Chapter 15 Exercises
15.1 Chapter 3
15.1.1 Exercise 1
A chemical analysis was performed in triplicate, with the following results: 125, 169 and 142 ng/g. Calculate mean, sum of squares, mean square, standard deviation and coefficient of variation.
15.1.2 Exercise 2
Download the Excel file ‘rimsulfuron.csv’ from https://www.casaonofri.it/_datasets/rimsulfuron.csv. This is a dataset relating to a field experiment to compare 15 herbicides with 4 replicates. The response variables are maize yield and height. Describe the dataset and show the results on a barplot, including some measure of variability. Check whether yield correlates to height and comment on the results.
15.1.3 Exercise 3
Load the csv file ‘students.csv’ from https://www.casaonofri.it/_datasets/students.csv. This dataset relates to a number of students, their votes in several undergraduate exams and information on high school. Determine whether (i) the average votes depend on the exam subject and (ii) the average votes depend on high school type.
15.2 Chapter 4
15.2.1 Exercise 1
A xenobiotic substance degrades in soil following a first-order kinetic, which is described by the following equation:
\[Y = 100 \, e^{-0.07 \, t}\]
where Y is the concentration at time \(t\). After spraying this substance in soil, what is the probability that 50 days later we observe a concentration below the toxicity threshold for mammalians (2 ng/g)? Please, consider that all the unknown sources of experimental error can be regarded as gaussian, with a coefficient of variability equal to 20%.
15.2.2 Exercise 2
Crop yield is a function of its density, according to the following function:
\[ Y = 8 + 8 \, X - 0.07 \, X^2\]
Draw the graph and find the required density to obtain the highest yield (use a simple graphical method). What is the probability of obtaining a yield level between 2.5 and 3 t/ha, by using the optimal density? Consider that random variability is 12%.
15.2.3 Exercise 3
The toxicity of a compound changes with the dose, according to the following expression:
\[ Y = \frac{1}{1 + exp\left\{ -2 \, \left[log(X) - log(15)\right] \right\}}\]
where \(Y\) is the proportion of dead animals and \(X\) is the dose. If we treat 150 animals with a dose of 35 g, what is the probability of finding more than 120 dead animals? The individual variability can be approximated by using a gaussian distribution, with a standard error equal to 10.
15.2.4 Exercise 4
Consider the sample C = [140 - 170 - 155], which was drawn by a gaussian distribution. Calculate the probability of drawing an individual value from the same pupulation in the following intervals:
- higher than 170
- lower than 140
- within the range from 170 and 140
15.2.5 Exercise 5
Consider a gaussian population with \(\mu\) = 23 and \(\sigma\) = 1. Calculate the probability of drawing three individuals with a mean
- higher than 25
- lower than 21
- between 21 and 25
15.3 Chapter 5
15.3.1 Exercise 1
A chemical analysis was repeated three times, with the following results: 125, 169 and 142 ng/g. Calculate mean, deviance, variance, standard deviation, standard error and confidence intervals (P = 0.95 and P = 0.99).
15.3.2 Exercise 2
An experiment was carried out, comparing the yield of four wheat genotypes (in tons per hectar). The results are as follows:
| Genotype | Rep-1 | Rep-2 | Rep-3 | Rep4 |
|---|---|---|---|---|
| A | 4.72 | 5.45 | 5.13 | 5.19 |
| B | 6.29 | 6.79 | 7.55 | 5.86 |
| C | 5.54 | 4.44 | 5.16 | 5.92 |
| D | 6.68 | 6.30 | 6.70 | 7.77 |
For each genotype, calculate the mean, deviance, variance, standard deviation, standard error and confidence interval (P = 0.95).
15.3.3 Exercise 3
We have measured the length of 30 maize seedlings, treated with selenium in water solution. The observed lengths are:
length <- c(2.07, 2.23, 2.04, 2.16, 2.12, 2.33, 2.21, 2.22, 2.29, 2.28,
2.44, 2.04, 2.02, 1.49, 2.12, 2.38, 2.51, 2.27, 2.55, 2.44, 2.28,
2.2, 2.03, 2.35, 2.34, 2.34, 1.99, 2.44, 2.44, 1.91)For the above sample, calculate the mean, deviance, variance, standard deviation, standard error and confidence interval (P = 0.95).
15.3.4 Exercise 4
A sample of 400 insects was sprayed with an insecticide and 136 individuals survived the treatment. Determine the efficacy of the insecticide, in terms of proportion of dead insects, together with 95% confidence limits.
15.4 Chapter 6
15.4.1 Exercise 1
We have compared two herbicides for weed control in maize. With the first herbicide (A), we observed the following weed coverings: 9.3, 10.2, 9.7 %. With the second herbicide, we observedd: 12.6, 12.3 e 12.5 %. Are the means for the two herbicides significantly different (P < 0.05)?
15.4.2 Exercise 2
We have made an experiment to compare two fungicides A and B. The first fungicide was used to treat 200 fungi colonies and the number of surviving colonies was 180. B was used to treat 100 colonies and 50 of those survived. Is there a significant difference between the efficiacies of A and B (P < 0.05)?
15.4.3 Exercise 3
A plant pathologist studied the crop performances with (A) and without (NT) a fungicide treatment. The results are as follows:
| A | NT |
|---|---|
| 65 | 54 |
| 71 | 51 |
| 68 | 59 |
Was the treatment effect significant (P < 0.05)?
15.4.4 Exercise 4
In this year, an assay showed that 600 olive drupes out of 750 were attacked by Daucus olee. In a close field, under the same environmental conditions, the count of attacked drupes was 120 on 750. Is the the observed difference statistically significant (P < 0.05) or is it just due to random fluctuation?
15.4.5 Exercise 5
In a hospital, blood cholesterol level was measured for eight patients, before and after a three months terapy. The observed values were:
| Patient | Before | After |
|---|---|---|
| 1 | 167.3 | 126.7 |
| 2 | 186.7 | 154.2 |
| 3 | 105.0 | 107.9 |
| 4 | 214.5 | 209.3 |
| 5 | 148.5 | 138.5 |
| 6 | 171.5 | 121.3 |
| 7 | 161.5 | 112.4 |
| 8 | 243.6 | 190.5 |
Can we say that this terapy is effective, or not?
15.4.6 Exercise 6
A plant breeder organised an experiment to compare three wheat genotypes, i.e. GUERCINO, ARNOVA and BOLOGNA, according to a completely randomised design with 10 replicates. The observed yields are:
| guercino | arnova | bologna |
|---|---|---|
| 53.2 | 53.1 | 43.5 |
| 59.1 | 51.0 | 41.0 |
| 62.3 | 51.9 | 41.2 |
| 48.6 | 55.3 | 44.8 |
| 59.7 | 58.8 | 40.2 |
| 60.0 | 54.6 | 37.2 |
| 55.7 | 53.0 | 45.3 |
| 55.8 | 51.4 | 38.9 |
| 55.7 | 51.7 | 42.9 |
| 54.4 | 64.7 | 39.3 |
- Describe the three samples, by using the appropriate statistics of central tendency and spread
- Infere the means of the pupulations from where the samples were drawn
- For each of the three possible couples (GUERCINO vs ARNOVA, GUERCINO vs BOLOGNA and ARNOVA vs BOLOGNA), test the hypothesis that the two means are significantly different.
15.4.7 Exercise 7
A botanist counted the number of germinated seeds for oilseed rape at two different temperatures (15 and 25°C). At 15°C, 358 germinations were counted out of 400 seeds. At 25°C, 286 germinations were counted out of 380 seeds.
- Describe the proportions of germination for the three samples
- Infere the proportion of germinated seeds in the two populations, from where the samples of seeds were extracted (remember that the variance for a proportion is calculated as \(p \times (1- p)\).
- Test the hypothesis that temperature had a significant effect on the germinability of oilseed rape seeds.
15.5 Chapters 7 to 9
15.5.1 Exercise 1
An experiment was conducted with a completely randomised design to compare the yield of 5 wheat genotypes. The results (in bushels per acre) are as follows:
| Variety | 1 | 2 | 3 |
|---|---|---|---|
| A | 32.4 | 34.3 | 37.3 |
| B | 20.2 | 27.5 | 25.9 |
| C | 29.2 | 27.8 | 30.2 |
| D | 12.8 | 12.3 | 14.8 |
| E | 21.7 | 24.5 | 23.4 |
- Write the linear model for this study and explain the model components
- Compute the ANOVA
- Check for the basic assumptions
- Compare the means
- Present the results and comment on them
The example is taken from: Le Clerg et al. (1962)
15.5.2 Exercise 2
Cell cultures of tomato were grown by using three types of media, based on glucose, fructose and sucrose. The experiment was conducted with a completely randomised design with 5 replicates and a control was also added to the design. Cell growths are reported in the table below:
| Control | Glucose | Fructose | Sucrose |
|---|---|---|---|
| 45 | 25 | 28 | 31 |
| 39 | 28 | 31 | 37 |
| 40 | 30 | 24 | 35 |
| 45 | 29 | 28 | 33 |
| 42 | 33 | 27 | 34 |
- Write the linear model for this study and explain the model components
- Compute the ANOVA
- Check for the basic assumptions
- Compare the means
- Present the results and comment on them
15.5.3 Exercise 3
The failure time for a heating system was assessed, to discover the effect of the operating temperature. Four temperatures were tested with 6 replicates, according to a completely randomised design and the number of hours before failure were measured. The results are as follows:
| Temp. | Hours to failure |
|---|---|
| 1520 | 1953 |
| 1520 | 2135 |
| 1520 | 2471 |
| 1520 | 4727 |
| 1520 | 6134 |
| 1520 | 6314 |
| 1620 | 1190 |
| 1620 | 1286 |
| 1620 | 1550 |
| 1620 | 2125 |
| 1620 | 2557 |
| 1620 | 2845 |
| 1660 | 651 |
| 1660 | 837 |
| 1660 | 848 |
| 1660 | 1038 |
| 1660 | 1361 |
| 1660 | 1543 |
| 1708 | 511 |
| 1708 | 651 |
| 1708 | 651 |
| 1708 | 652 |
| 1708 | 688 |
| 1708 | 729 |
Please, note that these are the only possible values for temperature. Determine the best operating temperature, in order to delay failure.
15.5.4 Exercise 4
An entomologist counted the number of eggs laid from a lepidopter on three tobacco genotypes. 15 females were tested for each genotype and the results are as follows:
| Female | Field | Resistant | USDA |
|---|---|---|---|
| 1 | 211 | 0 | 448 |
| 2 | 276 | 9 | 906 |
| 3 | 415 | 143 | 28 |
| 4 | 787 | 1 | 277 |
| 5 | 18 | 26 | 634 |
| 6 | 118 | 127 | 48 |
| 7 | 1 | 161 | 369 |
| 8 | 151 | 294 | 137 |
| 9 | 0 | 0 | 29 |
| 10 | 253 | 348 | 522 |
| 11 | 61 | 0 | 319 |
| 12 | 0 | 14 | 242 |
| 13 | 275 | 21 | 261 |
| 14 | 0 | 0 | 566 |
| 15 | 153 | 218 | 734 |
Which is the most resistant genotype?
15.6 Chapter 10
15.6.1 Exercise 1
Data were collected about 5 types of irrigation on orange trees in Spain. The experiment was laid down as complete randomised blocks with 5 replicates and the results are as follows:
| Method | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Localised | 438 | 413 | 375 | 127 | 320 |
| Surface | 413 | 398 | 348 | 112 | 297 |
| Sprinkler | 346 | 334 | 281 | 43 | 231 |
| Sprinkler + localised | 335 | 321 | 267 | 33 | 219 |
| Submersion | 403 | 380 | 336 | 101 | 293 |
- Write the linear model for this study and explain the model components
- Compute the ANOVA
- Check for the basic assumptions
- Compare the means
- Present the results and comment on them
15.6.2 Exercise 2
A fertilisation trial was conducted according to a RCBD with five replicates. One value is missing for the second treatment in the fifth block. The observed data are percentage contents in P2 O5 in leaf samples:
| Treatment | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Unfertilised | 5.6 | 6.1 | 5.3 | 5.9 | 9.4 |
| 50 lb N | 7.3 | 6.0 | 7.7 | 7.7 | NA |
| 100 lb N | 6.9 | 6.0 | 5.6 | 7.4 | 8.2 |
| 50 lb N + 75 lb P2O5 | 10.8 | 11.2 | 8.8 | 10.4 | 12.9 |
| 100 lb N + 75 lb P205 | 9.6 | 9.3 | 12 | 10.6 | 11.6 |
- Calculate arithmetic means
- Calculate the ANOVA
- Check for the basic assumptions
- Calculate expected marginal means and compare to arithmetic means
- The addition of P2 O5 is a convenient practice, in terms of agronomic effect?
15.6.3 Exercise 3
A latin square experiment was planned to assess effect of four different fertilisers on lettuce yield. The observed data are as follows:
| Fertiliser | Row | Column | Yield |
|---|---|---|---|
| A | 1 | 1 | 104 |
| B | 1 | 2 | 114 |
| C | 1 | 3 | 90 |
| D | 1 | 4 | 140 |
| A | 2 | 4 | 134 |
| B | 2 | 3 | 130 |
| C | 2 | 1 | 144 |
| D | 2 | 2 | 174 |
| A | 3 | 3 | 146 |
| B | 3 | 4 | 142 |
| C | 3 | 2 | 152 |
| D | 3 | 1 | 156 |
| A | 4 | 2 | 147 |
| B | 4 | 1 | 160 |
| C | 4 | 4 | 160 |
| D | 4 | 3 | 163 |
- Write the linear model for this study and explain the model components
- Compute the ANOVA
- Check for the basic assumptions
- Compare the means
- Present the results and comment on them: what is the best fertiliser?
15.7 Chapters 11 and 12
15.7.1 Exercise 1
A pot experiment was planned to evaluate the best timing for herbicide application against rhizome Sorghum halepense. Five timings were compared (2-3, 4-5, 6-7 and 8-9 leaves), including a splitted treatment in two timings (3-4/8-9 leaves) and the untreated control. In order to understand whether the application is effective against plants coming from rhizomes of different sizes, a second factor was included in the experiment, i.e. rhizome size (2, 4, six nodes). The design was a fully crossed two-way factorial, laid down as completely randomised with four replicates. The results (plant weights three weeks after the herbicide application) are as follows:
| Sizes ↓ / Timing → | 2-3 | 4-5 | 6-7 | 8-9 | 3-4/8-9 | Untreated |
|---|---|---|---|---|---|---|
| 2-nodes | 34.03 | 0.10 | 30.91 | 33.21 | 2.89 | 41.63 |
| 22.31 | 6.08 | 35.34 | 43.44 | 19.06 | 22.96 | |
| 21.70 | 3.73 | 24.23 | 44.06 | 0.10 | 52.14 | |
| 14.90 | 9.15 | 28.27 | 35.34 | 0.68 | 59.81 | |
| 4-nodes | 42.19 | 14.86 | 52.34 | 39.06 | 8.62 | 68.15 |
| 51.06 | 36.03 | 43.17 | 61.59 | 0.05 | 42.75 | |
| 43.77 | 21.85 | 57.28 | 48.89 | 0.10 | 57.77 | |
| 31.74 | 8.71 | 29.71 | 49.14 | 9.65 | 44.85 | |
| 6-nodes | 20.84 | 11.37 | 55.00 | 41.77 | 9.80 | 43.20 |
| 26.12 | 2.24 | 28.46 | 37.38 | 0.10 | 40.68 | |
| 35.24 | 14.17 | 21.81 | 39.55 | 1.42 | 34.11 | |
| 13.32 | 23.93 | 60.72 | 48.37 | 6.83 | 32.21 |
- Write the linear model for this study and explain the model components
- Compute the ANOVA
- Check for the basic assumptions
- Calculate marginal means and cell means
- Present the results and comment on them: what type of means should you report?
15.7.2 Exercise 2
Six faba bean genotypes were tested in two sowing times, according to a plit-plot design in 4 complete blocks. Sowing times were randomised to main-plots within blocks and genotypes were randomised to sub-plots within main-plots and blocks. Results are:
| Sowing Time | Genotype | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Autum | Chiaro | 4.36 | 4.00 | 4.23 | 3.83 |
| Collameno | 3.01 | 3.32 | 3.27 | 3.40 | |
| Palombino | 3.85 | 3.85 | 3.68 | 3.98 | |
| Scuro | 4.97 | 3.98 | 4.39 | 4.14 | |
| Sicania | 4.38 | 4.01 | 3.94 | 2.99 | |
| Vesuvio | 3.94 | 4.47 | 3.93 | 4.21 | |
| Spring | Chiaro | 2.76 | 2.64 | 2.25 | 2.38 |
| Collameno | 2.50 | 1.79 | 1.57 | 1.77 | |
| Palombino | 2.24 | 2.21 | 2.50 | 2.05 | |
| Scuro | 3.45 | 2.94 | 3.12 | 2.69 | |
| Sicania | 3.24 | 3.60 | 3.16 | 3.08 | |
| Vesuvio | 2.34 | 2.44 | 1.71 | 2.00 |
- Write the linear model for this study and explain the model components
- Compute the ANOVA
- Check for the basic assumptions
- Calculate marginal means and cell means
- Present the results and comment on them: what type of means should you report?
15.7.3 Exercise 3
Four crops were sown in soil 20 days after the application of three herbicide treatments, in order to evaluate possible carry-over effects of residuals. The untreated control was also added for comparison and the weight of plants was assessed four weeks after sowing. The experiment was laid down as strip-plot and, within each block, the herbicide were randomised to rows and crops to columns. The weight of plants is reported below:
| Herbidicide | Block | sorghum | rape | soyabean | sunflower |
|---|---|---|---|---|---|
| Untreated | 1 | 180 | 157 | 199 | 201 |
| 2 | 236 | 111 | 257 | 358 | |
| 3 | 287 | 217 | 346 | 435 | |
| 4 | 350 | 170 | 211 | 327 | |
| Imazethapyr | 1 | 47 | 10 | 193 | 51 |
| 2 | 43 | 1 | 113 | 4 | |
| 3 | 0 | 20 | 187 | 13 | |
| 4 | 3 | 21 | 122 | 15 | |
| primisulfuron | 1 | 271 | 8 | 335 | 379 |
| 2 | 182 | 0 | 201 | 201 | |
| 3 | 283 | 22 | 206 | 307 | |
| 4 | 147 | 24 | 240 | 337 | |
| rimsulfuron | 1 | 403 | 238 | 226 | 290 |
| 2 | 227 | 169 | 195 | 494 | |
| 3 | 400 | 364 | 257 | 397 | |
| 4 | 171 | 134 | 137 | 180 |
- Write the linear model for this study and explain the model components
- Compute the ANOVA
- Check for the basic assumptions
- Calculate marginal means and cell means
- Present the results and comment on them: what type of means should you report?
15.7.4 Exercise 4
A field experiment was conducted to evaluate the effect of fertilisation timing (early, medium, late) on two genotypes. The experiment was designed as a randomised complete block design and the data represent the amount of absorbed nitrogen by the plant:
| Genotype | Block | Early | Med | Late |
|---|---|---|---|---|
| A | 1 | 21.4 | 50.8 | 53.2 |
| 2 | 11.3 | 42.7 | 44.8 | |
| 3 | 34.9 | 61.8 | 57.8 | |
| B | 1 | 54.8 | 56.9 | 57.7 |
| 2 | 47.9 | 46.8 | 54.0 | |
| 3 | 40.1 | 57.9 | 62.0 |
- Write the linear model for this study and explain the model components
- Compute the ANOVA
- Check for the basic assumptions
- Calculate marginal means and cell means
- Present the results and comment on them: what type of means should you report?
15.7.5 Exercise 5
A study was carried out to evaluate the effect of washing temperature on the reduction of length for four types of fabric. Results are expressed as percentage reduction and the experiment was completely randomised, with two replicates:
| Fabric | 210 °F | 215 °F | 220 °F | 225 °F |
|---|---|---|---|---|
| A | 1.8 | 2.0 | 4.6 | 7.5 |
| 2.1 | 2.1 | 5.0 | 7.9 | |
| B | 2.2 | 4.2 | 5.4 | 9.8 |
| 2.4 | 4.0 | 5.6 | 9.2 | |
| C | 2.8 | 4.4 | 8.7 | 13.2 |
| 3.2 | 4.8 | 8.4 | 13.0 | |
| D | 3.2 | 3.3 | 5.7 | 10.9 |
| 3.6 | 3.5 | 5.8 | 11.1 |
Consider the temperature as a factor and:
- Write the linear model for this study and explain the model components
- Compute the ANOVA
- Check for the basic assumptions
- Calculate marginal means and cell means
- Present the results and comment on them: what type of means should you report?
15.7.6 Exercise 6
A chemical process requires one alcohol and one base. A study is organised to evaluate the factorial combinations of three alcohols and two bases on the efficiency of the process, expressed as a percentage. The experiment is designd as completely randomised.
| Base | Alcohol 1 | Alcohol 2 | Alcohol 3 |
|---|---|---|---|
| A | 91.3 | 89.9 | 89.3 |
| 88.1 | 89.5 | 87.6 | |
| 90.7 | 91.4 | 90.4 | |
| 91.4 | 88.3 | 90.3 | |
| B | 87.3 | 89.4 | 92.3 |
| 91.5 | 93.1 | 90.7 | |
| 91.5 | 88.3 | 90.6 | |
| 94.7 | 91.5 | 89.8 |
- Write the linear model for this study and explain the model components
- Compute the ANOVA
- Check for the basic assumptions
- Calculate marginal means and cell means
- Present the results and comment on them: what type of means should you report?
15.8 Chapter 13
15.8.1 Exercise 1
A study was conducted to evaluate the effect of nitrogen fertilisation in lettuce. The experiment is completely randomised with 4 replicates and the yield results are as follows:
| N level | B1 | B2 | B3 | B4 |
|---|---|---|---|---|
| 0 | 124 | 114 | 109 | 124 |
| 50 | 134 | 120 | 114 | 134 |
| 100 | 146 | 132 | 122 | 146 |
| 150 | 157 | 150 | 140 | 163 |
| 200 | 163 | 156 | 156 | 171 |
- Write the linear model for this study and explain the model components
- Estimate model parameters
- Check for the basic assumptions
- What yield might be obtained by using 120 kg N ha-1?
- Present the results and comment on them?
15.8.2 Exercise 2
A study was conducted to evaluate the effect of increasing densities of a weed (Sinapis arvensis) on sunflower yield. The experiment was completely randomised. Assuming that the yield response is linear, parameterise the model, check the goodness of fit and find the economical threshold level of weed density, considering that the yield worths 150 Euros per ton and the herbicide treatment costs 40 Euros per hectar. The observed results are:
| density | Rep y | ield |
|---|---|---|
| 0 | 1 | 36.63 |
| 14 | 1 | 29.73 |
| 19 | 1 | 32.12 |
| 28 | 1 | 30.61 |
| 32 | 1 | 27.7 |
| 38 | 1 | 27.43 |
| 54 | 1 | 24.79 |
| 0 | 2 | 36.11 |
| 14 | 2 | 34.72 |
| 19 | 2 | 30.12 |
| 28 | 2 | 30.8 |
| 32 | 2 | 26.53 |
| 38 | 2 | 27.6 |
| 54 | 2 | 23.31 |
| 0 | 3 | 38.35 |
| 14 | 3 | 32.16 |
| 19 | 3 | 31.72 |
| 28 | 3 | 28.69 |
| 32 | 3 | 25.88 |
| 38 | 3 | 28.43 |
| 54 | 3 | 30.26 |
| 0 | 4 | 36.74 |
| 14 | 4 | 32.566 |
| 19 | 4 | 29.57 |
| 28 | 4 | 33.663 |
| 32 | 4 | 28.751 |
| 38 | 4 | 27.114 |
| 54 | 4 | 24.664 |
15.9 Chapter 14
15.9.1 Exercise 1
Two soil samples were treated with two herbicides and put in a climatic chamber at 20°C. Sub-samples were collected from both samples in different times and the concentration of herbicide residues was measured. The results are as follows:
| Time | Herbicide A | Herbicide B |
|---|---|---|
| 0 | 100.00 | 100.00 |
| 10 | 50.00 | 60.00 |
| 20 | 25.00 | 40.00 |
| 30 | 15.00 | 23.00 |
| 40 | 7.00 | 19.00 |
| 50 | 3.50 | 11.00 |
| 60 | 2.00 | 5.10 |
| 70 | 1.00 | 3.00 |
Assuming that the degradation follows an exponential decay trend, determine the half-life for both herbicides.
15.9.2 Exercise 2
A microbial population grows exponentially over time. Considering the following data, determine the relative rate of growth, by fitting the exponential growth model.
| Time | Cells |
|---|---|
| 0 | 2 |
| 10 | 3 |
| 20 | 5 |
| 30 | 9 |
| 40 | 17 |
| 50 | 39 |
| 60 | 94 |
| 70 | 201 |
15.9.3 Exercise 3
An experiment was conducted to determine the absorption of nitrogen by roots of Lemna minor in hydroponic colture. Results (N content) are the following:
| Conc | Rate |
|---|---|
| 2.86 | 14.58 |
| 5.00 | 24.74 |
| 7.52 | 31.34 |
| 22.10 | 72.97 |
| 27.77 | 77.50 |
| 39.20 | 96.09 |
| 45.48 | 96.97 |
| 203.78 | 108.88 |
Use nonlinear least squares to estimate the parameters for the rectangular hyperbola (Michaelis-Menten model):
\[Y = \frac{a X} {b + X}\]
and make sure that model fit is good enough.
15.9.4 Exercise 4
An experiment was conducted to determine the yield of sunflower at increasing densities of a weed (Ammi majus). Based on the following results, parameterise a rectangular hyperbola (\(Y = (a \, X)/(b + X)\) and test for possible lack of fit. The results are:
| Weed density | Yield Loss (%) |
|---|---|
| 0 | 0 |
| 23 | 17.9 |
| 31 | 21.6 |
| 39 | 26.9 |
| 61 | 29.5 |
15.9.5 Exercise 5
An experiment was conducted in a pasture, to determine the effect of sampling area on the number of plant species (in general, the higher the sampling area and the higher the number of sampled species). The results are as follows:.
| Area | N. of species |
|---|---|
| 1 | 4 |
| 2 | 5 |
| 4 | 7 |
| 8 | 8 |
| 16 | 10 |
| 32 | 14 |
| 64 | 19 |
| 128 | 22 |
| 256 | 22 |
By using the above data, parameterise a power curve \(Y = a \, X^b\) and test for lack of fit.
15.9.6 Exercise 6
Crop growth can be often described by using a Gompertz model. The data below refer to an experiment were sugarbeet was grown either weed free, or weed infested; the weight of the crop per unit area was measured after six different numbers of Days After Emergence (DAE). The experiment was conducted by using a completely randomised design with three replicates and the results are reported below:
| DAE | Infested | Weed Free |
|---|---|---|
| 21 | 0.06 | 0.07 |
| 21 | 0.06 | 0.07 |
| 21 | 0.11 | 0.07 |
| 27 | 0.20 | 0.34 |
| 27 | 0.20 | 0.40 |
| 27 | 0.21 | 0.25 |
| 38 | 2.13 | 2.32 |
| 38 | 3.03 | 1.72 |
| 38 | 1.27 | 1.22 |
| 49 | 6.13 | 11.78 |
| 49 | 5.76 | 13.62 |
| 49 | 7.78 | 12.15 |
| 65 | 17.05 | 33.11 |
| 65 | 22.48 | 24.96 |
| 65 | 12.66 | 34.66 |
| 186 | 21.51 | 38.83 |
| 186 | 26.26 | 27.84 |
| 186 | 27.68 | 37.72 |
Parameterise two Gompertz growth models (one for the weed-free crop and one for the infested crop) and evalaute which of the parameters are most influenced by the competition. The Gompertz growth model is:
\[Y = d \cdot exp\left\{- exp \left[ - b (X - e)\right] \right\}\]
15.9.7 Exercise 7
Plants of Tripleuspermum inodorum in pots were treated with a sulphonylurea herbicide (tribenuron-methyl) at increasing rates. Three weeks after the treatment the weight per pot was recorded, with the following results:
| Dose (g a.i. ha\(^{-1}\)) | Fresh weight (g pot \(^{-1}\)) |
|---|---|
| 0 | 115.83 |
| 0 | 102.90 |
| 0 | 114.35 |
| 0.25 | 91.60 |
| 0.25 | 103.23 |
| 0.25 | 133.97 |
| 0.5 | 98.66 |
| 0.5 | 92.51 |
| 0.5 | 124.19 |
| 1 | 93.92 |
| 1 | 49.21 |
| 1 | 49.24 |
| 2 | 21.85 |
| 2 | 23.77 |
| 2 | 22.46 |
Assuming that the dose-response relationship can be described by using the following log-logistic model:
\[Y = c + \frac{d - c}{1 + exp \left\{ - b \left[ log (X) - log (e) \right] \right\}}\]
Parameterise the model and evaluate the goodnes of fit.