brianjmurphy.files.wordpress.com · Web view2011. 4. 21. · Data has been gathered in order to address the controversy over equity in public school expenditures. The provided data

Excel Project

1

Excel Project - 3

Running Head: Excel Project

Excel ProjectBrian MurphySeattle Pacific University

EDU 6976 Interpreting and Applying Educational ResearchFall Quarter: 2009

Data has been gathered in order to address the controversy over equity in public school expenditures. The provided data will allow for an analysis of the relationship between expenditures and academic performance. Data has been included for all fifty states plus the District of Columbia in the following categories: expenditure per pupil, average student/teacher ratio, average teacher salary, percent of graduates taking the SAT, average verbal SAT score, average math SAT score, average writing SAT score, percent of students eligible for free or reduced lunch, number of students eligible for free or reduced lunch, percent of students with disabilities and total revenues. The data has also been divided by region to aid in further analysis.

Part 1

By analyzing the histograms created using the available data general trends can be determined based on the distribution of the information.

Expenditures per pupil:

The graphic to the left is a histogram showing the frequency distribution of the current expenditure per pupil in average daily attendance for 2005-2006. This histogram shows that the information has a positive skew, as the majority of the data falls on the left side of the chart. This conclusion is further reinforced by finding the mean, which is $10327.78, and the median, which is $9805. As the median is the exact midpoint of the values, and the mean will always be pulled towards the extremes, we are assured that the distribution is positively skewed.

The box plot at right further categorizes this data by region. From this graphic, we can not only determine the differences between expenditure per pupil based on the region, but it also shows that the District of Columbia (#9 on the box plot) is a significant outlier. This information could account for part of the skewed distribution represented by the histogram for expenditures per pupil.

Average pupil/teacher ratio:

The graphic to the right is a histogram showing the frequency distribution of the average pupil to teacher ratio for the fall of 2005. This histogram shows that the information has a slight positive skew, for while much of the data resides to the left, it is also visually reminiscent of a normal distribution. This conclusion is further reinforced by finding the mean which is 15.23, and the median which is 14.80. While this information reinforces the conclusions that the distribution is positively skewed, the numbers are relatively close (they have a difference of less that 1).

The box plot at right further categorizes this data by region. This graphic shows two important details. First; this average pupil per teacher in 2005 is not evenly distributed between regions. Second; given the preceding conclusion, none of the points can be justified as outliers within the population as a whole (although Virginia (#47) is an outlier in this case for the South only).

Estimated average salary:

The graph to the right represents a topic of contention near and dear to every teacher’s heart: salary. More specifically, this histogram is a frequency distribution of the estimated average salary in 2005-2006 by state. Once again, a visual analysis of this histogram leads to the conclusion that this data represents a positively skewed distribution. Comparisons of the population mean ($47679.08) and median ($45575.00) values further prove this conclusion.

A visual comparison of the box plots (at right) could lead to the conclusion that the District of Columbia (9), Delaware (8) and Maryland (21) are distinct outliers (which they are when taking only the southern states into account) but notice that those values fall within the upper-ranges of the other three regions. Due to this comparison, it is difficult to conclude that these are actually outlying values.

Average verbal SAT score:

The histogram to the right is a frequency distribution of the average verbal SAT scores in 2005-2006 as reported by state. A visual analysis leads to the conclusion that the distribution in not normal, but has a distinct positive skew. According to the histogram there are 11 states with an average verbal SAT score between 490 and 500, with the next highest frequency being 6 states with scores between 510 and 520. Comparisons of the population mean (534.94) and median (523.00) values further prove this conclusion.

A visual comparison of the box plots (at left) could lead to the conclusion that Ohio (36) and Indiana (15) are distinct outliers (which they are when taking only the Midwest states into account) but notice that those values fall within the ranges of at least two of the other three regions. Due to this comparison, it is difficult to conclude that these are actually outlying values. More concerning is the higher-skewed Midwest region, the entire range of which fall outside the range of each other box plot except the upper-range of the South.

Average writing SAT scores:

The histogram to the right is a frequency distribution of the average writing SAT scores in 2005-2006 as reported by state. A visual analysis leads to the conclusion that the distribution in not normal, but has a distinct positive skew. According to the histogram there are 11 states with an average verbal SAT score between 480 and 490, with the next highest frequency being 6 states with scores between 500 and 510, and then another six states with scores between 570 and 580. Comparisons of the population mean (525.37) and median (511.00) values further prove this conclusion.

A visual comparison of the box plots (at right) could lead to the conclusion that Ohio (36) and Indiana (15) are distinct outliers (which they are when taking only the Midwest states into account) but notice that those values fall within the ranges of at least two of the other three regions. Due to this comparison, it is difficult to conclude that these are actually outlying values. More concerning is the higher-skewed Midwest region, the entire range of which fall outside the range of each other box plot except the upper-range of the South.

Average Math SAT scores:

The histogram to the left is a frequency distribution of the average math SAT scores in 2005-2006 as reported by state. A visual analysis leads to the conclusion that the distribution in not normal, but has a distinct positive skew. According to the histogram there are 11 states with an average verbal SAT score between 500 and 510, with the next highest frequency being 9 states with scores between 510 and 520. The rest of the values (while occurring up to average scores of 620) have a frequency of no higher than 5. Comparisons of the population mean (540.59) and median (529.00) values further prove this conclusion.

A visual comparison of the box plots (at right) could lead to the conclusion that Ohio (36) and Indiana (15) are distinct outliers (which they are when taking only the Midwest states into account) but notice that those values fall within the ranges of at least two of the other three regions. Due to this comparison, it is difficult to conclude that these are actually outlying values. More concerning is the higher-skewed Midwest region, the entire range of which fall outside the range of each other box plot.

Percentage of students taking the SAT:

The histogram at left is a frequency distribution of the percentage of eligible students per state who took the SAT in 2006-2007. A visual analysis leads to the simple conclusion the distribution has a positive skew, which is reinforced by comparing the population mean (39.33%) and median (32.00%) values.

The box plots at right represent the same data arranged by region. By comparison we see that while a high percentage of students from the Northeast took the sat in 2006-2007, and surprisingly low percentage of students in the Midwest did likewise. Once again we see that Ohio (36) and Indiana (15) are outliers in their region, but these values fall within the range of at least one other region as well.

Percent of students eligible for free and reduced lunch in 2006-2007:

The histogram at left is a frequency distribution of the percentage of students eligible for free and reduced lunch per state in 2006-2007. A visual analysis leads to the simple conclusion the distribution has a positive skew, which is reinforced by comparing the population mean (39.82%) and median (37.35%) values.

The corresponding box plot (at left) represents the same data rearrange by region. The only concerning outlier in this case is New Hampshire (30), which has a percentage of students eligible for free and reduced lunch far below the range of any region (including the Northeast). This is not the reason for the positive skew in the corresponding histogram, but should provide an interesting comparison as we analyze further.

Percent of students with disabilities:

The histogram at right is a frequency distribution of the percentage of students eligible for free and reduced lunch per state in 2006-2007. Of all the histograms analyzed so far, this appears to be the most normal. A comparison of the population mean (14.22%) and median (14.30%) values proves that this is the case, although the distribution does have a slightly negative skew.

An analysis of the corresponding box plot (below) represents the same data arranged by region. In this case there are no identified outliers, and while the box plots for each region do not match up perfectly, there is considerable overlap between each one suggesting there is a certain amount of consistency.

Total revenues:

The first histogram on the following page (on the right) is a frequency distribution of the total revenues by state for the 2005-2006 school year (represented here in thousands). More than any of the other histograms, it is obvious that this data has a decidedly positive skew. When we compare the mean of the population (10208705.02) to the median (6346033.00) of the population, it becomes evident that this is indeed the case.

An analysis of the corresponding box plots, which are once again arranged by region, shows that there is at least one distinct outlier contained within this data. California (# 5) is not only an outlier in the Western region but has a total revenue value (63,785,872.00) nearly 17,000,000 points (represented in thousands) higher than the next highest value (New York, at 46,776,452.00). Considering that the population mean alone is only approximately 10,000,000, this is a very significant outlier to take into account when analyzing this information.

Enrollment:

Finally, the histogram below represents the total enrollment per state for the year 2005. This distribution, like the histogram above for total revenue, has an obvious positive skew. When we compare the mean of the population (963005.84) to the median (654526.00) of the population, it becomes evident that this is indeed the case.

The corresponding box plot (on the following page, right) is once again arranged by region. This graphic shows us that while the majority of the data (arranged by region) is relatively uniform, there are a couple of outliers that must be taken into consideration. These two outliers are California (5) and Texas (44) with enrollments of 6,406,821 and 4,599,509 respectively. Looking back at the previous information, we are reminded that the mean of the population is 963,005.84 while the median is only 654,526.00, a point that illustrates just how extreme these two outliers are, and shows why the frequency distribution has a decidedly positive skew.

Frequency distribution by region:

The 51 states are categorized into the following four regions: West, Midwest, South and Northeast. These four regions have a geographical basis. This categorization allows for further, more specific analysis of the data. This frequency distribution is represented in tabular form below.

region

Frequency

Percent

Valid Percent

Cumulative Percent

Valid

West

13

25.5

25.5

25.5

Midwest

12

23.5

23.5

49.0

South

17

33.3

33.3

82.4

Northeast

9

17.6

17.6

100.0

Total

51

100.0

100.0

Part 2

Expenditures per pupil:

In order to determine if the regions differ in terms of expenditures per pupil, we analyze the reported data from the computed Analysis of Variance (ANOVA) which is shown below.

Tests of Between-Subjects Effects

Dependent Variable: current expenditure per pupil in average daily attendance in public elem and sec schools 2005-06

Source

Type III Sum of Squares

df

Mean Square

F

p

Partial Eta Squared

region

120,097,648.80

3

40,032,549.60

9.75

.00

.38

Error

192,953,101.83

47

4,105,385.15

Total

5,752,870,321.00

51

Corrected Total

313,050,750.63

50

In this case, the null hypothesis being tested is that all the regions have the same expenditure per pupil (or Ho: µ1 = µ2 = µ3 = µ4). For this analysis there are 3 degrees of freedom within groups and 47 degrees of freedom between groups, so the critical values of F are 2.805 at the .05 level, and 4.23 at the .01 level. The obtained value of F (F = 9.75) in the table above is greater than both of these values, allowing us to reject the null hypothesis. A one-way ANOVA was computed comparing expenditure per pupil from four different regions of the United States. A significant difference was found between the groups (F(3,47) = 9.75, p < .01). The null hypothesis was rejected.

Given that the F-ratio was significant, we can check the output of the Levene’s test (below left). Since the P-value is greater than .05 (in this case P = .24) we can assume homogeneity.

Levene's Test of Equality of Error Variancesa

Dependent Variable:current expenditure per pupil in average daily attendance in public elem and sec schools 2005-06

F

df1

df2

P

1.45

3

47

.24

From there, we are free to look at the Tukey’s HSD output in the following table. The significant differences are flagged with an asterisk, and from that information we can see that there is a significant difference between the Northeast and the other three regions.

Multiple Comparisons

current expenditure per pupil in average daily attendance in public elem and sec schools 2005-06

Tukey HSD

(I) region

(J) region

Mean Difference (I-J)

Std. Error

P

95% Confidence Interval

Lower Bound

Upper Bound

West

Midwest

-660.49

811.12

.85

-2820.82

1499.83

South

-475.96

746.52

.92

-2464.23

1512.31

Northeast

-4356.52*

878.61

.00

-6696.60

-2016.45

Midwest

West

660.49

811.12

.85

-1499.83

2820.82

South

184.53

763.94

1.00

-1850.14

2219.21

Northeast

-3696.03*

893.46

.00

-6075.66

-1316.40

South

West

475.96

746.52

.92

-1512.31

2464.23

Midwest

-184.53

763.94

.995

-2219.21

1850.14

Northeast

-3880.56*

835.25

.000

-6105.17

-1655.96

Northeast

West

4356.52*

878.61

.00

2016.45

6696.60

Midwest

3696.03*

893.46

.00

1316.40

6075.66

South

3880.56*

835.25

.00

1655.96

6105.17

*. The mean difference is significant at the .05 level.

Pupil/Teacher Ratio:

In order to determine if the regions differ in terms of pupil/teacher ratio, we analyze the reported data from the computed Analysis of Variance (ANOVA) which is shown below.


Dependent Variable:average pupil/teacher ratio Fall 2005

Source


df

Mean Square

F

P

Partial Eta Squared

region

146.88

3

48.96

13.08

.00

.46

Error

175.92

47

3.74

Total

12145.39

51

Corrected Total

322.80

50

a. R Squared = .455 (Adjusted R Squared = .420)

In this case, the null hypothesis being tested is that all the regions have the same pupil teacher ratio (or Ho: µ1 = µ2 = µ3 = µ4). For this analysis there are 3 degrees of freedom within groups and 47 degrees of freedom between groups, so the critical values of F are 2.805 at the .05 level, and 4.23 at the .01 level. The obtained value of F (F = 13.08) in the table above is greater than both of these values, allowing us to reject the null hypothesis. A one-way ANOVA was computed comparing average pupil teacher ratio from four different regions of the United States. A significant difference was found between the groups (F(3,47) = 13.08, p < .01). The null hypothesis was rejected.

Given that the F-ratio was significant, we can check the output of the Levene’s test (below left). Since the P-value is not greater than .05 (in this case P = .01) we cannot assume homogeneity.


Dependent Variable:average pupil/teacher ratio Fall 2005

F

df1

df2

P

4.92

3

47

.01

From there, we are free to look at the Tukey’s HSD output in the following table. The significant differences are flagged with an asterisk, and from that information we can see that there is a significant difference between the West and the other three regions, and between the Northeast and the South.


average pupil/teacher ratio Fall 2005

Dunnett C

(I) region

(J) region


Std. Error


Lower Bound

Upper Bound

West

Midwest

3.00*

.94

.19

5.81

South

2.94*

.86

.40

5.49

Northeast

5.07*

.94

2.23

7.92

Midwest

West

-3.00*

.94

-5.81

-.19

South

-.06

.58

-1.78

1.67

Northeast

2.08

.69

-.08

4.22

South

West

-2.94*

.86

-5.49

-.40

Midwest

.06

.58

-1.67

1.78

Northeast

2.13*

.58

.34

3.92

Northeast

West

-5.07*

.94

-7.92

-2.23

Midwest

-2.080

.69

-4.22

.08

South

-2.13*

.58

-3.92

-.34

Teacher Average Salary:

In order to determine if the regions differ in terms of teachers average salaries, we analyze the reported data from the computed Analysis of Variance (ANOVA) which is shown below.


Dependent Variable:estimated ave salary 2005-2006

Source


df

Mean Square

F

P

Partial Eta Squared

region

434910474.84

3

144970158.28

3.45

.02

.18

Error

1974666324.85

47

42014177.12

Total

118347597323.00

51

Corrected Total

2409576799.69

50

In this case, the null hypothesis being tested is that all the regions have the same average teacher salary (or Ho: µ1 = µ2 = µ3 = µ4). For this analysis there are 3 degrees of freedom within groups and 47 degrees of freedom between groups, so the critical values of F are 2.805 at the .05 level, and 4.23 at the .01 level. The obtained value of F (F = 3.45) in the table above is greater than the critical value at the .05 level, allowing us to reject the null hypothesis at that level. A one-way ANOVA was computed comparing average teacher salaries from four different regions of the United States. A significant difference was found between the groups (F(3,47) = 3.45, p < .05). The null hypothesis was rejected.

Given that the F-ratio was significant, we can check the output of the Levene’s test (below left). Since the P-value is greater than .05 (in this case P = .66) we can assume homogeneity.


Dependent Variable:estimated ave salary 2005-2006

F

df1

df2

P

.53

3

47

.66

From there, we are free to look at the Tukey’s HSD output in the following table. The significant differences are flagged with an asterisk, and from that information we can see that the only significant difference that exists here is between the Northeast and the South.


estimated ave salary 2005-2006

Tukey HSD

(I) region

(J) region


Std. Error

Sig.


Lower Bound

Upper Bound

West

Midwest

910.88

2594.81

.99

-6000.11

7821.88

South

1506.03

2388.15

.92

-4854.55

7866.62

Northeast

-6641.50

2810.72

.10

-14127.52

844.52

Midwest

West

-910.88

2594.81

.99

-7821.87

6000.11

South

595.15

2443.89

1.00

-5913.89

7104.18

Northeast

-7552.39

2858.22

.05

-15164.94

60.16

South

West

-1506.03

2388.15

.92

-7866.62

4854.55

Midwest

-595.15

2443.89

1.00

-7104.18

5913.89

Northeast

-8147.54*

2672.01

.02

-15264.15

-1030.92

Northeast

West

6641.50

2810.71

.10

-844.52

14127.52

Midwest

7552.39

2858.22

.05

-60.16

15164.94

South

8147.54*

2672.01

.02

1030.92

15264.15

Percentage of students taking the SAT:

In order to determine if the regions differ in terms of percentage of all eligible students taking the SAT in 2006-2007, we analyze the reported data from the computed Analysis of Variance (ANOVA) which is shown below.


Dependent Variable: percentage of all eligible students taking the SAT 2006-07

Source


df

Mean Square

F

P

Partial Eta Squared

region

24959.33

3

8319.78

16.66

.00

.52

Error

23472.00

47

499.40

Total

127334.00

51

Corrected Total

48431.33

50


In this case, the null hypothesis being tested is that all the regions have the same percentage of eligible students taking the SAT (or Ho: µ1 = µ2 = µ3 = µ4). For this analysis there are 3 degrees of freedom within groups and 47 degrees of freedom between groups, so the critical values of F are 2.805 at the .05 level, and 4.23 at the .01 level. The obtained value of F (F = 16.66) in the table above is greater than the critical value at both levels, allowing us to reject the null hypothesis. A one-way ANOVA was computed comparing the percentage of all eligible students taking the SAT from four different regions of the United States. A significant difference was found between the groups (F(3,47) = 16.66, p < .01). The null hypothesis was rejected.

Given that the F-ratio was significant, we can check the output of the Levene’s test (below). Since the P-value is not greater than .05 (in this case P = .00) we can not assume homogeneity.


Dependent Variable: percentage of all eligible students taking the SAT 2006-07

F

df1

df2

P

16.53

3

47

.00

From there, we are free to look at the Tukey’s HSD output in the following table. The significant differences are flagged with an asterisk, and from that information we can see that significant differences exist between the Northeast and the other three regions and the South and the Midwest.


percentage of all eligible students taking the SAT 2006-07

Dunnett C

(I) region

(J) region


Std. Error


Lower Bound

Upper Bound

West

Midwest

20.80

7.12

-.48

42.07

South

-6.89

9.14

-33.35

19.57

Northeast

-47.98*

6.26

-67.02

-28.95

Midwest

West

-20.80

7.12

-42.07

.48

South

-27.69*

8.92

-53.59

-1.79

Northeast

-68.78*

5.94

-87.03

-50.53

South

West

6.89

9.14

-19.57

33.35

Midwest

27.69*

8.92

1.79

53.59

Northeast

-41.09*

8.25

-65.17

-17.01

Northeast

West

47.98*

6.26

28.95

67.02

Midwest

68.78*

5.94

50.53

87.03

South

41.09*

8.25

17.01

65.17


Verbal SAT Scores:

In order to determine if the regions differ in terms of Verbal SAT Scores, we analyze the reported data from the computed Analysis of Variance (ANOVA) which is shown below.


Dependent Variable:average verbal SAT score 2005-06

Source


df

Mean Square

F

P

Partial Eta Squared

region

30986.00

3

10328.67

12.00

.00

.43

Error

40450.83

47

860.66

Total

14665702.00

51

Corrected Total

71436.82

50


In this case, the null hypothesis being tested is that all the regions have the same performance on verbal SAT scores (or Ho: µ1 = µ2 = µ3 = µ4). For this analysis there are 3 degrees of freedom within groups and 47 degrees of freedom between groups, so the critical values of F are 2.805 at the .05 level, and 4.23 at the .01 level. The obtained value of F (F = 12.00) in the table above is greater than the critical value at both levels, allowing us to reject the null hypothesis. A one-way ANOVA was computed comparing the percentage of all verbal SAT results from four different regions of the United States. A significant difference was found between the groups (F(3,47) = 12.00, p < .01). The null hypothesis was rejected.


Dependent Variable:average verbal SAT score 2005-06

F

df1

df2

P

6.57

3

47

.00

Given that the F-ratio was significant, we can check the output of the Levene’s test (left). Since the P-value is not greater than .05 (in this case P = .00) we can not assume homogeneity.

From there, we are free to look at the Tukey’s HSD output in the following table. The significant differences are flagged with an asterisk, and from that information we can see that significant differences exist between the Midwest and the three other regions, and the Northeast and the West.


average verbal SAT score 2005-06

Dunnett C

(I) region

(J) region


Std. Error


Lower Bound

Upper Bound

West

Midwest

-47.81*

11.31

-81.68

-13.94

South

1.93

11.26

-30.74

34.60

Northeast

24.69*

7.73

1.37

48.02

Midwest

West

47.81*

11.31

13.94

81.68

South

49.74*

12.63

12.65

86.82

Northeast

72.50*

9.62

43.31

101.69

South

West

-1.93

11.26

-34.60

30.74

Midwest

-49.74*

12.63

-86.82

-12.65

Northeast

22.77

9.56

-5.02

50.55

Northeast

West

-24.69*

7.73

-48.02

-1.37

Midwest

-72.50*

9.62

-101.69

-43.31

South

-22.77

9.56

-50.55

5.02


Math SAT Scores:

In order to determine if the regions differ in terms of Math SAT Scores, we analyze the reported data from the computed Analysis of Variance (ANOVA) which is shown below.


Dependent Variable:average math SAT score 2005-06

Source


df

Mean Square

F

P

Partial Eta Squared

region

36447.57

3

12149.19

16.94

.00

.52

Error

33708.78

47

717.21

Total

14974174.00

51

Corrected Total

70156.35

50



Dependent Variable:average math SAT score 2005-06

F

df1

df2

P

5.82

3

47

.00

In this case, the null hypothesis being tested is that all the regions have the same performance on math SAT scores (or Ho: µ1 = µ2 = µ3 = µ4). For this analysis there are 3 degrees of freedom within groups and 47 degrees of freedom between groups, so the critical values of F are 2.805 at the .05 level, and 4.23 at the .01 level. The obtained value of F (F = 16.94) in the table above is greater than the critical value at both levels, allowing us to reject the null hypothesis. A one-way ANOVA was computed comparing the percentage of all math SAT results from four different regions of the United States. A significant difference was found between the groups (F(3,47) = 16.94, p < .01). The null hypothesis was rejected.

Given that the F-ratio was significant, we can check the output of the Levene’s test (right). Since the P-value is not greater than .05 (in this case P = .00) we cannot assume homogeneity.

From there, we are free to look at the Tukey’s HSD output in the following table. The significant differences are flagged with an asterisk, and from that information we can see that significant differences exist between the Midwest and the three other regions, and the Northeast and the West.


average math SAT score 2005-06

Dunnett C

(I) region

(J) region


Std. Error


Lower Bound

Upper Bound

West

Midwest

-51.84*

10.39

-83.01

-20.67

South

8.02

9.67

-19.96

36.00

Northeast

22.75*

6.11

4.23

41.26

Midwest

West

51.84*

10.39

20.67

83.01

South

59.86*

12.14

24.15

95.57

Northeast

74.58*

9.548

45.67

103.50

South

West

-8.02

9.67

-35.99

19.96

Midwest

-59.86*

12.14

-95.57

-24.14

Northeast

14.73

8.75

-10.71

40.16

Northeast

West

-22.74*

6.11

-41.26

-4.23

Midwest

-74.58*

9.54

-103.50

-45.67

South

-14.73

8.75

-40.16

10.71


Writing SAT Scores:

In order to determine if the regions differ in terms of Writing SAT Scores, we analyze the reported data from the computed Analysis of Variance (ANOVA) which is shown below.


Dependent Variable:average writing SAT score 2005-06

Source


df

Mean Square

F

P

Partial Eta Squared

region

26942.23

3

8980.74

9.62

.00

.38

Error

43857.69

47

933.14

Total

14147632.00

51

Corrected Total

70799.92

50


In this case, the null hypothesis being tested is that all the regions have the same performance on writing SAT scores (or Ho: µ1 = µ2 = µ3 = µ4). For this analysis there are 3 degrees of freedom within groups and 47 degrees of freedom between groups, so the critical values of F are 2.805 at the .05 level, and 4.23 at the .01 level. The obtained value of F (F = 9.62) in the table above is greater than the critical value at both levels, allowing us to reject the null hypothesis. A one-way ANOVA was computed comparing the percentage of all writing SAT results from four different regions of the United States. A significant difference was found between the groups (F(3,47) = 9.62, p < .01). The null hypothesis was rejected.

Given that the F-ratio was significant, we can check the output of the Levene’s test (right). Since the P-value is not greater than .05 (in this case P = .00) we cannot assume homogeneity.


Dependent Variable:average writing SAT score 2005-06

F

df1

df2

P

8.01

3

47

.00

From there, we are free to look at the Tukey’s HSD output in the following table. The significant differences are flagged with an asterisk, and from that information we can see that significant differences exist between the Midwest and the three other regions.


average writing SAT score 2005-06

Dunnett C

(I) region

(J) region


Std. Error


Lower Bound

Upper Bound

West

Midwest

-49.17*

11.38

-83.24

-15.10

South

-5.82

11.81

-40.07

28.42

Northeast

17.78

7.95

-6.24

41.80

Midwest

West

49.17*

11.389

15.10

83.24

South

43.34*

13.06

5.06

81.63

Northeast

66.94*

9.71

37.43

96.46

South

West

5.82

11.81

-28.42

40.07

Midwest

-43.34*

13.06

-81.63

-5.06

Northeast

23.60

10.22

-6.10

53.30

Northeast

West

-17.78

7.95

-41.80

6.24

Midwest

-66.94*

9.71

-96.46

-37.43

South

-23.60

10.22

-53.30

6.10


Percentage of students on free and reduced lunch:

In order to determine if the regions differ in terms of percentage of students on free and reduced lunch, we analyze the reported data from the computed Analysis of Variance (ANOVA) which is shown below.


Dependent Variable:% of students eligible for free/reduced lunch 2006-07

Source


df

Mean Square

F

P

Partial Eta Squared

region

2732.83

3

910.94

14.87

.00

.49

Error

2817.70

46

61.25

Total

84848.08

50

Corrected Total

5550.53

49


In this case, the null hypothesis being tested is that all the regions have the same percentage of students on free and reduced lunch (or Ho: µ1 = µ2 = µ3 = µ4). For this analysis there are 3 degrees of freedom within groups and 46 degrees of freedom between groups, so the critical values of F are 2.81 at the .05 level, and 4.24 at the .01 level. The obtained value of F (F = 14.87) in the table above is greater than the critical value at both levels, allowing us to reject the null hypothesis. A one-way ANOVA was computed comparing the percentage of students on free and reduced lunch from four different regions of the United States. A significant difference was found between the groups (F(3,46) = 14.87, p < .01). The null hypothesis was rejected.

Given that the F-ratio was significant, we can check the output of the Levene’s test (right). Since the P-value is greater than .05 (in this case P = .29) we can assume homogeneity.


Dependent Variable:% of students eligible for free/reduced lunch 2006-07

F

df1

df2

P

1.30

3

46

.29

From there, we are free to look at the Tukey’s HSD output in the following table. The significant differences are flagged with an asterisk, and from that information we can see that significant differences exist between the South and the three other regions, and the West and the Northeast.


% of students eligible for free/reduced lunch 2006-07

Tukey HSD

(I) region

(J) region


Std. Error

P


Lower Bound

Upper Bound

West

Midwest

5.14

3.20

.38

-3.38

13.66

South

-9.57*

2.95

.01

-17.43

-1.70

Northeast

9.79*

3.45

.03

.59

18.99

Midwest

West

-5.14

3.20

.38

-13.66

3.38

South

-14.71*

2.95

.00

-22.58

-6.84

Northeast

4.65

3.45

.54

-4.55

13.85

South

West

9.57*

2.95

.01

1.70

17.43

Midwest

14.71*

2.95

.00

6.84

22.58

Northeast

19.36*

3.23

.00

10.76

27.96

Northeast

West

-9.79*

3.45

.03

-18.99

-.59

Midwest

-4.65

3.45

.54

-13.85

4.55

South

-19.36*

3.23

.00

-27.96

-10.76


Percent of students with disabilities:

In order to determine if the regions differ in terms of percent of students with disabilities, we analyze the reported data from the computed Analysis of Variance (ANOVA) which is shown below.


Dependent Variable:% of students with disabilities 2006-07

Source


df

Mean Square

F

P

Partial Eta Squared

region

89.78

3

29.93

10.25

.00

.40

Error

137.19

47

2.92

Total

10533.34

51

Corrected Total

226.97

50


In this case, the null hypothesis being tested is that all the regions have the same percentage of students with disabilities (or Ho: µ1 = µ2 = µ3 = µ4). For this analysis there are 3 degrees of freedom within groups and 47 degrees of freedom between groups, so the critical values of F are 2.805 at the .05 level, and 4.23 at the .01 level. The obtained value of F (F = 10.25) in the table above is greater than the critical value at both levels, allowing us to reject the null hypothesis. A one-way ANOVA was computed comparing the percentage of students with disabilities from four different regions of the United States. A significant difference was found between the groups (F(3,47) = 10.25, p < .01). The null hypothesis was rejected.

Given that the F-ratio was significant, we can check the output of the Levene’s test (right). Since the P-value is greater than .05 (in this case P = .07) we can assume homogeneity.


Dependent Variable:% of students with disabilities 2006-07

F

df1

df2

P

2.55

3

47

.07

From there, we are free to look at the Tukey’s HSD output in the following table. The significant differences are flagged with an asterisk, and from that information we can see that significant differences exist between the Northeast and the South, the Northeast and the West and the Midwest and the West.


% of students with disabilities 2006-07

Tukey HSD

(I) region

(J) region


Std. Error

P


Lower Bound

Upper Bound

West

Midwest

-2.49*

.68

.00

-4.31

-.67

South

-1.58

.63

.07

-3.26

.10

Northeast

-3.94*

.74

.00

-5.91

-1.96

Midwest

West

2.49*

.68

.00

.67

4.31

South

.91

.64

.50

-.80

2.63

Northeast

-1.4

.75

.24

-3.45

.56

South

West

1.58

.63

.07

-.10

3.26

Midwest

-.91

.64

.50

-2.63

.80

Northeast

-2.36*

.70

.01

-4.23

-.48

Northeast

West

3.94*

.74

.00

1.96

5.91

Midwest

1.44

.75

.24

-.56

3.45

South

2.36*

.70

.01

.48

4.23


Revenues:

In order to determine if the regions differ in terms of average revenues, we analyze the reported data from the computed Analysis of Variance (ANOVA) which is shown below.


Dependent Variable:Total revenues for the year 2005-06 (in thousands)

Source


df

Mean Square

F

P

Partial Eta Squared

region

139,274,225,326,152.02

3

46,424,741,775,384.01

.30

.83

.02

Error

7,323,334,065,723,234.00

47

155,815,618,419,643.28

Total

12,777,708,858,095,064.00

51

Corrected Total

7,462,608,291,049,387.00

50

In this case, the null hypothesis being tested is that all the regions have the same average revenues (or Ho: µ1 = µ2 = µ3 = µ4). For this analysis there are 3 degrees of freedom within groups and 47 degrees of freedom between groups, so the critical values of F are 2.805 at the .05 level, and 4.23 at the .01 level. The obtained value of F (F = .30) in the table above is less than the critical value at both levels, allowing us to accept the null hypothesis. A one-way ANOVA was computed comparing the average revenues from four different regions of the United States. No significant difference was found between the groups (F(3,47) = 0.30, p > .05). The null hypothesis was accepted.

While it is clear that at least two of the regions differ in each of the preceding comparisons besides that of average revenue, it is difficult to summarize the practical significance of this information as of yet as the ways in which the values correspond to each other remain ambiguous. We now know how the regions differ from one another given these variables, but no how the variable impact or influence each other in the first place.

Part 3

In order to find the correlations between variables we will be graphing scatter plots of paired information, and calculating the correlation coefficients in order to determine whether or not there are any statistically significant relationships between the variables.

Expenditures and SAT scores:

The scatter plot at right shows an apparent negative correlation between average verbal SAT score and expenditure per pupil.

The scatter plot to the right show a similar correlation, this time for average math SAT scores and expenditure per pupil. This also appears to have a negative correlation.

(verbal SAT score 2005-06math SAT score 2005-06writing SAT score 2005-06Expenditure/ pupil 2005-06 Pearson r-0.42**-0.39**-0.40**p0.000.000.00n515151)The table in the text box to the right gives the calculated Pearson r values for the correlations between expenditure and SAT scores. This information shows that the Pearson correlation coefficients are significant for all three correlations at the .01 level (flagged by a double asterisk). It is important to note that all three correlations are negative. We conclude that this relationship in not likely to be a result of chance.

Finally, the two scatter plots here both have lines of best fit that can be calculated based on the coefficient data output in the tables below. Using the standard equation for the line of best fit (Y = bX + a) we plug in the variables to get the correct regression equation.

Yverba; = 599.76X - 0.01

Ymath = 601.42X - 0.01

Coefficientsa

Model

Unstandardized

Standardized

t

Sig.

B

Std. Error

Beta

1

(Constant)

599.76

20.85

28.77

.00

Expenditure/pupil 2005-06

-.01

.00

-.42

-3.20

.00

a. Dependent Variable: average verbal SAT score 2005-06

Model

Coefficients

Coefficients

t

Sig.

B

Std. Error

Beta

1

(Constant)

601.42

20.88

28.80

.00

Expenditure/ pupil 2005-06

-.01

.00

-.39

-3.00

.00

a. Dependent Variable: average math SAT score 2005-06

Pupil/teacher ratio and SAT scores:

The scatter plot at right shows an apparent negative correlation between average verbal SAT score and pupil/teacher ratio.

The scatter plot below that shows a similar correlation, this time for average math SAT scores and pupil/teacher ratio. This also appears to have a negative correlation.

The table in the text box to the right gives the calculated Pearson r values for the correlations between pupil/teacher ratio and SAT scores. This information shows that the Pearson correlation coefficients are not significant for any of the three correlations). It is important to note that all three correlations are negative.


(writing SAT score verbal SAT score math SAT score average pupil/teacher ratio Fall 2006Pearson r-.07-.03-.03p.64.82.85n515151)Yverba; = 542.32 - 0.49

Ymath = 546.76 - 0.41

Coefficientsa

Model

Unstand.

Standard.

t

Sig.

B

Std. Error

Beta

1

(Constant)

542.32

32.86

16.50

.00

pupil/teacher ratio

-.49

2.14

-.033

-.23

.82


Model

Unstandardized

Standardized

t

Sig.

B

Std. Error

Beta

1

(Constant)

546.76

32.57

16.79

.00

average pupil/teacher ratio Fall 2006

-.41

2.12

-.03

-.19

.85


Estimated average salary and SAT scores:

The scatter plot at right shows an apparent negative correlation between average math SAT scores and estimated salary.

The scatter plot below that shows a similar correlation, this time for average verbal SAT scores and estimated salary. This also appears to have a negative correlation.

The table in the text box to the right gives the calculated Pearson r values for the correlations between estimated salary and SAT scores. This information shows that the Pearson correlation coefficients are significant for all three correlations at the .01 level (flagged by a double asterisk). It is important to note that all three correlations are negative. We can conclude that this relationship in not likely to be a result of chance.


Yverba; = 658.2 - 0.00

(writing SAT score verbal SAT score math SAT score estimated ave salary 2005-2006Pearson r-.45**-.48**-.41**p.00.00.00n515151)Ymath = 645.24 - 0.00

Given that these regression equations have no slope, and the standard error of the estimate is fairly high (below) the effectiveness of this correlation can be called into question.

Model

R

R Square

Std. Error of the Estimate

1

.48a

.23

33.61

Coefficientsa

Model

Unstandardized Coefficients

Standardized Coefficients

t

Sig.

B

Std. Error

Beta

1

(Constant)

645.24

33.92

19.02

.00

estimated ave salary 2005-2006

-.00

.00

-.41

-3.12

.00


Coefficientsa

Model

Unstandardized Coefficients

Standardized Coefficients

t

Sig.

B

Std. Error

Beta

1

(Constant)

658.20

32.98

19.96

.00

ave salary 2005-2006

-.00

.00

-.48

-3.78

.00


Average revenue and SAT scores:

The scatter plot at left shows an apparent negative correlation between average verbal SAT scores and total revenue.

The scatter plot below that shows a similar correlation, this time for average math SAT scores and total revenue. This also appears to have a negative correlation.

( writing SAT score verbal SAT score math SAT score Revenues for 2005-06 (in 1000)Pearson r-.25-.29*-.20p.07.04.16N515151)The table in the text box to the right gives the calculated Pearson r values for the correlations between total revenues and SAT scores. This information shows that the Pearson correlation coefficient is only significant for verbal SAT scores at the .05 level (flagged by a single asterisk). It is important to note that all three correlations are negative. We can conclude that this relationship in not likely to be a result of chance.


Yverba; = 544.20 - 9.070E-7

Ymath = 546.80 - 6.089E-7

Coefficientsa

Model

Unstandardized

Standardized

t

Sig.

B

Std. Error

Beta

1

(Constant)

544.20

6.69

81.36

.00

Total revenues 2005-06 (in thousands)

-9.070E-7

.00

-.29

-2.15

.04


Model

Unstandardized

Standardized

t

Sig.

B

Std. Error

Beta

1

(Constant)

546.80

6.80

80.47

.00

Total revenues 2005-06 (in thousands)

-6.089E-7

.00

-.20

-1.42

.16


Percentage of students eligible for free and reduced lunch and SAT scores:

The scatter plot at right shows an apparent negative correlation between average math SAT scores and the percentage of students eligible for free and reduced lunch.

The scatter plot at right on the following page shows a similar correlation, this time for average verbal SAT scores and the percentage of students eligible for free and reduced lunch. This also appears to have a negative correlation.

The table in the text box on the right of the following page gives the calculated Pearson r values for the correlations between the percentages of students eligible for free and reduced lunch and SAT scores. This information shows that the Pearson correlation coefficient is not significant for SAT scores at any level.

( writing SAT score verbal SAT score math SAT score % eligible for free/reduced lunch Pearson r.08.02-.08p.58.87.57n505050)Finally, the two scatter plots here both have lines of best fit that can be calculated based on the coefficient data output in the tables below. Using the standard equation for the line of best fit (Y = bX + a) we plug in the variables to get the correct regression equation.

Yverba; = 532.29 +.09

Ymath = 552.85 - .292

Note that the regression line for the verbal data is in fact positive, proving that my original assumption (of a negative correlation) was incorrect.

Coefficientsa

Model

Unstandardized

Standardized

t

Sig.

B

Std. Error

Beta

1

(Constant)

552.85

20.90

26.45

.00


-.292

.51

-.08

-.58

.57


Model

Unstandardized

Standardized

t

Sig.

B

Std. Error

Beta

1

(Constant)

532.29

21.12

25.21

.00


.09

.51

.02

.17

.87


Students with disabilities and SAT scores:

The scatter plot at left shows an apparent negative correlation between average verbal SAT scores and percentage of students with disabilities.

The scatter plot below that shows a similar correlation, this time for average math SAT scores and percentage of students with disabilities. This also appears to have a negative correlation.

The table in the text box to the right gives the calculated Pearson r values for the correlations between percentage of students with disabilities and SAT scores. This information shows that the Pearson correlation coefficient is not significant for SAT scores at the any level.


Yverba; = 551.39 – 1.16

(writing SAT score verbal SAT score math SAT score % of students with disabilities 2006-07Pearson r-.05-.07-.07Sig. (2-tailed).71.65.61N515151)Ymath = 558.66 – 1.27

Coefficientsa

Model

Unstandardized

Standardized

t

Sig.

B

Std. Error

Beta

1

(Constant)

551.39

36.35

15.17

.00

% of students with disabilities

-1.16

2.53

-.07

-.46

.65


Model

Unstandardized

Standardized

t

Sig.

B

Std. Error

Beta

1

(Constant)

558.66

36.00

15.52

.00

% of students with disabilities

-1.27

2.51

-.07

-.51

.61


Discussion:

The findings of the various correlations above were disappointing. I was hoping for some blatant significant correlation with a positive line of regression to compare it to. Unfortunately many of the correlations were insignificant or their subsequent regression lines were negative rather than positive. For example the correlation coefficient between estimated average salaries is significant for all SAT scores based on the calculated Pearson coefficients, but the lines of regression for both verbal SAT scores and math SAT scores have no slope. This makes any sort of accurate prediction of the Y-value impossible, especially since the standard error of the estimate (for all of the calculated correlations, not just this one) was a surprisingly large number.

The most surprising conclusion regards the correlation between expenditures and SAT scores. While the Pearson correlation coefficient calculation revealed that the relationship was significant for all SAT scores, the significance was negative, just as the slope of the regression equations were. This negative slope suggests that in using the regression equations to find the value of Y, an increase in expenditures results in a decrease in SAT scores (hence the negative relationship).

In the end, there is no significant relationship between the data sets that allows for a decisive conclusion to be drawn about either a positive or negative correlation between SAT scores and the other variables, or even consistent differences between regions.

Documents

brianjmurphy.files.wordpress.com · Web view2011. 4. 21. · Data has been gathered in order to address the controversy over equity in public school expenditures. The provided data