homogeneous.
## Variance Ratio test: F = 18802648.590/8868151.227 = 2.120, df = 546;689, p-value = 0.000
## Levene's test, Brown-Forsythe: t = 1.805, df = 1235, p-value = 0.071
##
##
## ------ Inference ------
##
## --- Assume equal population variances of Annual.Salary19 for each Job.Title
##
## t-cutoff: tcut = 1.962
## Standard Error of Mean Difference: SE = 208.469
##
## Hypothesis Test of 0 Mean Diff: t = 82.374, df = 1235, p-value = 0.000
##
## Margin of Error for 95% Confidence Level: 408.993
## 95% Confidence Interval for Mean Difference: 16763.375 to 17581.361
##
##
## --- Do not assume equal population variances of Annual.Salary19 for each Job.Title
##
## t-cutoff: tcut = 1.963
## Standard Error of Mean Difference: SE = 217.317
##
## Hypothesis Test of 0 Mean Diff: t = 79.020, df = 927.836, p-value = 0.000
##
## Margin of Error for 95% Confidence Level: 426.489
## 95% Confidence Interval for Mean Difference: 16745.879 to 17598.857
##
##
## ------ Effect Size ------
##
## --- Assume equal population variances of Annual.Salary19 for each Job.Title
##
## Standardized Mean Difference of Annual.Salary19, Cohen's d: 4.716
##
##
## ------ Practical Importance ------
##
## Minimum Mean Difference of practical importance: mmd
## Minimum Standardized Mean Difference of practical importance: msmd
## Neither value specified, so no analysis
##
##
## ------ Graphics Smoothing Parameter ------
##
## Density bandwidth for Job.Title Custodial Worker: 1400.246
## Density bandwidth for Job.Title Lunchroom Attendant: 918.098
Results
The mean salary for the job title ‘Lunchroom Attendant’ is 18,189.778. The standard deviation for the job title ‘Lunchroom Attendant’ is 2977.944. The mean salary for the job title ‘Custodial Worker’ is 35,362.146. The standard deviation for the job title ‘Custodial Worker’ is 4336.202. The degrees of freedom is 1235. The t-value is 82.374. The p-value is 0.000
Background
The null hypothesis is that the two groups have the same mean. We want to disprove the null hypothesis using the results from the t-test. If the p is less than 0.05, then we reject the null hypothesis. If the p is greater than 0.05, then we don’t have enough infomration to be statistically different.
Interpretation
The p-value of 0.000 is less than 0.05, so we disprove the null hypothesis. This means that Lunchroom Attendants are not paid the same as Custodial Workers on average. Or more accurately, Custodial Workers are paid more than Lunchroom Attendants on average.
Assumptions
Null hypothesis, for each group, is a normal distribution of Annual.Salary19. Group Custodial Worker: Sample mean assumed normal because n>30, so no test needed. Group Lunchroom Attendant: Sample mean assumed normal because n>30, so no test needed.
Null hypothesis is equal variances of Annual.Salary19, i.e., homogeneous. Variance Ratio test: F = 18802648.590/8868151.227 = 2.120, df = 546;689, p-value = 0.000 Levene’s test, Brown-Forsythe: t = 1.805, df = 1235, p-value = 0.071
APA Style Answer
Lunchroom Attendants did have significantly different Annual Salaries (M = 18,189.78, SD = 2977.94) than Custodial Workers (M = 35,362.15, SD = 4336.20), t(1235) = 82.374, ns (two-tailed).
Paired Sample t-tests
Run the following paired sample t-tests:
Compare Annual Salary in 18 and 19 for Regular Teachers
Compare Annual Salary in 18 and 19 for Principals
Compare Annual Salary in 18 and 19 for Assistant Principals
- Report the results of each of these three analyses and provide an interpretation of the results (including tests of the key assumptions). Your results should be written as if you were preparing them for inclusion in a research paper.
Part 1
Compare Annual Salary in 18 and 19 for Regular Teachers
# Code
paireddata1 <- subset(Proj3, Job.Title == "Regular Teacher")
ttest(Annual.Salary18, Annual.Salary19, data = paireddata1, paired = TRUE)##
##
## ------ Description ------
##
## Difference: n.miss = 0, n = 10032, mean = 1240.013, sd = 1649.945
##
##
## ------ Normality Assumption ------
##
## Sample mean assumed normal because n>30, so no test needed.
##
##
## ------ Inference ------
##
## t-cutoff: tcut = 1.960
## Standard Error of Mean: SE = 16.473
##
## Hypothesized Value H0: mu = 0
## Hypothesis Test of Mean: t-value = 75.275, df = 10031, p-value = 0.000
##
## Margin of Error for 95% Confidence Level: 32.291
## 95% Confidence Interval for Mean: 1207.723 to 1272.304
##
##
## ------ Effect Size ------
##
## Distance of sample mean from hypothesized: 1240.013
## Standardized Distance, Cohen's d: 0.752
##
##
## ------ Graphics Smoothing Parameter ------
##
## Density bandwidth for 297.540
## --------------------------------------------------
Results
The n is 10,032. The mean is 1240.013. The standard deviation is 1,649.945 The degrees of freedom is 10,031. The t-value is 75.275. The p-value is 0.000.
Background
The null hypothesis is that the two seperate times have the same mean. We want to disprove the null hypothesis using the results from the t-test. If the p is less than 0.05, then we reject the null hypothesis. If the p is greater than 0.05, then we don’t have enough information to be statistically different.
Interpretation
The mean difference from 2018 to 2019 is $1240.01, so Regular Teachers made 1240.01 more dollars in 2019 than 2018.
The p-value of 0.000 is less than 0.05, so we disprove the null hypothesis. This means that Regular Teachers made a statistically different annual salary in 2019 than 2018. More accurately, Regular Teachers made $1240.01 more in 2019 than they did in 2018.
Assumptions
Normality Assumption: The sample mean is assumed normal because n>30. This means that no test is needed.
APA Style Answer
Regular Teachers recieved significantly higher annual salaries in 2019 (M = 1240.01, SD = 1,649.95) than they did in 2018, t(10,031) = 75.275, p < 0.05 (two-tailed).
Part 2
Compare Annual Salary in 18 and 19 for Principals
# Code
paireddata2 <- subset(Proj3, Job.Title == "Principal")
ttest(Annual.Salary18, Annual.Salary19, data = paireddata2, paired = TRUE)##
##
## ------ Description ------
##
## Difference: n.miss = 0, n = 377, mean = 2416.337, sd = 1163.865
##
##
## ------ Normality Assumption ------
##
## Sample mean assumed normal because n>30, so no test needed.
##
##
## ------ Inference ------
##
## t-cutoff: tcut = 1.966
## Standard Error of Mean: SE = 59.942
##
## Hypothesized Value H0: mu = 0
## Hypothesis Test of Mean: t-value = 40.311, df = 376, p-value = 0.000
##
## Margin of Error for 95% Confidence Level: 117.864
## 95% Confidence Interval for Mean: 2298.473 to 2534.201
##
##
## ------ Effect Size ------
##
## Distance of sample mean from hypothesized: 2416.337
## Standardized Distance, Cohen's d: 2.076
##
##
## ------ Graphics Smoothing Parameter ------
##
## Density bandwidth for 404.880
## --------------------------------------------------
Results
The n is 377. The mean is 2,416.337. The standard deviation is 1,163.865. The degrees of freedom is 376. The t-value is 40.311. The p-value is 0.000.
Background
The null hypothesis is that the two seperate times have the same mean. We want to disprove the null hypothesis using the results from the t-test. If the p is less than 0.05, then we reject the null hypothesis. If the p is greater than 0.05, then we don’t have enough information to be statistically different.
Interpretation
The mean difference from 2018 to 2019 is $2,416.38, so Principals made 2,416.38 more dollars in 2019 than 2018.
The p-value of 0.000 is less than 0.05, so we disprove the null hypothesis. This means that Principals made a statistically different annual salary in 2019 than 2018. More accurately, Regular Teachers made $2,416.38 more in 2019 than they did in 2018.
Assumptions
Normality Assumption: The sample mean is assumed normal because n>30. This means that no test is needed.
APA Style Answer
Principles recieved significantly higher annual salaries in 2019 (M = 2416.38, SD = 1,163.87) than they did in 2018, t(376) = 40.311, p < 0.05 (two-tailed).
Part 3
Compare Annual Salary in 18 and 19 for Assistant Principals
# Code
paireddata3 <- subset(Proj3, Job.Title == "Assistant Principal")
ttest(Annual.Salary18, Annual.Salary19, data = paireddata3, paired = TRUE)##
##
## ------ Description ------
##
## Difference: n.miss = 0, n = 413, mean = 2316.087, sd = 1240.096
##
##
## ------ Normality Assumption ------
##
## Sample mean assumed normal because n>30, so no test needed.
##
##
## ------ Inference ------
##
## t-cutoff: tcut = 1.966
## Standard Error of Mean: SE = 61.021
##
## Hypothesized Value H0: mu = 0
## Hypothesis Test of Mean: t-value = 37.955, df = 412, p-value = 0.000
##
## Margin of Error for 95% Confidence Level: 119.952
## 95% Confidence Interval for Mean: 2196.136 to 2436.039
##
##
## ------ Effect Size ------
##
## Distance of sample mean from hypothesized: 2316.087
## Standardized Distance, Cohen's d: 1.868
##
##
## ------ Graphics Smoothing Parameter ------
##
## Density bandwidth for 423.601
## --------------------------------------------------
Results
The n is 413. The mean is 2,316.087. The standard deviation is 1240.096. The degrees of freedom is 412. The t-value is 37.955. The p-value is 0.000.
Background
The null hypothesis is that the two seperate times have the same mean. We want to disprove the null hypothesis using the results from the t-test. If the p is less than 0.05, then we reject the null hypothesis. If the p is greater than 0.05, then we don’t have enough information to be statistically different.
Interpretation
The mean difference from 2018 to 2019 is $2,316.09, so Assistant Principals made 2,316.09 more dollars in 2019 than 2018.
The p-value of 0.000 is less than 0.05, so we disprove the null hypothesis. This means that Assistant Principals made a statistically different annual salary in 2019 than 2018. More accurately, Assistant Principals made $2,316.09 more in 2019 than they did in 2018.
Assumptions
Normality Assumption: The sample mean is assumed normal because n>30. This means that no test is needed.
APA Style Answer
Assistant Principals recieved significantly higher annual salaries in 2019 (M = 2,316.09, SD = 1,240.10) than they did in 2018, t(412) = 37.955, p < 0.05 (two-tailed).
One-way ANOVA
Run the following one-way ANOVA test:
- DV = Annual Salary 19, IV = Principals, Asst. Principals, Regular Teacher, Special Education Teacher, Custodial Worker
- Report the results of this analysis and provide an interpretation of the results (including tests of the key assumptions). Your results should be written as if you were preparing them for inclusion in a research paper.
Part 1
DV = Annual Salary 19, IV = Principals, Asst. Principals, Regular Teacher, Special Education Teacher, Custodial Worker
# Code
anovadata1 <- subset(Proj3, Job.Title == "Principal" | Job.Title == "Assistant Principal" | Job.Title == "Regular Teacher" | Job.Title == "Special Education Teacher" | Job.Title == "Custodial Worker")
ANOVA(Annual.Salary19 ~ Job.Title, data = anovadata1)##
## >>> Note: Converting Job.Title to a factor for this analysis only.
## BACKGROUND
##
## Response Variable: Annual.Salary19
##
## Factor Variable: Job.Title
## Levels: Assistant Principal Custodial Worker Principal Regular Teacher Special Education Teacher
##
## Number of cases (rows) of data: 14125
## Number of cases retained for analysis: 14125
##
##
## DESCRIPTIVE STATISTICS
##
## n mean sd min max
## Assistant Principal 413 116027.00 8310.19 62139.00 137409.00
## Custodial Worker 547 35362.15 4336.20 28323.00 50509.00
## Principal 377 147641.18 8947.85 128750.00 167417.00
## Regular Teacher 10032 80068.26 14752.45 11533.00 149329.00
## Special Education Teacher 2756 79522.61 14703.75 11663.00 117648.00
##
## Grand Mean: 81085.459
##
##
## BASIC ANALYSIS
##
## df Sum Sq Mean Sq F-value p-value
## Job.Title 4 3334900909599.48 833725227399.87 4134.15 0.0000
## Residuals 14120 2847547864600.48 201667695.79
##
##
## R Squared: 0.54
## R Sq Adjusted: 0.54
## Omega Squared: 0.54
##
## Cohen's f: 1.08
##
##
## TUKEY MULTIPLE COMPARISONS OF MEANS
##
## Family-wise Confidence Level:
## -----------------------------------------------------------------------------------
## diff lwr upr p adj
## Custodial Worker-Assistant Principal -80664.85 -83190.36 -78139.34 0.00
## Principal-Assistant Principal 31614.19 28854.56 34373.81 0.00
## Regular Teacher-Assistant Principal -35958.73 -37903.95 -34013.51 0.00
## Special Education Teacher-Assistant Principal -36504.38 -38548.61 -34460.16 0.00
## Principal-Custodial Worker 112279.03 109685.72 114872.34 0.00
## Regular Teacher-Custodial Worker 44706.12 43005.07 46407.17 0.00
## Special Education Teacher-Custodial Worker 44160.47 42347.02 45973.91 0.00
## Regular Teacher-Principal -67572.92 -69605.38 -65540.45 0.00
## Special Education Teacher-Principal -68118.57 -70245.99 -65991.15 0.00
## Special Education Teacher-Regular Teacher -545.65 -1378.85 287.55 0.38
##
##
## RESIDUALS
##
## Fitted Values, Residuals, Standardized Residuals
## [sorted by Standardized Residuals, ignoring + or - sign]
## [res_rows = 20, out of 14125 cases (rows) of data, or res_rows="all"]
## -----------------------------------------------------------------------------
## Job.Title Annual.Salary19 fitted residual z-resid
## 6671 Regular Teacher 149329 80068.26 69260.74 4.88
## 10699 Regular Teacher 11533 80068.26 -68535.26 -4.83
## 10702 Regular Teacher 11773 80068.26 -68295.26 -4.81
## 24674 Special Education Teacher 11663 79522.61 -67859.61 -4.78
## 8095 Regular Teacher 12386 80068.26 -67682.26 -4.77
## 9544 Regular Teacher 12386 80068.26 -67682.26 -4.77
## 8096 Regular Teacher 12753 80068.26 -67315.26 -4.74
## 8097 Regular Teacher 12883 80068.26 -67185.26 -4.73
## 9368 Regular Teacher 12883 80068.26 -67185.26 -4.73
## 9545 Regular Teacher 12883 80068.26 -67185.26 -4.73
## 9546 Regular Teacher 12883 80068.26 -67185.26 -4.73
## 9874 Regular Teacher 14316 80068.26 -65752.26 -4.63
## 24675 Special Education Teacher 14495 79522.61 -65027.61 -4.58
## 24676 Special Education Teacher 15361 79522.61 -64161.61 -4.52
## 24677 Special Education Teacher 16117 79522.61 -63405.61 -4.47
## 24673 Special Education Teacher 16166 79522.61 -63356.61 -4.46
## 11101 Regular Teacher 17652 80068.26 -62416.26 -4.40
## 9119 Regular Teacher 17733 80068.26 -62335.26 -4.39
## 10230 Regular Teacher 17750 80068.26 -62318.26 -4.39
## 9367 Regular Teacher 18061 80068.26 -62007.26 -4.37
##
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## ----------------------------------------
## Plot 1: Scatterplot with Cell Means
## Plot 2: 95% family-wise confidence level
## ----------------------------------------Results
The descriptive statistics for the group are listed above. The degrees of freedom is 4. The sum of squaes is 3334900909599.48. The mean squares is 833725227399.87. The f-value is 4134.15. The p-value is 0.0000.
Background
The null hypothesis is that the jobs Principals, Asst. Principals, Regular Teacher, Special Education Teacher, and Custodial Worker have the same mean. We want to disprove the null hypothesis using the results ANOVA. If the p is less than 0.05, then we reject the null hypothesis. If the p is greater than 0.05, then we don’t have enough information to be statistically different.
If p is less than 0.05, then we need to find out why. This is done by doing a Tukey post hoc comparison test for each of the groups. If any of the groups have a greater than 0.05, then we reject the null hypothesis for those groups.
Interpretation
A one-way analysis of variance was done between the jobs of Assistant Principal, Custodial Worker, Principal, Regular Teacher, and Special Education Teacher. The anaysis resulted in an F-value of 4134.15. This is the ratio of variance between groups and variance within groups.
The p value is 0.0000, so we reject the null hypothesis. This means that there is a significant difference between at least two of the groups. This is important because now we need to identify where the difference lies.
The Tukey multiple comparison of means tests indicates that every group has a p value of 0.000 except for Special Education Teacher vs. Regular Teacher.
Special Education Teacher vs. Regular Teacher has a p-value of 0.38. This is higher than 0.05, so we reject the null hypothesis. This means that there is a significant difference between Special Education Teachers and Regular Teachers. This difference is $-545.65. The upper limit is 287.55, and the lower limit is -1378.85.
Assumptions
“The distribution is unimodal” (From Chapter 13 slides)
“The distribution has a lower limit of 0 which means it’s positively skewed” (From Chapter 13 slides)
“Has two difference degrees of freedom” (From Chapter 13 slides)
APA Style Answer
A one-way analysis of variance was used to test the annual salary between five treatments. Significant differences were observed between means of the five groups (F(4,14120) = 4134.15, p<0.05). Tukey post hoc comparisons of the groups indicated that both the Special Education Teachers and the Regular Teachers had signficantly different annual salaries than the other groups.