Data Analysis Assignment #2 (75 points total)
Dukum, Dian
Instructions
R markdown is a plain-text file format for integrating text and R code, and creating transparent, reproducible and interactive reports. An R markdown file (.Rmd) contains metadata, markdown and R code “chunks”, and can be “knit” into numerous output types. Answer the test questions by adding R code to the fenced code areas below each item. There are questions that require a written answer that also need to be answered. Enter your comments in the space provided as shown below:
Answer: (Enter your answer here.)
Once completed, you will “knit” and submit the resulting .html document and the .Rmd file. The .html will present the output of your R code and your written answers, but your R code will not appear. Your R code will appear in the .Rmd file. The resulting .html document will be graded and a feedback report returned with comments. Points assigned to each item appear in the template.
Before proceeding, look to the top of the .Rmd for the (YAML) metadata block, where the title, author and output are given. Please change author to include your name, with the format ‘lastName, firstName.’
If you encounter issues with knitting the .html, please send an email via Canvas to your TA.
Each code chunk is delineated by six (6) backticks; three (3) at the start and three (3) at the end. After the opening ticks, arguments are passed to the code chunk and in curly brackets. Please do not add or remove backticks, or modify the arguments or values inside the curly brackets. An example code chunk is included here:
# Comments are included in each code chunk, simply as prompts
#...R code placed here
#...R code placed hereR code only needs to be added inside the code chunks for each assignment item. However, there are questions that follow many assignment items. Enter your answers in the space provided. An example showing how to use the template and respond to a question follows.
Example Problem with Solution:
Use rbinom() to generate two random samples of size 10,000 from the binomial distribution. For the first sample, use p = 0.45 and n = 10. For the second sample, use p = 0.55 and n = 10. Convert the sample frequencies to sample proportions and compute the mean number of successes for each sample. Present these statistics.
set.seed(123)
sample.one <- table(rbinom(10000, 10, 0.45)) / 10000
sample.two <- table(rbinom(10000, 10, 0.55)) / 10000
successes <- seq(0, 10)
round(sum(sample.one*successes), digits = 1) # [1] 4.5## [1] 4.5round(sum(sample.two*successes), digits = 1) # [1] 5.5## [1] 5.5Question: How do the simulated expectations compare to calculated binomial expectations?
Answer: The calculated binomial expectations are 10(0.45) = 4.5 and 10(0.55) = 5.5. After rounding the simulated results, the same values are obtained.
Submit both the .Rmd and .html files for grading. You may remove the instructions and example problem above, but do not remove the YAML metadata block or the first, “setup” code chunk. Address the steps that appear below and answer all the questions. Be sure to address each question with code and comments as needed. You may use either base R functions or ggplot2 for the visualizations.
##Data Analysis #2
## 'data.frame': 1036 obs. of 10 variables:
## $ SEX : Factor w/ 3 levels "F","I","M": 2 2 2 2 2 2 2 2 2 2 ...
## $ LENGTH: num 5.57 3.67 10.08 4.09 6.93 ...
## $ DIAM : num 4.09 2.62 7.35 3.15 4.83 ...
## $ HEIGHT: num 1.26 0.84 2.205 0.945 1.785 ...
## $ WHOLE : num 11.5 3.5 79.38 4.69 21.19 ...
## $ SHUCK : num 4.31 1.19 44 2.25 9.88 ...
## $ RINGS : int 6 4 6 3 6 6 5 6 5 6 ...
## $ CLASS : Factor w/ 5 levels "A1","A2","A3",..: 1 1 1 1 1 1 1 1 1 1 ...
## $ VOLUME: num 28.7 8.1 163.4 12.2 59.7 ...
## $ RATIO : num 0.15 0.147 0.269 0.185 0.165 ...Test Items starts from here - There are 10 sections - total of 75 points
#### Section 1: (5 points) ####
(1)(a) Form a histogram and QQ plot using RATIO. Calculate skewness and kurtosis using ‘rockchalk.’ Be aware that with ‘rockchalk’, the kurtosis value has 3.0 subtracted from it which differs from the ‘moments’ package.
## [1] 0.7147056## [1] 1.667298(1)(b) Tranform RATIO using log10() to create L_RATIO (Kabacoff Section 8.5.2, p. 199-200). Form a histogram and QQ plot using L_RATIO. Calculate the skewness and kurtosis. Create a boxplot of L_RATIO differentiated by CLASS.
## [1] -0.09391548## [1] 0.5354309
(1)(c) Test the homogeneity of variance across classes using bartlett.test() (Kabacoff Section 9.2.2, p. 222).
##
## Bartlett test of homogeneity of variances
##
## data: RATIO by CLASS
## Bartlett's K-squared = 21.49, df = 4, p-value = 0.0002531##
## Bartlett test of homogeneity of variances
##
## data: L_RATIO by CLASS
## Bartlett's K-squared = 3.1891, df = 4, p-value = 0.5267Essay Question: Based on steps 1.a, 1.b and 1.c, which variable RATIO or L_RATIO exhibits better conformance to a normal distribution with homogeneous variances across age classes? Why?
Answer: (It seems the L_RATIO’s histogram and qqplot have a more normal distribution with homogeneous variances across age classes. The distribution of the L_RATIO is less skewed in its histogram whereas, the RATIO histogram is more skewed to the right. The L_RATIO also has a lower Kurtosis. Lastly, while comparing L_RATIO’s chi-squared to the df, degrees of freedom; we can conclude that the chi-squared is less than the degrees of freedom which proves there is no rejection of the null hypothesis.)
#### Section 2 (10 points) ####
(2)(a) Perform an analysis of variance with aov() on L_RATIO using CLASS and SEX as the independent variables (Kabacoff chapter 9, p. 212-229). Assume equal variances. Perform two analyses. First, fit a model with the interaction term CLASS:SEX. Then, fit a model without CLASS:SEX. Use summary() to obtain the analysis of variance tables (Kabacoff chapter 9, p. 227).
## Df Sum Sq Mean Sq F value Pr(>F)
## CLASS 4 1.055 0.26384 38.370 < 2e-16 ***
## SEX 2 0.091 0.04569 6.644 0.00136 **
## CLASS:SEX 8 0.027 0.00334 0.485 0.86709
## Residuals 1021 7.021 0.00688
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## Df Sum Sq Mean Sq F value Pr(>F)
## CLASS 4 1.055 0.26384 38.524 < 2e-16 ***
## SEX 2 0.091 0.04569 6.671 0.00132 **
## Residuals 1029 7.047 0.00685
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Essay Question: Compare the two analyses. What does the non-significant interaction term suggest about the relationship between L_RATIO and the factors CLASS and SEX?
Answer: (A non-significant interaction refers to an interaction effect that does not have a significant effect on the response variable. For example, the CLASS:SEX has a low F value, which means such a combination is not a good predictor. Therefore, separating CLASS and SEX, is a better mode of predication since the F value is higher, this means there is a higher correlation and a better predictor.)
(2)(b) For the model without CLASS:SEX (i.e. an interaction term), obtain multiple comparisons with the TukeyHSD() function. Interpret the results at the 95% confidence level (TukeyHSD() will adjust for unequal sample sizes).
## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = L_RATIO ~ CLASS + SEX, data = mydata)
##
## $CLASS
## diff lwr upr p adj
## A2-A1 -0.01248831 -0.03876038 0.013783756 0.6919456
## A3-A1 -0.03426008 -0.05933928 -0.009180867 0.0018630
## A4-A1 -0.05863763 -0.08594237 -0.031332896 0.0000001
## A5-A1 -0.09997200 -0.12764430 -0.072299703 0.0000000
## A3-A2 -0.02177176 -0.04106269 -0.002480831 0.0178413
## A4-A2 -0.04614932 -0.06825638 -0.024042262 0.0000002
## A5-A2 -0.08748369 -0.11004316 -0.064924223 0.0000000
## A4-A3 -0.02437756 -0.04505283 -0.003702280 0.0114638
## A5-A3 -0.06571193 -0.08687025 -0.044553605 0.0000000
## A5-A4 -0.04133437 -0.06508845 -0.017580286 0.0000223
##
## $SEX
## diff lwr upr p adj
## I-F -0.015890329 -0.031069561 -0.0007110968 0.0376673
## M-F 0.002069057 -0.012585555 0.0167236691 0.9412689
## M-I 0.017959386 0.003340824 0.0325779478 0.0111881Additional Essay Question: first, interpret the trend in coefficients across age classes. What is this indicating about L_RATIO? Second, do these results suggest male and female abalones can be combined into a single category labeled as ‘adults?’ If not, why not?
Answer: (The Tukey test indicated that the classes A2-A1 has a higher than .5 p-value which results in a strong correlation when comparing the L_RATIO. The other classes indicate there are significant differences. The results of the L_RATIO suggest that there is a high correlation between Male and Female due to a p-value higher than .5 and can suggest the two sexes can be combined into one, which means there is a failure in rejecting the null hypothesis. The infant versus either Female or Male sex, indicates that the infants need to be in their own category since the p-values are lower than .5, which means we reject the null hypothesis and there are alternative hypothesis.)
#### Section 3: (10 points) ####
(3)(a1) Here, we will combine “M” and “F” into a new level, “ADULT”. The code for doing this is given to you. For (3)(a1), all you need to do is execute the code as given.
##
## ADULT I
## 707 329(3)(a2) Present side-by-side histograms of VOLUME. One should display infant volumes and, the other, adult volumes.
Essay Question: Compare the histograms. How do the distributions differ? Are there going to be any difficulties separating infants from adults based on VOLUME?
Answer: (The adult distribution is more normal than the infant distributions, which is skewed to the right with lower values. The highest of the adult distribution occurred at 300, but began to decline after 200 (A > 300; I < 300). This suggests that the volume is a means to help seperate the infants from the adult abalones.)
(3)(b) Create a scatterplot of SHUCK versus VOLUME and a scatterplot of their base ten logarithms, labeling the variables as L_SHUCK and L_VOLUME. Please be aware the variables, L_SHUCK and L_VOLUME, present the data as orders of magnitude (i.e. VOLUME = 100 = 10^2 becomes L_VOLUME = 2). Use color to differentiate CLASS in the plots. Repeat using color to differentiate by TYPE.
Additional Essay Question: Compare the two scatterplots. What effect(s) does log-transformation appear to have on the variability present in the plot? What are the implications for linear regression analysis? Where do the various CLASS levels appear in the plots? Where do the levels of TYPE appear in the plots?
Answer: (For the scatterplots on the left, as the weight of the abalones increases so do the variablities. The L_SHUCK scatterplots shows there are less variablities as the weight of the abalones increase. If the linear regression was used to analyze all scatterplots, the linear regression will provide better results for the L_SHUCKS scatterplots. The various class levels and types on the SHUCK scatterplots are densely grouped to the bottom left of the graph, and the L_SHUCK scatterplots showcases the various class levels and types to the top right of the graph. )
#### Section 4: (5 points) ####
(4)(a1) Since abalone growth slows after class A3, infants in classes A4 and A5 are considered mature and candidates for harvest. You are given code in (4)(a1) to reclassify the infants in classes A4 and A5 as ADULTS.
##
## ADULT I
## 747 289(4)(a2) Regress L_SHUCK as the dependent variable on L_VOLUME, CLASS and TYPE (Kabacoff Section 8.2.4, p. 178-186, the Data Analysis Video #2 and Black Section 14.2). Use the multiple regression model: L_SHUCK ~ L_VOLUME + CLASS + TYPE. Apply summary() to the model object to produce results.
##
## Call:
## lm(formula = L_SHUCK ~ L_VOLUME + CLASS + TYPE, data = mydata)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.270634 -0.054287 0.000159 0.055986 0.309718
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.796418 0.021718 -36.672 < 2e-16 ***
## L_VOLUME 0.999303 0.010262 97.377 < 2e-16 ***
## CLASSA2 -0.018005 0.011005 -1.636 0.102124
## CLASSA3 -0.047310 0.012474 -3.793 0.000158 ***
## CLASSA4 -0.075782 0.014056 -5.391 8.67e-08 ***
## CLASSA5 -0.117119 0.014131 -8.288 3.56e-16 ***
## TYPEI -0.021093 0.007688 -2.744 0.006180 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.08297 on 1029 degrees of freedom
## Multiple R-squared: 0.9504, Adjusted R-squared: 0.9501
## F-statistic: 3287 on 6 and 1029 DF, p-value: < 2.2e-16Essay Question: Interpret the trend in CLASS level coefficient estimates? (Hint: this question is not asking if the estimates are statistically significant. It is asking for an interpretation of the pattern in these coefficients, and how this pattern relates to the earlier displays).
Answer: (The trend noticed in the CLASS level coefficient estimates is, as the CLASS levels increased from A2 - A5; the estimates would decrease, which concludes to have an inverse correlation. The type I indicates there is a positive correlation, as the abalone is younger (Type I), the larger the volume of Shuck. These results are consistent with Data Analysis 1.)
Additional Essay Question: Is TYPE an important predictor in this regression? (Hint: This question is not asking if TYPE is statistically significant, but rather how it compares to the other independent variables in terms of its contribution to predictions of L_SHUCK for harvesting decisions.) Explain your conclusion.
Answer: (Type will not have a strong effect on the predictions of the L_SHUCK. This is proven by p value of 2.2e-16 which is less than .5.)
The next two analysis steps involve an analysis of the residuals resulting from the regression model in (4)(a) (Kabacoff Section 8.2.4, p. 178-186, the Data Analysis Video #2).
#### Section 5: (5 points) ####
(5)(a) If “model” is the regression object, use model$residuals and construct a histogram and QQ plot. Compute the skewness and kurtosis. Be aware that with ‘rockchalk,’ the kurtosis value has 3.0 subtracted from it which differs from the ‘moments’ package.
## [1] -0.05945234## [1] 0.3433082(5)(b) Plot the residuals versus L_VOLUME, coloring the data points by CLASS and, a second time, coloring the data points by TYPE. Keep in mind the y-axis and x-axis may be disproportionate which will amplify the variability in the residuals. Present boxplots of the residuals differentiated by CLASS and TYPE (These four plots can be conveniently presented on one page using par(mfrow..) or grid.arrange(). Test the homogeneity of variance of the residuals across classes using bartlett.test() (Kabacoff Section 9.3.2, p. 222).
##
## Bartlett test of homogeneity of variances
##
## data: RESIDUALS by CLASS
## Bartlett's K-squared = 3.6882, df = 4, p-value = 0.4498Essay Question: What is revealed by the displays and calculations in (5)(a) and (5)(b)? Does the model ‘fit’? Does this analysis indicate that L_VOLUME, and ultimately VOLUME, might be useful for harvesting decisions? Discuss.
Answer: (The model does ‘fit’. Yes, it seems that the L_VOLUME and VOLUME seems to be very useful in harvesting decisions. The value averages below .5, roughly around 0, which indicates a high correlation between VOLUME and harvesting decisions.)
Harvest Strategy:
There is a tradeoff faced in managing abalone harvest. The infant population must be protected since it represents future harvests. On the other hand, the harvest should be designed to be efficient with a yield to justify the effort. This assignment will use VOLUME to form binary decision rules to guide harvesting. If VOLUME is below a “cutoff” (i.e. a specified volume), that individual will not be harvested. If above, it will be harvested. Different rules are possible.The Management needs to make a decision to implement 1 rule that meets the business goal.
The next steps in the assignment will require consideration of the proportions of infants and adults harvested at different cutoffs. For this, similar “for-loops” will be used to compute the harvest proportions. These loops must use the same values for the constants min.v and delta and use the same statement “for(k in 1:10000).” Otherwise, the resulting infant and adult proportions cannot be directly compared and plotted as requested. Note the example code supplied below.
#### Section 6: (5 points) ####
(6)(a) A series of volumes covering the range from minimum to maximum abalone volume will be used in a “for loop” to determine how the harvest proportions change as the “cutoff” changes. Code for doing this is provided.
(6)(b) Our first “rule” will be protection of all infants. We want to find a volume cutoff that protects all infants, but gives us the largest possible harvest of adults. We can achieve this by using the volume of the largest infant as our cutoff. You are given code below to identify the largest infant VOLUME and to return the proportion of adults harvested by using this cutoff. You will need to modify this latter code to return the proportion of infants harvested using this cutoff. Remember that we will harvest any individual with VOLUME greater than our cutoff.
## [1] 526.6383## [1] 0.2476573## [1] 0(6)(c) Our next approaches will look at what happens when we use the median infant and adult harvest VOLUMEs. Using the median VOLUMEs as our cutoffs will give us (roughly) 50% harvests. We need to identify the median volumes and calculate the resulting infant and adult harvest proportions for both.
## [1] 0.4982699## [1] 0.9330656## [1] 0.02422145## [1] 0(6)(d) Next, we will create a plot showing the infant conserved proportions (i.e. “not harvested,” the prop.infants vector) and the adult conserved proportions (i.e. prop.adults) as functions of volume.value. We will add vertical A-B lines and text annotations for the three (3) “rules” considered, thus far: “protect all infants,” “median infant” and “median adult.” Your plot will have two (2) curves - one (1) representing infant and one (1) representing adult proportions as functions of volume.value - and three (3) A-B lines representing the cutoffs determined in (6)(b) and (6)(c).
Essay Question: The two 50% “median” values serve a descriptive purpose illustrating the difference between the populations. What do these values suggest regarding possible cutoffs for harvesting?
Answer: (The adult median is much higher than the infant median, which indicates utilizing the median cutoff at the adult median (384.56 cm^3) is the best cutoff point as to not over harvest the abalones. If the infant median (133.82 cm^3) cutoff was used to dictate a harvesting point, then that will cause an over harvesting of both the infant and adult abalones before they could continue to reproduce.)
More harvest strategies:
This part will address the determination of a cutoff volume.value corresponding to the observed maximum difference in harvest percentages of adults and infants. In other words, we want to find the volume value such that the vertical distance between the infant curve and the adult curve is maximum. To calculate this result, the vectors of proportions from item (6) must be used. These proportions must be converted from “not harvested” to “harvested” proportions by using (1 - prop.infants) for infants, and (1 - prop.adults) for adults. The reason the proportion for infants drops sooner than adults is that infants are maturing and becoming adults with larger volumes.
Note on ROC:
There are multiple packages that have been developed to create ROC curves. However, these packages - and the functions they define - expect to see predicted and observed classification vectors. Then, from those predictions, those functions calculate the true positive rates (TPR) and false positive rates (FPR) and other classification performance metrics. Worthwhile and you will certainly encounter them if you work in R on classification problems. However, in this case, we already have vectors with the TPRs and FPRs. Our adult harvest proportion vector, (1 - prop.adults), is our TPR. This is the proportion, at each possible ‘rule,’ at each hypothetical harvest threshold (i.e. element of volume.value), of individuals we will correctly identify as adults and harvest. Our FPR is the infant harvest proportion vector, (1 - prop.infants). We can think of TPR as the Confidence level (ie 1 - Probability of Type I error and FPR as the Probability of Type II error. At each possible harvest threshold, what is the proportion of infants we will mistakenly harvest? Our ROC curve, then, is created by plotting (1 - prop.adults) as a function of (1 - prop.infants). In short, how much more ‘right’ we can be (moving upward on the y-axis), if we’re willing to be increasingly wrong; i.e. harvest some proportion of infants (moving right on the x-axis)?
#### Section 7: (10 points) ####
(7)(a) Evaluate a plot of the difference ((1 - prop.adults) - (1 - prop.infants)) versus volume.value. Compare to the 50% “split” points determined in (6)(a). There is considerable variability present in the peak area of this plot. The observed “peak” difference may not be the best representation of the data. One solution is to smooth the data to determine a more representative estimate of the maximum difference.
## Warning in geom_vline(xintercept = med_adl_vol, linetpe = 2): Ignoring unknown
## parameters: `linetpe`## Warning in annotate("text", label = paste("Median Adult = ", round(med_adl_vol,
## : Ignoring unknown parameters: `angel`
(7)(b) Since curve smoothing is not studied in this course, code is supplied below. Execute the following code to create a smoothed curve to append to the plot in (a). The procedure is to individually smooth (1-prop.adults) and (1-prop.infants) before determining an estimate of the maximum difference.
(7)(c) Present a plot of the difference ((1 - prop.adults) - (1 - prop.infants)) versus volume.value with the variable smooth.difference superimposed. Determine the volume.value corresponding to the maximum smoothed difference (Hint: use which.max()). Show the estimated peak location corresponding to the cutoff determined.
Include, side-by-side, the plot from (6)(d) but with a fourth vertical A-B line added. That line should intercept the x-axis at the “max difference” volume determined from the smoothed curve here.
## [1] 262.143
(7)(d) What separate harvest proportions for infants and adults would result if this cutoff is used? Show the separate harvest proportions. We will actually calculate these proportions in two ways: first, by ‘indexing’ and returning the appropriate element of the (1 - prop.adults) and (1 - prop.infants) vectors, and second, by simply counting the number of adults and infants with VOLUME greater than the vlume threshold of interest.
Code for calculating the adult harvest proportion using both approaches is provided.
## [1] 0.7416332## [1] 0.06827309## [1] 262.143There are alternative ways to determine cutoffs. Two such cutoffs are described below.
#### Section 8: (10 points) ####
(8)(a) Harvesting of infants in CLASS “A1” must be minimized. The smallest volume.value cutoff that produces a zero harvest of infants from CLASS “A1” may be used as a baseline for comparison with larger cutoffs. Any smaller cutoff would result in harvesting infants from CLASS “A1.”
Compute this cutoff, and the proportions of infants and adults with VOLUME exceeding this cutoff. Code for determining this cutoff is provided. Show these proportions. You may use either the ‘indexing’ or ‘count’ approach, or both.
## [1] 206.786## [1] 0.2871972## [1] 0.8259705(8)(b) Next, append one (1) more vertical A-B line to our (6)(d) graph. This time, showing the “zero A1 infants” cutoff from (8)(a). This graph should now have five (5) A-B lines: “protect all infants,” “median infant,” “median adult,” “max difference” and “zero A1 infants.”
#### Section 9: (5 points) ####
(9)(a) Construct an ROC curve by plotting (1 - prop.adults) versus (1 - prop.infants). Each point which appears corresponds to a particular volume.value. Show the location of the cutoffs determined in (6), (7) and (8) on this plot and label each.
(9)(b) Numerically integrate the area under the ROC curve and report your result. This is most easily done with the auc() function from the “flux” package. Areas-under-curve, or AUCs, greater than 0.8 are taken to indicate good discrimination potential.
## [1] 0.8666894#### Section 10: (10 points) ####
(10)(a) Prepare a table showing each cutoff along with the following: 1) true positive rate (1-prop.adults, 2) false positive rate (1-prop.infants), 3) harvest proportion of the total population
To calculate the total harvest proportions, you can use the ‘count’ approach, but ignoring TYPE; simply count the number of individuals (i.e. rows) with VOLUME greater than a given threshold and divide by the total number of individuals in our dataset.
| Volume | True Positive | False Positive | totalHarvest | |
|---|---|---|---|---|
| Protect All Infants | 526.6383 | 0.2476573 | 0.0000000 | 0.1785714 |
| Median Infants | 133.8214 | 0.9330656 | 0.4982699 | 0.8117761 |
| Median Adults | 384.5584 | 0.0000000 | 0.0242215 | 0.0067568 |
| Max Difference | 262.1430 | 0.7416332 | 0.0682731 | 0.5537943 |
| Zero A1 Infants | 206.7860 | 0.8259705 | 0.2871972 | 0.6756757 |
Essay Question: Based on the ROC curve, it is evident a wide range of possible “cutoffs” exist. Compare and discuss the five cutoffs determined in this assignment.
Answer: (There is a range of cutoff points from the most conservative to the most aggressive. The most conservative form of harvesting is the value based on Protecting All Infants (526.64 cm^3), followed by the Median Adults (384.56 cm^3), Max Difference (262.14 cm^3), Zero A1 Infants (206.79 cm^3), and finally the most aggressive, Median Infants (133.82 cm^3). The cutoff points are merely suggest options for harvesting the abalones. The Max Difference (262.14 cm^3) suggests this could be the best means of harvesting adult and infant abalones without over harvesting infant abalones.)
Final Essay Question: Assume you are expected to make a presentation of your analysis to the investigators How would you do so? Consider the following in your answer:
- Would you make a specific recommendation or outline various choices and tradeoffs?
- What qualifications or limitations would you present regarding your analysis?
- If it is necessary to proceed based on the current analysis, what
suggestions would you have for implementation of a cutoff?
- What suggestions would you have for planning future abalone studies of this type?
Answer: (1. I would outline various choices and tradeoffs of harvesting abalones. After providing my various outlined choices , I would then suggest a specific recommendation of using the Max Difference in order to provide a best practice in harvesting abalones. 2. Based off of the Volume findings, we can better identify the abalones class and better identify which abalones to harvest for studies. The limitations of the abalone study could be limited on abalones habitat that could lead to non-random sampling. It is also notable that the infant abalones have a greater amount of variables versus the adult abalones. 3. Some implementation of cutoffs are as follows, minimizing the harvest of infant abalones, which means avoiding A1 class of infant abalones, to allow reproduction and maintaining the habitat. Also, harvesting more of the adult abalones could be utilized in the analysis. As mentioned before, I would siggest utilizing the Max Difference as the cutoff point in harvesting abalones for the study. 4. My suggestion for planning future abalone studies would be to use random sampling from different abalone habitats, environmental effects on the abalones, each abalone habitat must have a break down of abalone classes, type, volume, sex, etc. which should be compared against each of the abalone habitats.)