Chapter 9, Problem 9SE

The article “Combined Analysis of Real-Time Kinematic GPS Equipment and Its Users for Height Determination” (W. Featherstone and M. Stewart, Journal of Surveying Engineering. 2001:31–51) presents a study of the accuracy of global positioning system (GPS) equipment in measuring heights. Three types of equipment were studied, and each was used to make measurements at four different base stations (in the article a fifth station was included, for which the results differed considerably from the other four). There were 60 measurements made with each piece of equipment at each base. The means and standard deviations of the measurement errors (in mm) are presented in the following table for each combination of equipment type and base station.

1. a. Construct an ANOVA table. You may give ranges for the P-values.
2. b. The question of interest is whether the mean error differs among instruments. It is not of interest to determine whether the error differs among base stations. For this reason, a surveyor suggests treating this as a randomized complete block design, with the base stations as the blocks. Is this appropriate? Explain.

To determine

Construct the ANOVA.

Answer to Problem 9SE

The ANOVA table is,

Source	DF	SS	MS	F	P
Base	3	13,495	4498.3	7.5308	0.000
Instrument	2	90,990	45,495	76.164	0.000
Interaction	6	12,050	2,008.3	3.3622	0.003
Error	708	422,912	597.33
Total	719	539,447

Explanation of Solution

Calculation:

The given information is based on the accuracy of the global positioning system (GPS) which has 3 instruments that have to make measurements at four different base stations have 60 measurements with each piece of instrument at each base.

Let us denote the main effects of Instruments (I) with ${\alpha }^{\prime }s$, the main effects of Base (J) with ${\beta }^{\prime }s$ and measurements (K) denotes the number of replicates .

The aim is to find the ANOVA.

State the hypothesis:

Main effect of Instruments:

Null hypothesis:

$H_0:{\alpha }_1={\alpha }_2={\alpha }_3=0$

Alternative hypothesis:

$H_1:\text{ At least one }{\alpha }_i\ne 0$

Main effect of Base:

Null hypothesis:

$H_0:{\beta }_1={\beta }_2={\beta }_3={\beta }_4=0$

Alternative hypothesis:

$H_1:\text{ At least one }{\beta }_i\ne 0$

Interaction:

Null hypothesis:

$H_0:{\gamma }_{11}={\gamma }_{12}=\mathrm{...}={\gamma }_{34}=0$

Alternative hypothesis:

$H_1:\text{At least one}{\gamma }_{ij}\ne 0$

ANOVA table:

Source	DF	SS	MS	F
Treatment	$\left(I-1\right)$	$JK{\displaystyle \sum _{i=1}^I{\left({\overline{X}}_{i\mathrm{..}}-{\overline{X}}_{\mathrm{...}}\right)}^2}$	$\frac{SSTr}{DF\left(SSTr\right)}$	$\frac{MSR}{MSE}$
Blocks	$\left(J-1\right)$	$IK{\displaystyle \sum _{j=1}^J{\left({\overline{X}}_{.j.}-{\overline{X}}_{\mathrm{...}}\right)}^2}$	$\frac{SSB}{DF\left(SSB\right)}$	$\frac{MSC}{MSE}$
Interaction	$\left(I-1\right)\left(J-1\right)$	$\left\{\begin{array}{l}K{\displaystyle \sum _{i=1}^I{\displaystyle \sum _{j=1}^J{\overline{X}}_{ij.}^2}}-JK{\displaystyle \sum _{i=1}^I{\overline{X}}_{i\mathrm{..}}^2}-\\ IK{\displaystyle \sum _{i=1}^I{\overline{X}}_{.j.}^2}+IJK{\overline{X}}_{\mathrm{...}}^2\end{array}\right\}$	$\frac{SSAB}{DF\left(SSAB\right)}$	$\frac{MSI}{MSE}$
Error	$IJ\left(K-1\right)$	${\displaystyle \sum _{i=1}^I{\displaystyle \sum _{j=1}^J{\displaystyle \sum _{k=1}^K{\left({\overline{X}}_{ijk}-{\overline{X}}_{ij.}\right)}^2}}}$	$\frac{{\displaystyle \sum _{i=1}^I{\displaystyle \sum _{j=1}^J{s}_{ij.}{}^2}}}{IJ}$
Total	$IJK-1$

Where, ${\overline{X}}_{\mathrm{...}}=\frac{{\displaystyle \sum _{i=1}^I{\displaystyle \sum _{j=1}^J{\displaystyle \sum _{k=1}^K{\overline{X}}_{ijk}}}}}{N}$

Where, N is the sample size, I denotes the number of treatments, $J$ denotes the number of blocks of the i^th and j^th groups respectively, ${\overline{X}}_{i\mathrm{..}}$ be means of the i^th treatment mean, ${\overline{X}}_{.j.}$ be the j^th block mean respectively, ${\overline{X}}_{\mathrm{...}}$ denotes the overall mean, $s_{ij}$ be the standards deviation of the i^th and j^th groups, MSTr, MSB, MSI and MSE are the mean square of treatments, blocks, interaction and error respectively obtained from the analysis and $\alpha$ is the level of significance at which the test is performed.

Here, the number of treatments (I) is 3, the number of blocks (J) is 4 and the number of replicates (K) is 60.

Test the hypothesis on 5% level of significance:

Here, Instrument is the treatments and Base is the blocks.

Level of significance:

The level of significance is $\alpha =0.05$.

Degrees of freedom:

Base degrees of freedom:

$\begin{array}{c}df=I-1\\ =4-1\\ =3\end{array}$

Instrument degrees of freedom:

$\begin{array}{c}df=J-1\\ =3-1\\ =2\end{array}$

Interaction degrees of freedom:

$\begin{array}{c}df=\left(I-1\right)\left(J-1\right)\\ =\left(4-1\right)\left(3-1\right)\\ =6\end{array}$

Error degrees of freedom:

$\begin{array}{c}df=IJ\left(K-1\right)\\ =4\left(3\right)\left(60-1\right)\\ =708\end{array}$

Total degrees of freedom:

$\begin{array}{c}df=IJK-1\\ =4\left(3\right)\left(60\right)-1\\ =720-1\\ =719\end{array}$

The treatment and block means are tabulated below:

Base	Instruments			Block mean ${\overline{X}}_{i\mathrm{..}}$
	Instrument A	Instrument B	Instrument C
0	3	–24	–6	–9
1	14	–13	–2	–0.33
2	1	–22	4	–5.667
3	8	–17	15	2
Treatment mean ${\overline{X}}_{.j.}$	6.5	–19	2.75	${\overline{X}}_{\mathrm{...}}=-3.25$

By observing the data, the values of ${\overline{X}}_{i\mathrm{..}}$,${\overline{X}}_{.j.}$ and ${\overline{X}}_{\mathrm{...}}$ has been found.

For Base:

The value of SSB (Base) is:

Substitute ${\overline{X}}_{\mathrm{.1.}}=-9$, ${\overline{X}}_{\mathrm{.2.}}=-0.33$,${\overline{X}}_{\mathrm{.3.}}=-5.667$ and ${\overline{X}}_{\mathrm{.4.}}=2$ for $j=1,2,3,4$,$I=3$,$K=60$,${\overline{X}}_{\mathrm{...}}=-3.25$ to get the  value of SSB(Base):

$\begin{array}{c}SSB\left(\text{Base}\right)=IK{\displaystyle \sum _{j=1}^4{\left({\overline{X}}_{.j.}-{\overline{X}}_{\mathrm{...}}\right)}^2}\\ =\left(3\times 60\right)\left\{{\left({\overline{X}}_{\mathrm{1..}}-{\overline{X}}_{\mathrm{...}}\right)}^2+{\left({\overline{X}}_{\mathrm{2..}}-{\overline{X}}_{\mathrm{...}}\right)}^2+{\left({\overline{X}}_{\mathrm{3..}}-{\overline{X}}_{\mathrm{...}}\right)}^2+{\left({\overline{X}}_{\mathrm{4..}}-{\overline{X}}_{\mathrm{...}}\right)}^2\right\}\\ =\left(3\times 60\right)\left\{\begin{array}{l}{\left(-9-\left(-3.25\right)\right)}^2+{\left(-0.3333-\left(-3.25\right)\right)}^2+\\ {\left(-5.6667-\left(-3.25\right)\right)}^2+{\left(2-\left(-3.25\right)\right)}^2\end{array}\right\}\\ =180\left\{33.0625+8.5071+5.8404+27.5625\right\}\\ \approx 13,495.05\end{array}$

The value of MSB (Base) is:

Substitute $J=4$,$SSB\left(\text{Base}\right)=13,495.05$ to get the  value of MSA( Base):

$\begin{array}{c}MSB\left(\text{Base}\right)=\frac{SSB\left(\text{Base}\right)}{J-1}\\ =\frac{13,495.05}{4-1}\\ =\frac{13,495.05}{3}\\ =4,498.35\end{array}$

For Instruments:

The value of SSTr ( Instruments) is:

Substitute ${\overline{X}}_{\mathrm{1..}}=6.5$, ${\overline{X}}_{\mathrm{2..}}=-19$ and ${\overline{X}}_{\mathrm{3..}}=2.75$ for $i=1,2,3$,$J=4$,$K=60$,${\overline{X}}_{\mathrm{...}}=-3.25$ to get the  value of SSTr ( Instruments):

$\begin{array}{c}SSR\left(\text{Instruments}\right)=JK{\displaystyle \sum _{j=1}^3{\left({\overline{X}}_{i\mathrm{..}}-{\overline{X}}_{\mathrm{...}}\right)}^2}\\ =\left(4\times 60\right)\left\{{\left({\overline{X}}_{\mathrm{.1.}}-{\overline{X}}_{\mathrm{...}}\right)}^2+{\left({\overline{X}}_{\mathrm{.2.}}-{\overline{X}}_{\mathrm{...}}\right)}^2+{\left({\overline{X}}_{\mathrm{.3.}}-{\overline{X}}_{\mathrm{...}}\right)}^2\right\}\\ =\left(4\times 60\right)\left\{\begin{array}{l}{\left(6.5-\left(-3.25\right)\right)}^2+{\left(-19-\left(-3.25\right)\right)}^2+\\ {\left(2.75-\left(-3.25\right)\right)}^2\end{array}\right\}\\ =240\left\{95.0625+248.0625+36\right\}\\ \approx 90,990\end{array}$

The value of MSTr (Instruments) is:

Substitute $I=3$,$SSTr\left(\text{Instruments}\right)=90,990$ to get the  value of MSR( Instruments):

$\begin{array}{c}MSR\left(\text{Instruments}\right)=\frac{SSTr\left(\text{Instruments}\right)}{I-1}\\ =\frac{90,990}{3-1}\\ =\frac{90,990}{2}\\ =45,495\end{array}$

For Interaction Factor (AB):

The value of SSAB is:

Substitute ${\overline{X}}_{\mathrm{.1.}}=-9$, ${\overline{X}}_{\mathrm{.2.}}=-0.33$,${\overline{X}}_{\mathrm{.3.}}=-5.667$ and ${\overline{X}}_{\mathrm{.4.}}=2$ for $j=1,2,3,4$,${\overline{X}}_{\mathrm{1..}}=6.5$, ${\overline{X}}_{\mathrm{2..}}=-19$ and ${\overline{X}}_{\mathrm{3..}}=2.75$ for $i=1,2,3$,$I=3$,$K=60$,${\overline{X}}_{\mathrm{...}}=-3.25$ to get the  value of SSAB:

$\begin{array}{c}SSAB=\left\{K{\displaystyle \sum _{i=1}^3{\displaystyle \sum _{j=1}^4{\overline{X}}_{ij.}^2}}-JK{\displaystyle \sum _{i=1}^3{\overline{X}}_{i\mathrm{..}}^2}-IK{\displaystyle \sum _{j=1}^4{\overline{X}}_{.j.}^2}+IJK{\overline{X}}_{\mathrm{...}}^2\right\}\\ =\left\{\begin{array}{l}60\left\{\begin{array}{l}{\overline{X}}_{11.}^2+{\overline{X}}_{12.}^2+{\overline{X}}_{13.}^2+{\overline{X}}_{14.}^2+{\overline{X}}_{21.}^2+{\overline{X}}_{22.}^2+\\ {\overline{X}}_{23.}^2+{\overline{X}}_{24.}^2+{\overline{X}}_{31.}^2+{\overline{X}}_{32.}^2+{\overline{X}}_{33.}^2+{\overline{X}}_{34.}^2\end{array}\right\}-\\ \left(3\times 60\right)\left\{{\overline{X}}_{\mathrm{.1.}}^2+{\overline{X}}_{\mathrm{.2.}}^2+{\overline{X}}_{\mathrm{.3.}}^2+{\overline{X}}_{\mathrm{.4.}}^2\right\}-\\ \left(4\times 60\right)\left\{{\overline{X}}_{\mathrm{1..}}^2+{\overline{X}}_{\mathrm{2..}}^2+{\overline{X}}_{\mathrm{3..}}^2\right\}+\left(4\times 60\right){\overline{X}}_{\mathrm{...}}^2\end{array}\right\}\\ =\left\{\begin{array}{l}60\left\{\begin{array}{l}{3}^2+{14}^2+1^2+8^2+{\left(-24\right)}^2+{\left(-13\right)}^2+\\ {\left(-22\right)}^2+{\left(-17\right)}^2+{\left(-6\right)}^2+{\left(-2\right)}^2+4^2+{15}^2\end{array}\right\}-\\ \left(3\times 60\right)\left\{{\left(-9\right)}^2+{\left(-0.333\right)}^2+{\left(-5.667\right)}^2+2^2\right\}-\\ \left(4\times 60\right)\left\{{\left(6.5\right)}^2+{\left(-19\right)}^2+{\left(2.75\right)}^2\right\}+\left(3\times 4\times 60\right){\left(-3.25\right)}^2\end{array}\right\}\\ =\left\{\begin{array}{l}60\left\{2,069\right\}-180\left\{117.2258\right\}-\\ 240\left\{410.8125\right\}+\left(720\right)10.5625\end{array}\right\}\end{array}$

$\begin{array}{c}=\left\{\begin{array}{l}124,140-21,100.64-\\ 98,595+7,605\end{array}\right\}\\ \simeq 12,050\end{array}$

The value of MSAB is:

Substitute $I=3$,$J=4$,$SSAB=12,050$ to get the  value of MSAB:

$\begin{array}{c}MSAB=\frac{SSAB}{\left(I-1\right)\left(J-1\right)}\\ =\frac{12,050}{\left(3-1\right)\left(4-1\right)}\\ =\frac{12,050}{6}\\ =2,008.33\end{array}$

The value of MSE is:

Substitute $J=4$,$I=3$, corresponding values of $s_{ij}$ to get the  value of MSE:

$\begin{array}{c}MSE=\frac{{\displaystyle \sum _{i=1}^3{\displaystyle \sum _{j=1}^4{s}_{ij.}{}^2}}}{IJ}\\ =\frac{\left\{\begin{array}{l}{s}_{11.}^2+s_{12.}^2+s_{13.}^2+s_{14.}^2+s_{21.}^2+s_{22.}^2+\\ s_{23.}^2+s_{24.}^2+s_{31.}^2+s_{32.}^2+s_{33.}^2+s_{34.}^2\end{array}\right\}}{\left(4\times 3\right)}\\ =\frac{\left\{\begin{array}{l}{15}^2+{26}^2+{26}^2+{34}^2+{18}^2+{13}^2+\\ {39}^2+{26}^2+{18}^2+{16}^2+{29}^2+{18}^2\end{array}\right\}}{12}\\ =\frac{7,168}{12}\\ \approx 597.33\end{array}$

The value of SSE is:

$\begin{array}{c}SSE=MSE\times IJ\left(K-1\right)\\ =597.33\times \left(3\times 4\times \left(60-1\right)\right)\\ =597.33\times \left(12\times 59\right)\\ =597.33\times 708\\ =422,909.64\end{array}$

The value of SST is:

$\begin{array}{c}SST=SSB+SSTr+SSAB+SSE\\ =13,495+90,990+12,050+422,912\\ =539,447\end{array}$

The value of F statistic is:

For Base:

Substitute $MSB=4,498.356\text{and}MSE=597.33$,

$\begin{array}{c}F-\text{statistic}=\frac{MSB}{MSE}\\ =\frac{4,498.356}{597.33}\\ =7.5308\end{array}$

Thus, the value of F statistic for Base is 7.5308.

For Instruments:

Substitute $MSTr=45,495\text{and}MSE=597.33$,

$\begin{array}{c}F-\text{statistic}=\frac{MSTr}{MSE}\\ =\frac{45,495}{597.33}\\ =76.164\end{array}$

Thus, the value of F statistic for Instruments is 76.164.

For Interaction factor (AB):

Substitute $MSAB=2008.3\text{and}MSE=597.33$,

$\begin{array}{c}F-\text{statistic}=\frac{MSAB}{MSE}\\ =\frac{2,008.33}{597.33}\\ =3.3622\end{array}$

Thus, the value of F statistic for interaction is 3.3622.

The ranges of P-values are:

For Base $F_{3,708,0.05}$

Software procedure:

Step by step procedure to obtain the critical-value using the MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘F’ distribution.
• Enter Numerator Df as 3.
• Enter Denominator Df as 708.
• Click the Shaded Area tab.
• Choose X Value and Right Tail for the region of the curve to shade.
• Enter the Data value as 7.5308.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the value of $F_{3,708,0.05}$ is 0.000.

For Instruments $F_{2,708,0.05}$

Software procedure:

Step by step procedure to obtain the critical-value using the MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘F’ distribution.
• Enter Numerator Df as 2.
• Enter Denominator Df as 708.
• Click the Shaded Area tab.
• Choose X Value and Right Tail for the region of the curve to shade.
• Enter the Data value as 76.164.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the value of $F_{2,708,0.05}$ is 0.000.

For interaction $F_{6,708,0.05}$

Software procedure:

Step by step procedure to obtain the critical-value using the MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘F’ distribution.
• Enter Numerator Df as 6.
• Enter Denominator Df as 708.
• Click the Shaded Area tab.
• Choose X Value and Right Tail for the region of the curve to shade.
• Enter the Data value as 3.3622.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the value of $F_{6,708,0.05}$ is 0.003.

The ANOVA table is,

Source	DF	SS	MS	F	P
Base	3	13,495	4498.3	7.5308	0.000
Instrument	2	90,990	45,495	76.164	0.000
Interaction	6	12,050	2,008.3	3.3622	0.003
Error	708	422,912	597.33
Total	719	539,447

Conclusion:

Base (Block):

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(0.000\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

Thus, it can be concluded that there is a significant difference between block effects.

Instrument (Treatment):

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(0.000\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

Thus, it can be concluded that there is a significant difference between treatment effects.

Interaction:

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(0.003\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

Thus, it can be concluded that there is a significant effect of the interaction between the base (block) and instrument (treatment).

To determine

Decide whether it is appropriate to treat the randomized complete block design with base stations as blocks to determine the interest that the mean error differs among the instruments.

Answer to Problem 9SE

No, it is not appropriate to treat the data with a randomized complete block design with base stations as blocks.

Explanation of Solution

In randomized complete block design, there must be no the interaction between the treatment factor and the blocking factor, so that the treatment factor may be interpreted in RBD. However, here, the interaction effect is significant, suggesting that it is not possible to use randomized complete block design.

Thus, the suggestion given by the surveyor to treat the randomized complete block design with base station as blocks is not appropriate.