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Flashcards do conjunto (26)
- 11. In plant breeding, especially in the later phases of a breeding program, trials with breeding material (the genotypes) are often done on multiple locations. Often per location multiple replicates of each genotype are trialled. Suppose we conduct a field experiment on three locations where every location has each of twelve genotypes in two replicates. The field is divided into two blocks, one block containing a complete replicate of all the genotypes, the other block containing the other replicate. The experimental design is a randomized complete block design. A single plot of a genotype within a block consists of 20 plantsf
- 2Why is replication of genotypes important ? Give four reasons for replication of the genotypes1. To obtain estimates of the variability between experimental units, required for statistical testing and to quantify data quality
2. To increase the statistical power of the experiment: an average over more replicates is more accurate and will allow discovery of smaller size effects with a higher degree of significance. In addition power will be increased by the larger number of degrees of freedom for the residual variance. - 3Why is replication of genotypes important ? Give four reasons for replication of the genotypes3. Increase the robustness of the statistical test, for example, by having a variability estimate closer to the true variability in the population; in addition, to have less problems with the assumption of normality for the averages
4. To allow generalization from the level of individuals to the population level. - 4Why is replication of genotypes important ? Give four reasons for replication of the genotypes5. To be protected against losing information when some plots get lost, for example to bad germination of seeds, or damage by rabbits or whatever.
6. To be protected against specific environmental conditions of a single plot: by having the same genotype on more than one plots, we are likely cancelling out some of the environmental errors associated with the specific field plots - 5b. What is the experimental unit in this experiment? Explain.the plot is the experimental unit. The genotypes were randomized over the plots within a block. Individual plants are NOT the experimental unit, because the 20 plants within a plot were not random plants over all genotypes but they were all of the same genotype (per plot).
- 6c. How many experimental units are there in this multi-location experiment in total ?Twenty-four per location (12 genotypes x 2 replicates), so 72 in total over three locations.
- 7d. Describe in detail how genotypes should be assigned to the experimental units.Per location per block the 12 genotypes should be assigned, at random, to the 12 plots per block. This random assignment could be done in different ways: for example, writing the names of the 12 genotypes on 12 equal pieces of paper, putting these in a closed box and drawing, for each plot in a block, one of the pieces of paper containing the genotype that will then be assigned to that plot. Redo this for all blocks in all locations. Another way would be to give numbers 1 to 12 to the cultivars and then randomize the numbers 1 to 12 six times in a software package (Excel, Genstat, R), so we get 6 different orders of the values between 1 and 12. We can use these orders for the 12 genotypes in each of our blocks.
- 8e. What are the advantages of a randomized complete block design over a completely randomized designIt is possible to correct for variability due to the blocks, so block differences do not decrease the power of establishing differences between the genotypes, even in the case that there are huge block effects. In this way we can correct for environmental differences in as far as these coincide with the blocks.
- 9f. What could be possible disadvantages?Some more complexity in generating the design; laying out the blocks in an efficient way. Keeping the plots within a block close together could be difficult in some fields. Another disadvantage could be that we lose degrees of freedom, so if there are no clear block effects we will have some reduced power. (However, the data could still be analyzed as a completely randomized design in that case).
- 102. The presence and the size of Genotype*Environment interactions are important aspects to take into account when developing cultivars of a certain crop. One way of characterizing genotype*environment interactions is by doing an Analysis of variance. Another way of characterizing genotype*environment interactions is by doing a Finlay-Wilkinson regression. Both methods use a statistical model that includes the phenotype (P), the genotypic effects (G) and the environmental effects (E).x
- 11a. Give the statistical model for the phenotype in Anova, using the appropriate terms, and using ‘i’ to index the genotypes, and ‘j’ to index the enviromentsPij = Gi + Ej + GEij + eij , or, equivalently, Pij = μ + Gi + Ej +GEij + eij with Gi the genotype main effects, Ej the environment main effects, GEij the term for genotype*environment interaction and eij, the random errors, which are independent and identically distributed (normal distribution with equal variance).
- 12b. Give the statistical model for the phenotype in a Finlay-Wilkinson regression, using the appropriate terms, and using ‘i’ to index the genotypes, and ‘j’ to index the enviroments.Pij = Gi + βiEj + eij also with eij, the errors, are independent and identically distributed (normal distribution with equal variance). Here Ej is the environmental index of environment j, calculated as the average value of all cultivars in that environment minus the overall mean of the cultivars over all environments.
- 13c. Which method would you use to characterize the ‘adaptability’ of a set of genotypes? Explain.The Finlay-Wilkinson regression. In FW regression the slope βi is a direct estimate of the adaptability. In Anova, there is no estimation of adaptability
- 14d. How would differences in adaptibility be characterized according to the method mentioned in c ? Explain.Cultivars with different adaptability will have different slopes in the FW regression: a cultivar with a low adaptability would have a low value for the regression coefficient for the slope, a cultivar with a high adaptability would have a high value for the regression coefficient for the slope.
- 15e. How would you select the genotype with the highest adaptibility?They would have the highest value for the slope in the FW regression
- 16What would be advantages of an approach where you analyze only one treatment factor at a time, for example first an Anova with cultivars as the only treatment factor, then an analysis with only gender as a treatment factor ? Assume that males and females have evaluated the exact same cultivars. Tick all the correct answers:1. In a single-treatment analysis you would not need to worry about the statistical balance of the treatments with respect to one another
2. The analysis would be more simple to carry out and the interpretation would be straightforward
The advantages are that the analysis is simple and the interpretation of the results is straightforward and there is no need to worry about the balance of treatment factors with respect to one another. - 17Suppose that gender of the taster (in a taste trial) has a large influence on the taste scores, but we do not include gender in the statistical model: we just include cultivars as the only treatment factor and we test for cultivar differences. [Suppose also that all women have tasted all cultivars, the men also have tasted all cultivars]. In which part(s) of the Anova table will the variation in the data, caused by a systematic gender difference, end up, if gender is not included in the model and analysis ? Tick the correct answer.1. In the mean squares of the error/residual
* the sum of squares and also the mean squares of the residual/error would now include also the variation due to differences between the sexes, and therefore be larger than they should be (hat variation will end up in the residual variance or error variance, estimated from the Mean Squares Residual in the Anova table. The residual variance estimate will then be very large (larger than it should be) - 18What will be the likely consequences for the F-test of the cultivar differences in the Anova ? Tick the correct answer.The F-value will be lower and therefore the significance of the differences among the cultivars becomes lower
(The F-value will be lower and therefore the significance of the differences among the cultivars becomes lower. A lower F-value corresponds to lower significance, given the same number of degrees of freedom.
By leaving out gender from the analysis, the degrees of freedom for the residual would be larger by 1, but the error sum of squares would be much increased. The F statistic is the ratio between the Mean Squares for cultivar differences (in the numerator) and the residual/error mean squares (in the denominator). Since the residual mean squares is larger, the F value will be lower than it should be, and hence the significance for cultivar differences will go down, so less statistical power to detect cultivar differences) - 19What will be the consequences for the multiple-comparisons test using the Least Significant Difference (LSD) to compare pairs of cultivars in a follow-up analysis ? Explain. Tick the correct answer.The LSD will be higher and therefore there will likely be fewer significant differences among the cultivars (the LSD will be higher and it is likely that fewer differences among the cultivars will be detected as significant. The differences between the cultivar means are unchanged)
- 20What would be the consequences of an approach where you analyze effects of cultivars and genders on overall liking in a combined analysis ? Tick the two most important correct answers, considering the above.1. With both treatments included, the mean square residual will be lower and therefore the statistical power to detect true differences will be higher (There will be a reduction of the mean squares residual (the variation due to gender differences will now be under mean squares ‘gender’ and not in the mean squares residual anymore)
2. A possible interaction between the treatments can be taken into account in a combined analysis (Interaction between the treatment factors can be taken into account (not possible if only one treatment factor is included) - 21What are the assumptions for an analysis of variance, as performed by you ? Tick all the correct answers.1. Independence of errors
2. Equal variance of the errors
3. A normal distribution of the errors - 22Considering these assumptions, what would be reasons to leave out Elstar ? Tick all the correct answers.1. Elstar was given a constant value (always 5). This violates the assumption of independence of the errors (the errors would be all 0 for elstar therefore not independent)
2. Elstar was given a constant value (always 5). This violates the assumption of equal variance among the errors (The errors would be 0 for Elstar and the variance among these errors would also be 0, whereas they would not be 0 for the other cultivars, so not equal)
3. Elstar was given a constant value (always 5). This violates the assumption of normality of the errors (A normal distribution should have variation around the expected mean, decreasing with distance from the expected mean but in this case the errors for Elstar would be 0, so no variation around the expected mean of 0) - 23Do you consider the values for Elstar as actual observations in the taste trial ? Tick the correct answer.No. Elstar was given a constant value, as a kind of reference value, it was not an observed value
- 24What would happen to the residual variance (as estimated from the Mean Squares Residual) and to the F-test if you would include the values for Elstar ? Tick the correct answerThe residual variance would be under-estimated (too small) because the residual variance is a pooled variance over all cultivars. If Elstar is included it becomes smaller than in the case that Elstar is excluded
Feedback:
The variance among the Elstar values is 0. The combined error variance is a pooled estimate over all the cultivars and will become smaller is a 0 is included than in the situation that the estimate is pooled only over positive non-zero values. - 25Even though Elstar was left out of the Anova, it is still possible, after the Anova, to test whether a cultivar is significantly different from Elstar. Which statistical test or procedure can be used to evaluate whether a cultivar is significantly different from Elstar ? Tick all the correct answers1. One sample t-test (using a pooled estimate of the residual variance, from Anova) to test whether the expected mean of the cultivar could be equal to the value 5 (A one-sample t-test is meant to compare a mean of a cultivar to a constant value (here: 5), so this is appropriate in this case. The estimate of residual variance is taken from the mean square residual in the Anova)
- 26Even though Elstar was left out of the Anova, it is still possible, after the Anova, to test whether a cultivar is significantly different from Elstar. Which statistical test or procedure can be used to evaluate whether a cultivar is significantly different from Elstar ? Tick all the correct answers2. Computation of a confidence interval for the mean of the cultivar that needs to be compared with Elstar (using a pooled estimate of the residual variance, from Anova) and check if it includes the value 5 (If the value 5 is included in the confidence interval, this means that the cultivar could have an expected mean of 5 and therefore no significant difference from Elstar which has the ‘expected’ value 5. The residual variance is estimated from the residual mean squares in Anova)
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- Breeding for quality FlashCards sobre sample exam statistics, criado por Robetauli Simangunsong em 10-08-2017.
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