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.
- a. Construct an ANOVA table. You may give
ranges for the P-values. - 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.
a.
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
The aim is to find the ANOVA.
State the hypothesis:
Main effect of Instruments:
Null hypothesis:
Alternative hypothesis:
Main effect of Base:
Null hypothesis:
Alternative hypothesis:
Interaction:
Null hypothesis:
Alternative hypothesis:
ANOVA table:
Source | DF | SS | MS | F |
Treatment | ||||
Blocks | ||||
Interaction | ||||
Error | ||||
Total |
Where,
Where, N is the sample size, I denotes the number of treatments,
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
Degrees of freedom:
Base degrees of freedom:
Instrument degrees of freedom:
Interaction degrees of freedom:
Error degrees of freedom:
Total degrees of freedom:
The treatment and block means are tabulated below:
Base | Instruments | Block mean | ||
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 | 6.5 | –19 | 2.75 |
By observing the data, the values of
For Base:
The value of SSB (Base) is:
Substitute
The value of MSB (Base) is:
Substitute
For Instruments:
The value of SSTr ( Instruments) is:
Substitute
The value of MSTr (Instruments) is:
Substitute
For Interaction Factor (AB):
The value of SSAB is:
Substitute
The value of MSAB is:
Substitute
The value of MSE is:
Substitute
The value of SSE is:
The value of SST is:
The value of F statistic is:
For Base:
Substitute
Thus, the value of F statistic for Base is 7.5308.
For Instruments:
Substitute
Thus, the value of F statistic for Instruments is 76.164.
For Interaction factor (AB):
Substitute
Thus, the value of F statistic for interaction is 3.3622.
The ranges of P-values are:
For Base
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
For Instruments
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
For interaction
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
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,
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,
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,
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).
b.
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.
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