Final Report

IE 6308 - Design of Experiments

Pneumatic Cannon Experiment Final Report

Professor

Dr. Victoria Chen

Group Members Mewan Wijemanne

Ukesh Chawal Tejas Pawar

P a g e | 1

1. PROPOSAL

The following experiment is being conducted to analyze whether the response variable is affected by the factors taken into consideration. The experiment uses the following apparatus: a Pneumatic cannon with a pressure gauge, air pump, projectiles (balls) and a measuring tape.

Experiment conditions – The Pneumatic cannon will be in a fixed position with an angle of 45 degrees. Since the angle is not considered as one of the factors, it will be constant throughout the experiment. The experiment will be conducted in a closed environment to minimize the effect of wind on the observation.

Problem Statement - The problem is to analyze whether the distance travelled by the ball is being affected by the factors i.e. the air pressure used to launch the ball and the weight of the ball.

Response variable – The distance travelled by the ball is the response variable. The unit of the variable will be feet.

Factors – The two factors are listed below:

Weight of the ball: This factor will have three levels. The weight of the ball will be measured in grams (g). The balls will be of the same size (radius) and shape to further control the factors. The only difference will be the material of the balls which would be the means of getting different weights, which are being considered as the levels of the factor. The three factors will be 11 g (A1), 21 g (A2) and 31 g (A3).

Air pressure used to launch: This factor will have two levels. It will be measured by the pressure gauge on the Pneumatic cannon. The units of the factor is pound per square inch (psi). The two levels of the pressure will be 75 psi (B1) and 90 psi (B2).

The resulting set of treatment combinations are as follows: A1B1 A2B1 A3B1

A1B2 A2B2 A3B2

The experimental units and number of replications: In each treatment, we conduct 3 replications and measure the distance travelled by the ball. Thus, the experimental units are 6*3 = 18 samples of distance travelled by the ball corresponding to each level of pressure and weights.

Data Collection

The data is collected using the following method. The Pneumatic cannon will be grounded to the floor and would not be changed. The cannon would be at an angle of 45 degrees and would remain constant throughout the experiment. Then the projectile of choice (Factor 2) will be inserted into the barrel of the Pneumatic cannon and the compression tank will be pumped with air, to increase the pressure to the desired level (Factor 1). Once it has reached the desired level, all the air will be released, instantly, in to the barrel of the Pneumatic cannon, causing the projectile to launch out of the barrel.

Once the projectile has exited the barrel, it will be tracked till it hits the floor. The distance between the exit point and the first ground impact point will be collected for our experimentation (Figure1.1). Any further travel after the projectiles initial impact, would not be considered as distance traveled. This process would be repeated 3 times, for every combination of factors 1 and 2.

P a g e | 2

Figure1.1 – Data collection visual

Goal of the study: The goal of the study is to prove that the factors, the air pressure used to launch and the weight of the ball, affect the distance travelled by the ball.

The following are the randomization assignment of our treatment, with 18 experimental units.

A2B1 0.061528

A2B1 0.085538

A1B2 0.100582

A2B2 0.223706

A1B1 0.25512

A3B2 0.394642

A3B1 0.408395

A2B1 0.429218

A3B2 0.433996

A3B1 0.451546

A2B2 0.494628

A1B1 0.55602

A1B1 0.66778

A1B2 0.72701

A2B2 0.777258

A3B2 0.823183

A3B1 0.951996

A1B2 0.999628

The data was collected using randomization. We randomized the treatment combinations, resulting in the above mentioned order of data collection:

Factor A – Weight of the ball (Grams)

Factor B – Pressure of the launch (PSI)

Number of observation (n) = 18,

Replications (r) = 3

Treatment combinations (v) = 6

The data collected in the randomized order can be found in the appendix (Table A1)

The same person was responsible for launching the ball every time, to maintain consistency in the launches. This is important as every person might take different speeds to turn the value, to release the pressure for the launch. We conducted the experiment in the area between ERB and ELB to block the effect of wind. Each launch was made in minimal wind flow conditions. All the readings were taken on the same day. Thus we tried to eliminate the time series effect on our data. Each ball was checked for weight consistency before the launch. This was to ensure that the impact of the ball hitting the grass, was not changing the weight of the ball after every launch.

The final data arranged in an order of the two factors and their respective levels is mentioned in the appendix (Table A2).

P a g e | 3

2. PRELIMINARY ANALYSIS

Level of Weight

N Distance

Mean Std Dev

11 6 137.888333 28.6961310

21 6 130.346667 30.1925426

31 6 114.888333 78.6504315

The Factor - Weight of the ball has 3 levels, 11g (A1), 21g (A2) and 31g (A3). In the plot (Figure 2.1) we can see that as the factor level increases the mean is decreasing. The box plot (Figure 2.2) shows the decreasing trend on the mean as the weight of the ball increases. The distribution of the response is significantly higher for the A3 as compared to the other factor levels. This is seen in the initial plot of distance vs weight.

Figure 2.1 – Weight vs Distance Figure 2.2 – Weight vs Distance (Box plot)

Table 2.1 – Weight Means and Standard Deviations

Figure 2.3 – Pressure vs Distance Figure 2.4 – Pressure vs Distance (Box plot)

P a g e | 4

Level of Pressure

N Distance

Mean Std Dev

75 9 86.036667 32.4299676

90 9 169.378889 14.0717975

The factor- Pressure has two levels, 75 psi (B1) and 90 psi (B2). In the plot (Figure 2.3), we notice that with an increase in pressure, there is an increase in the means of the distance. Furthermore, we also see the distribution of the response decreases as the factor level increases from 75 psi to 90 psi.

The Pneumatic cannon project was designed as a complete factorial experiment. There are two factors, Weight of the ball (factor A) and Pressure of the launch (factor B), in the model. So the full interaction model form is;

Yijt = µij + ɛijt

ɛijt ~ N (0, σ2)

µij = µ.. + αi + βj + (αβ)ij

Where Yijt is the response variable of the model, distance travelled by the ball. µ.. is the overall mean, αi is the main treatment effect due factor A, the weight of the ball, βj is the main treatment effect due to factor B, pressure, (αβ)ij is the interaction treatment effect of pressure and the weight, ɛijt is the associate random error.

Residual Analysis

The purpose of conducting residual analysis is to verify model assumptions made on the complete factorial model. By verifying the model assumptions, we can check the validity of the model.

The following are the main model assumptions that are focused for this project:

Normality

Constant variance

No outliers

No serial correlation

For the above mentioned model assumptions, we are using an alpha level (α) of 0.05.

Table 2.2 – Pressure Means and Standard Deviations

P a g e | 5

Normality: In the normality section, residual analysis is performed to assess the Normality distribution of the data.

Normal probability plot (Figure 2.5) is constructed to check the normality of residuals. A slight longer left tail is seen in the plot but overall the distribution is pretty straight. Also no outliers were detected. Thus we conclude normality assumption is satisfied.

Test for Normality

H0: Normality is ok

H1: Normality is violated

Sample correlation – 0.98010

C (α, n) = C (0.05, 18) = 0.946

Since the sample correlation (0.9801) > C (α, n), we fail to reject H0. With 95% confidence, we conclude that the normality ok. Since the conclusion from the normality test agrees with the plot (Figure 2.5), we conclude that the model assumption of normality is satisfied

Pearson Correlation Coefficients, N = 18 Prob > |r| under H0: Rho=0

e enrm

e

1.00000

0.98010

Enrm

0.98010

1.00000

Figure 2.5– Normality Plot

Table 2.3 – Normality Correlation Coefficient

P a g e | 6

Constant variance: In the constant variance section, residual analysis is performed to assess the constant variance of the data.

Although a clear funnel shape is not seen in the plot (Figure 2.6), we can see in the two circled areas that the data does not have constant variance. Also no outliers were detected. Thus we conclude that the constant variance assumption is not satisfied.

Modified Levene Test (SAS Output):

Source DF Sum of Squares Mean Square F Value Pr > F

Model 5 31.8417167 6.3683433 0.49 0.7775

Error 12 155.8323333 12.9860278

Corrected Total 17 187.6740500

Source DF Anova SS Mean Square F Value Pr > F

TC 5 31.84171667 6.36834333 0.49 0.7775

Test for Constant Variance

H0: Means of the dit Population are equal

H1: Not all means are equal.

Conclusion: Since the p-value is greater than α=0.05, thus we fail to reject Ho. Thus, we can conclude that the means of the population are equal, hence the constant variance assumption is satisfied. This conclusion contradicts our plot (Figure 2.6) analysis, therefore given that the data is distributed normally, we are conducting the Hartley’s Constant Variance test.

Figure 2.6 – Variance Plot

P a g e | 7

Hartley’s Test: Since the replications are equal in the experiment and our normality is satisfied, we can conduct the Hartley’s test for non-constant variance.

H0: Variance is constant

H1: Variance is non-constant.

Test Statistic (H*) = max si2 / min si2 = 19.1874

H1-α, v, r-1 = H0.95, 6, 2 = 10.8

Since test statistic H* > H table value, we reject H0. Thus with 95% confidence we can conclude non constant variance assumption is satisfied.

Conclusion: We used two separate tests to conclude the interpretation of the constant variance assumption. The plot and Hartley’s test show that we have non constant variance and the modified Levene shows that we have constant variance. Under the conclusion that the data is normally distributed and the plot (Figure 2.6) analysis justifies the outcome, we decided to go with the Hartley test conclusion. Thus, we conclude that the variance in the data is non- constant.

Serial Correlation Test

In this experiment, the data was collected in a series. Thus, we need to assess if there is any serial correlation in the data. Here we can see there is no clear trend in the graph, we can only see random jaggedness. Also no outliers were detected. Thus we can conclude that there is no serial correlation in the data.

Std Dev Variance 4.223775 17.84028 5.500842 30.25926 1.835857 3.370371 3.004626 9.027777 1.964971 3.861111 8.041703 64.66899

Table 2.4 – Standard Deviation and Variance of data

18161412108642

5.0

2.5

0.0

-2.5

-5.0

-7.5

-10.0

Index

RES

I1

Time Series Plot of RESI1

Figure 2.7 – Time Series plot

P a g e | 8

Bonferroni Outlier Test

In order to determine if there are outliers present in the data, we use Bonferroni outlier test to check by comparing the absolute value of studentized deleted residual with the cutoff at significance 0.05.

The Deleted studentized residuals are mentioned in the appendix.

The cut off rule is, if abs(t) > tn-v-1,α/2n , then it is a outlier. In our project the tn-v-1,α/2n = t 11, 0.00138 = 3.837

By analyzing rstudent value in the appendix table (Table A3), with a 95% confidence, we can conclude that there are no Studentized deleted residuals greater than 3.837. Therefore no outliers are identified with using Bonferroni outlier test. As a result, we conclude the assumption of no outliers is satisfied.

Analysis of the model assumptions concluded that the normality, outlier and serial correlation assumptions were satisfied. However since the constant variance assumption was not satisfied, we decided to do a transformation on the data.

Transformation

Due to the presence of non-constant variance in the model, we are conducting a variance stabilizing transformations on the response (Distance). To best determine the transformation required, we inspected consistency among the transformation options.

Factor Level Std. Dev Yhat

1 4.223775 112

2 5.500842 163.67

3 1.835857 102.86

4 3.004626 157.833

5 1.964971 43.25

6 8.041703 186.53

Si2/ Yhat Si/ Yhat Si/ Yhat2

0.159288 0.037712 0.000337

0.18488 0.033609 0.000205

0.032767 0.017848 0.000174

0.057198 0.019037 0.000121

0.089274 0.045433 0.00105

0.346695 0.043112 0.000231

The most consistent values are of the transformation Si/ Yhat, based on the transformation values (Table 2.6). Thus we transform the data using log Y transformation.

Table 2.5 – Standard Deviation and Predicted value

Table 2.6 – Transformation Options

P a g e | 9

Normality (on transformed data)

From inspection of the normality plot (Figure 2.7) a shorter left tail is seen but overall the distribution is pretty straight. Also no outliers were detected. Thus we conclude normality assumption is satisfied.

Test for normality

H0: Normality is OK

H1: Normality is violated

sample correlation, =0.98207,

we find cutoff c (α =0.05, n=18)= 0.943.

Since the sample correlation (0.98207) > C (α, n), we fail to reject H0. With 95% confidence, we conclude that the normality ok. Since the conclusion from the normality test agrees with the plot (Figure 2.7), we conclude that the model assumption of normality is satisfied.

Pearson Correlation Coefficients, N = 18 Prob > |r| under H0: Rho=0

e enrm

e

1.00000

0.98207

Enrm

0.98207

1.00000

Figure 2.7 – Normality plot

Table 2.7 – Normality Correlation Coefficient

P a g e | 10

Constant variance (on transformed data)

The above constant variance plot (Figure 2.8) shows no clear shape and have similar variances. Thus we can conclude that the constant variance assumpition is satisfied.

Modified Levene Test (SAS Output):


Model 5 0.00015847 0.00003169 0.25 0.9303

Error 12 0.00150392 0.00012533


Source DF Anova SS Mean Square F Value Pr > F

TC 5 0.00015847 0.00003169 0.25 0.9303

H0 = Variance is Equal

H1 = Variance is Unequal.

The P-value obtained from the Modified-Levene test is 0.9303, which is greater than α=0.05, therefore it fails to reject H0. Thus with 95% we can conclude that the constant variance model assumption is satisfied, which is consistent with the conclusion of the plot (Figure2.8).

Figure 2.8 – Constant Variance plot

P a g e | 11

Bonferroni Outlier Test (On transformed data)

The rstudent residuals for the transformed data can be found in the appendix table (Table A4). The cut off rule is, if abs (t) > tn-v-1,α/2n , then it is a outlier. In our project the tn-v-1,α/2n = t 11, 0.00138 = 3.837

We can clearly see, from Table A4, that there are no Studentized deleted residuals greater than 3.837. Therefore no outliers are identified by using Bonferroni outlier test. As a result, with 95% confidence, we conclude the model assumption of no outlier, is satisfied.

Serial Correlation Test

In this experiment, the data was collected in a series. Thus, we need to assess if there is any serial correlation in the data. Here we can see there is no clear trend in the graph, we can only see random jaggedness. Also no outliers were detected. Thus we can conclude that there is no serial correlation in the data.

Thus, all The Model Assumptions have been satisfied after Log Y transformation.

18161412108642

0.2

0.1

0.0

-0.1

-0.2

Index

RES

I1

Time Series Plot of RESI1

Figure 2.9 – Time Series plot

P a g e | 12

3. ANALYSIS OF VARIANCE

Interaction: In this analysis, we start with the interaction effect of the factors. In this plot, we can see two distinct lines for factor A, weight of the ball, and factor B, Pressure of the launch. Thus we can conclude that Factor A and Factor B are main effects. As seen that the lines are not parallel (slopes not similar), thus the interaction effect is also significant.

The GLM Procedure

Dependent Variable: LogY (SAS Output)


Model 5 0.80853145 0.16170629 711.17 <.0001

Error 12 0.00272855 0.00022738


Figure 3.1 – Interaction plot

P a g e | 13

R-Square Coeff Var Root MSE LogY Mean

0.996637 0.730837 0.015079 2.063266

Source DF Type I SS Mean Square F Value Pr > F

Weight 2 0.11136659 0.05568329 244.89 <.0001

Pressure 1 0.48590448 0.48590448 2136.98 <.0001

Weight*Pressure 2 0.21126038 0.10563019 464.55 <.0001

Source DF Type III SS Mean Square F Value Pr > F

Weight 2 0.11136659 0.05568329 244.89 <.0001

Pressure 1 0.48590448 0.48590448 2136.98 <.0001

Weight*Pressure 2 0.21126038 0.10563019 464.55 <.0001

The SSTot = SSTr + SSE

SSTr = SSA +SSB + SSAB where SSA – sum of squares of the weight; SSB – sum of squares of pressure and SSAB – sum of squares of weight*pressure.

SSTr = 0.80853145

SSA = 0.48590448; SSB = 0.11136659 & SSAB = 0.21126038

SSA + SSB + SSAB = 0.80853145

Sum Squares of Error (SSE) represents the unexplained variability, in this case SSE = 0.00272855

Type I sum of squares is the sum of squares for all the independent variables corresponding to the predicted variable; test the main effect of each factor, followed by the interaction term after the main effects. Type III sum of squares which is the sum of squares added contribution after all other effects are included; tests the effect of the interaction term and then the main effects.

The interaction term of weight*pressure is significant as the p-value is <0.0001 which is less than α = 0.05. The p-value of pressure and weight individually are <0.0001, both of which are less than α = 0.05. Thus these findings are consistent with the results of the interaction plot.

The Type I SS and the Type III SS, both are the same, thus the design is orthogonal, which means that factor A and factor B are independent and are main effects.

P a g e | 14

Estimate the effects

αi + βj + (αβ)ij

The main effects of factor A; Weight, and factor B; Pressure, can be estimated by the following formula:

∝𝑖 ̂= µ̂𝑖. + µ̂.. where µi. is the mean of the factor level – Weight

�̂�1 = µ̂1. + µ̂.. = 1.90 – 2.06 = -0.16

Level of Weight

N LogY

Mean Std Dev

11 6 2.13155090 0.09146845

21 6 2.10517338 0.10209813

31 6 1.95307507 0.34812093

𝛽�̂� = µ̂.j + µ̂.. where µ̂.j is the mean of the factor level – Pressure

𝛽1̂ = µ̂.1 + µ̂.. = 2.13 – 2.06 = 0.07

Level of Pressure

N LogY

Mean Std Dev

75 9 1.89896583 0.19855950

90 9 2.22756707 0.03526423

The interaction effect due to factor A and factor B can be estimated by:

(𝛼�̂�)𝑖𝑗 = µ̂ij - µ̂i. - µ̂.j + µ̂..

(𝛼𝛽)̂11 = µ̂11 - µ̂1. - µ̂.1 + µ̂.. = 2.05 – 1.90 – 2.13 + 2.06 = 0.08

Parameter Estimate Standard Error t Value Pr > |t|

Pressure75 vs. Pressure90 -0.32860124 0.00710836 -46.23 <.0001

Weights<31 vs. Weight31 0.16528707 0.00753955 21.92 <.0001

Weight11 vs. Weights>11 0.10242667 0.00753955 13.59 <.0001

Table 3.1 – Estimates

Table 3.3 – Means and Standard Deviations of Pressure

Table 3.2 – Means and Standard Deviations of Weight

P a g e | 15

Level of TC

N LogY

Mean Std Dev

1 3 2.04901438 0.01624520

2 3 2.21408742 0.01467838

3 3 2.01220037 0.00775288

4 3 2.19814640 0.00824724

5 3 1.63568275 0.01993918

6 3 2.27046739 0.01895310

Main Treatment Effect Interaction Treatment Effect

𝛼1̂ -0.16 (𝛼𝛽)̂11 0.082

𝛼2̂ 0.16 (𝛼𝛽)̂12 -0.082

𝛽1̂ 0.07 (𝛼𝛽)̂21 0.071

𝛽2̂ 0.04 (𝛼𝛽)̂22 -0.071

𝛽3̂ -0.11 (𝛼𝛽)̂31 -0.153

(𝛼𝛽)̂32 0.153

Table 3.4 – Treatment effects with main effects and interaction effects

Table 3.3 – Means and Standard Deviations of the Treatment Combinations

P a g e | 16

4. ANALYSIS OF EFFECT

For pairwise comparisons, we have 3 methods i.e. Bonferroni, Scheffe and Tukey. We will be selecting the method which provides the narrowest confidence interval.

Bonferroi Multiplier (WB)= tn-v, α/2m = t 18-6, 0.05/36 = 3.749

Sceffe Multiplier (WS) = √(𝑣 − 1)𝐹𝑣−1,𝑛−𝑣,∝ = 3.943

Tukey Multiplier (WT) = 1

√2 𝑞𝑣,𝑛−𝑣,∝ = 3.359

Since the Tukey Coefficeint has the smallest value, thus we will get the narrowest confidence interval.

Tukey (C.I): �̂� + 𝑤𝑇 𝑠𝑒(�̂�) , se (𝐷)̂ = √𝜎2/𝑟

H0: D = 0 means treatments difference between i and j is negligible

H1: D ≠ 0 means treatments difference between i and j is not negligible

Fail to Reject H0 : In the Tukey confidence interval table, we see that there are two pairs of treatment combinations which have 0 in the confidence interval. Thus, with 95% confidence, we fail to reject H0. The insignificant pair of treatment combinations has been highlighted in the table. This means that

Treatment 1 (pressure = 75; weight = 11) and treatment 3 (pressure = 75; weight = 21) are not statistically different.

Treatment 2 (pressure = 90; weight =11) and treatment 4(pressure = 90; weight = 21) are not statistically different.

Reject Ho : In all the other cases, the treatment combinations are statistically different. For eg: Treatment 1(pressure = 75; weight = 11) and treatment 2(pressure =90; weight =11) are statistically different. Hence, with 95% confidence, we can conclude that treatment 1 and treatment 2 are statistically different.

Least Squares Means for Effect Weight*Pressure

i j Difference Between Means

Simultaneous 95% Confidence Limits for LSMean(i)-LSMean(j)

1 2 -0.165073 -0.206428 -0.123718

1 3 0.036814 -0.004541 0.078169

1 4 -0.149132 -0.190487 -0.107777

1 5 0.413332 0.371977 0.454687

1 6 -0.221453 -0.262808 -0.180098

2 3 0.201887 0.160532 0.243242

2 4 0.015941 -0.025414 0.057296

2 5 0.578405 0.537050 0.619760

Table 4.1 – Least Square Means for effect Weight*Pressure

P a g e | 17

Least Squares Means for Effect Weight*Pressure

i j Difference Between Means

Simultaneous 95% Confidence Limits for LSMean(i)-LSMean(j)

2 6 -0.056380 -0.097735 -0.015025

3 4 -0.185946 -0.227301 -0.144591

3 5 0.376518 0.335163 0.417873

3 6 -0.258267 -0.299622 -0.216912

4 5 0.562464 0.521109 0.603819

4 6 -0.072321 -0.113676 -0.030966

5 6 -0.634785 -0.676140 -0.593430

The Confidence Interval that contain zero, are statistically different.

Line plot

The line plot (Figure 4.1) shows that the treatment combinations of 1 & 3, 2 & 4 are statistically the same. All the other treatment combinations are statistically different.

Means Treatment Combinations

1.63568 5

2.0122 3

2.04901 1

2.19815 4

2.21409 2

2.27047 6

Figure 4.1 – Line Plot

Table 4.2 – Means difference

P a g e | 18

Dot plot

The above dot plot (Figure 4.2) illustrates that the treatment effects are divided from the line, thus we can conclude that the treatment effects are not negligible.

Bonferroni Method: The Bonferroni method is used to test the multiple comparisons of our pre-selected set of the differences and contrasts

µ̂.1 − µ̂.2 : - In this case, we determine if there is a significant difference between the two levels of pressure i.e. 75 psi and 90 psi.

H0:- Means of both the pressure levels (75 psi and 90 psi) are equal H1:- Means of both are not equal

I) µ̂1.+ µ̂2.

2 - µ̂3. :- In this case, we determine if there is a difference in distance travelled

between using balls of weights greater than or equal to 31g or less than 31g H0:- weight >= 31g H1:- weight< 31g

II) µ̂1. −µ̂2.+ µ̂3.

2 :- In this case, we determine if there is a difference in distance travelled

between using balls of weights less than or equal to 11 g and greater than 11g. H0:- weight <= 11g H1:- weight > 11g

Figure 4.2 – Dot Plot

P a g e | 19

�̂� ± 𝑤𝐵𝑠𝑒(�̂�)

𝑤𝐵 = 𝑡𝑑𝑓𝐸𝑟𝑟,𝛼 2(𝑚)⁄ = 𝑡12,0.052(3)⁄

= 2.779

MSE = 0.00022738

𝐷1̂ = µ̂.1 − µ̂.2 = 1.89896583 - 2.22756707 = - 0.32860124

𝐿2̂ = µ̂1.+ µ̂2.

2 - µ̂3. =

2.13155090 + 2.10517338

2 - 1.95307507 = 0.16528707

𝐿3̂ = µ̂1. −µ̂2.+ µ̂3.

2 =

1.95307507 + 2.10517338

2 - 2.13155090 = 0.10242667

Se(�̂�) = √𝑀𝑆𝐸 (1

𝑟1+

1

𝑟2) = √0.00022738 (

1

9+

1

9) = 0.00710837

Se(�̂�) = √𝑀𝑆𝐸 (∑𝑐𝑖

2

𝑟𝑖) = √0.00022738 [(

14⁄

6) + (

14⁄

6) +

1

6] = 0.00753957

𝐷1̂ ± 𝑤𝐵𝑠𝑒(�̂�) = 0.32860124 ± (2.779)( 0.00710837) = (0.308847 , 0.328601)

Reject Ho,

With 95% confidence we can conclude that projectiles launched with a pressure of 90 psi travels a greater distance, than projectiles launched with a pressure of 75 psi, with a difference, in mean Log of Distance (feet), between 0.308847 and 0.328601.

𝐿2̂ ± 𝑤𝐵𝑠𝑒(�̂�) = 0.16528707 ± (2.779)( 0.00753957) = (0.144335 , 0.165287)

Reject Ho,

With 95% confidence, we can conclude that using weights less than 31g will travel a greater distance as compared to weights that are 31 g, with a difference, in mean Log of Distance (feet), between 0.144335 & 0.165287.

𝐿3̂ ± 𝑤𝐵𝑠𝑒(�̂�) = 0.10242667 ± (2.779)( 0.00753957) = (0.081474 , 0.102427)

Reject Ho,

With 95% confidence, we can conclude that using weights that are 11g, travel greater distance than weight greater than 11gwith a difference, in mean Log of Distance (feet), betwee 0.081474 & 0.102427.

P a g e | 20

5. FINAL DISCUSSION

The objective of this project was to determine if the weight of a projectile and the pressure at which it is launched at, has an effect on the distance it travels. In order to determine this, we used 3 different weights and 2 different pressures. And these combinations were replicated 3 times. Given the findings from the data analysis, it is easy to confirm this theory.

The initial analysis on the data showed that the normality and constant variance conditions were satisfied. However after visual analysis of the 2 plots, it was concluded that the constant variance is violated. Therefore a decision was made to do a Log transformation on Y. This resulted in a much better constant variance plot and an acceptable normality plot. In order to determine if this can further be corrected, a square root transformation on Y was conducted. This resulted in the constant variance plot to violating the condition. Based on these results, the final conclusion was to use the Log Y transformation for our analysis.

Next a Bonferroni outlier test was conducted on the data to determine if there were any outliers in the dataset. The analysis concluded that all data points are within bounds and that there were no outliers in the dataset. In order to determine if there was serial correlation in our data, a time series plot was generated. This showed random jagged lines, which indicated that there is no serial correlation in the data.

Afterwards the main factor effects (Pressure and Weight) and the interaction effect was tested for significance. The results stated that both main factor effects and the interaction effects were significant. This concludes that both the main factors and the interaction factor has an effect in determining the distance travelled.

Finally the pairwise comparison and the estimates were calculated. The pairwise comparison was calculated using the Tukey method and showed that the treatment combinations 1 and 3 are statistically the same. The same was shown about the treatment combination 2 and 4. The estimated were calculated using Bonferroni method and showed that using a pressure at 95 PSI will give a greater distance than that of using 75 PSI, that using weights less than 31g will have a greater distance than that of using 31g, that using weights that are 11g will have a greater distance than that of using weights greater than 11g.

In conclusion the results from the analysis on the data confirmed the project hypothesis that the weight of a projectile and the pressure at which it is launched at, has an effect on the distance it travels. However a Log transformation on Y is needed to get the most accurate results. For future experimentation, adding 1 or more of the following factors should be considered; shape of the projectile, angle of the cannon and projectile surface texture or even adding a blocking factor. In terms of future work for analytics, a weighted model would be considered, also an optimal sample size would be used (Refer to next page for preliminary future work).

P a g e | 21

Future Work

A. Weighted Least Square

Weights Pressure N Mean Std. Dev Si Wi= 1/si2

11 75 3 112 4.22302 17.8339 0.003144

11 90 3 163.777 5.49714 30.21853 0.001095

21 75 3 102.86 1.83747 3.3763 0.087724

21 90 3 157.833 3.00786 9.047233 0.012217

31 75 3 43.25 1.96822 3.8739 0.066635

31 90 3 186.527 8.04361 64.69963 0.000239

ANOVA

Source DF Adj SS Adj MS F-Value P-Value

Pressure(psi) 1 994.82 994.821 53.44 0

Weights(g) 2 34.06 17.031 0.91 0.423

Error 14 260.63 18.617 Lack-of-Fit 2 259.35 129.676 1215.56 0

Pure Error 12 1.28 0.107 Total 17 1310.73

H0: All the means are equal.

H1: Atleast one mean is different.

F* = (SSTrw / v-1)/(SSEw/n-v) = 9.3157

F Table value = 53.44

Since F table > F statistics, thus we fail to reject Ho. In conclusion constant variance assumption is

satisfied.

B. Sample size calculation

As a part of our future work, we wanted to see how many observations would be needed to ensure that treatment-versus-control confidence intervals are of length less than or equal to 0.1 log of distance (feet).

N= 18 , v=6 , n-v=12, α = .05, SSE = 0.00272855, L<=0.1.

Upper bound 95% C.I estimate for 𝜎2 is (0, 𝑆𝑆𝐸

𝜒𝑛−𝑣,1−α2 ) = (0, 0.00052211)

Total number of replications for the observations needed, r = 4q2𝜎2 / L2 where q v, n-v, α = 4.37

Hence, r = 4.71.

P a g e | 22

If we choose r =5 that is n = 30 then q v, n-v, α = 4.37 and confidence interval length is ( 2q√𝜎2

√𝑟) =0.089

which is less than 0.1.


√𝑟) =0.103

which is greater than 0.1.


√𝑟) =0.080

which is less than 0.1.

Based on the above calculation the best solution is with r = 5 meaning 30 (n=v*r=6*5) sample observations would be needed to ensure that treatment-versus-control confidence intervals are of length less than or equal to 0.1 log of distance (feet).

P a g e | 23

APPENDIX:

Table A1- Raw Data: Randomization assignment of our treatment

Observation Weights(g) Pressure(psi) Distance(Feet)

1 21 75 101.08

2 21 75 104.75

3 11 90 165.33

4 21 90 161.17

5 11 75 116.75

6 31 90 190

7 31 75 43.75

8 21 75 102.75

9 31 90 192.25

10 31 75 44.92

11 21 90 155.33

12 11 75 110.58

13 11 75 108.67

14 11 90 157.67

15 21 90 157

16 31 90 177.33

17 31 75 41.08

18 11 90 168.33

P a g e | 24

Table A2 - Raw Data (Sorted)

Weights(g) Pressure(psi) Distance(Feet)

11 75 116.75

11 75 110.58

11 75 108.67

11 90 165.33

11 90 157.67

11 90 168.33

21 75 101.08

21 75 104.75

21 75 102.75

21 90 161.17

21 90 155.33

21 90 157

31 75 43.75

31 75 44.92

31 75 41.08

31 90 190

31 90 192.25

31 90 177.33

P a g e | 25

Table A3 - Deleted Studentized Residuals for Bonferroni Outlier Test

Obs Weight Pressure Distance TC yhat e rstudent

1 11 75 116.75 1 112 4.75 1.28844

2 11 75 110.58 1 112 -1.42 -0.36116

3 11 75 108.67 1 112 -3.33 -0.87048

4 11 90 165.33 2 163.777 1.55333 0.39553

5 11 90 157.67 2 163.777 -6.10667 -1.7446

6 11 90 168.33 2 163.777 4.55333 1.22761

7 21 75 101.08 3 102.86 -1.78 -0.45426

8 21 75 104.75 3 102.86 1.89 0.48291

9 21 75 102.75 3 102.86 -0.11 -0.02781

10 21 90 161.17 4 157.833 3.33667 0.87234

11 21 90 155.33 4 157.833 -2.50333 -0.6448

12 21 90 157 4 157.833 -0.83333 -0.21113

13 31 75 43.75 5 43.25 0.5 0.12651

14 31 75 44.92 5 43.25 1.67 0.42571

15 31 75 41.08 5 43.25 -2.17 -0.55633

16 31 90 190 6 186.527 3.47333 0.91071

17 31 90 192.25 6 186.527 5.72333 1.60826

18 31 90 177.33 6 186.527 -9.19667 -3.26103

P a g e | 26

Table A4 - Deleted Studentized Residuals for Bonferroni Outlier Test (After Transformation)

Obs Weight Pressure Distance TC LogY yhatY eY rstudentY

1 11 75 116.75 1 2.06726 2.04901 0.018243 1.56941

2 11 75 110.58 1 2.04368 2.04901 -0.005338 -0.41838

3 11 75 108.67 1 2.03611 2.04901 -0.012905 -1.05287

4 11 90 165.33 2 2.21835 2.21409 0.004264 0.33327

5 11 90 157.67 2 2.19775 2.21409 -0.016338 -1.37545

6 11 90 168.33 2 2.22616 2.21409 0.012074 0.97897

7 21 75 101.08 3 2.00467 2.01220 -0.007535 -0.59532

8 21 75 104.75 3 2.02015 2.01220 0.007954 0.62955

9 21 75 102.75 3 2.01178 2.01220 -0.000419 -0.03255

10 21 90 161.17 4 2.20728 2.19815 0.009138 0.72748

11 21 90 155.33 4 2.19126 2.19815 -0.006891 -0.54301

12 21 90 157.00 4 2.19590 2.19815 -0.002247 -0.17496

13 31 75 43.75 5 1.64098 1.63568 0.005295 0.41499

14 31 75 44.92 5 1.65244 1.63568 0.016757 1.41704

15 31 75 41.08 5 1.61363 1.63568 -0.022052 -2.00345

16 31 90 190.00 6 2.27875 2.27047 0.008286 0.65688

17 31 90 192.25 6 2.28387 2.27047 0.013399 1.09752

18 31 90 177.33 6 2.24878 2.27047 -0.021685 -1.95834

P a g e | 27

Documents

Final Report