Data Analysis Assignment #2 (75 points total)

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.

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.

## RATIO - Skewness: 0.7147056 Kurtosis: -1.332702

(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.

## L_RATIO - Skewness: -0.09391548 Kurtosis: -2.464569

(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 = 0.5267

Essay 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: Based on the statistical analysis, L_RATIO demonstrates superior conformance to a normal distribution with homogeneous variances across age classes compared to RATIO. The log transformation successfully normalized the distribution, reducing skewness from 0.715 (moderately right-skewed) to -0.094 (nearly symmetric). Most importantly, Bartlett’s test confirms that L_RATIO exhibits homogeneous variances across all age classes (p = 0.527), whereas RATIO shows significant variance heterogeneity (p = 0.0003). The combination of improved distributional symmetry and verified variance homogeneity makes L_RATIO the appropriate variable for parametric statistical analyses requiring normality assumptions.

#### 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 ' ' 1

Essay 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: The non-significant interaction term (CLASS:SEX, p = 0.867) indicates that the relationship between L_RATIO and the factors CLASS and SEX is additive rather than interactive. This means the effect of age class on meat-to-volume ratio remains consistent across different sex categories, and the effect of sex remains constant across all age classes. The nearly identical F-values for the main effects in both models confirm that removing the non-significant interaction does not meaningfully change the interpretation of CLASS and SEX as independent predictors.

(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).

##              diff         lwr          upr        p adj
## A2-A1 -0.01248831 -0.03876038  0.013783756 6.919456e-01
## A3-A1 -0.03426008 -0.05933928 -0.009180867 1.863018e-03
## A4-A1 -0.05863763 -0.08594237 -0.031332896 5.917850e-08
## A5-A1 -0.09997200 -0.12764430 -0.072299703 0.000000e+00
## A3-A2 -0.02177176 -0.04106269 -0.002480831 1.784128e-02
## A4-A2 -0.04614932 -0.06825638 -0.024042262 1.520829e-07
## A5-A2 -0.08748369 -0.11004316 -0.064924223 0.000000e+00
## A4-A3 -0.02437756 -0.04505283 -0.003702280 1.146380e-02
## A5-A3 -0.06571193 -0.08687025 -0.044553605 0.000000e+00
## A5-A4 -0.04133437 -0.06508845 -0.017580286 2.233949e-05

##             diff          lwr           upr      p adj
## I-F -0.015890329 -0.031069561 -0.0007110968 0.03766729
## M-F  0.002069057 -0.012585555  0.0167236690 0.94126890
## M-I  0.017959386  0.003340824  0.0325779478 0.01118812

Additional 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 trend in age class coefficients shows a consistent decrease in L_RATIO values from younger to older classes, with all pairwise comparisons between successive classes being statistically significant (except A2-A1). This indicates that as abalones mature, their meat-to-volume ratio systematically declines, meaning older abalones contain relatively less meat per unit volume. Regarding combining males and females, the direct comparison shows no significant difference (M-F, p = 0.94), and the significant differences are actually between infants versus both males (p = 0.011) and infants versus females (p = 0.038). This pattern suggests that males and females are statistically similar to each other in their meat-to-volume ratios, while both differ from infants. Therefore, it would be appropriate to combine males and females into a single “adult” category for practical management purposes, as this simplification would not mask meaningful biological differences while making harvesting decisions more straightforward to implement.

#### 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 histograms reveal distinct volume distributions between infants and adults. Infant volumes are concentrated at lower values (mostly below 200), showing a right-skewed distribution, while adult volumes are more spread out with a broader range extending to much higher values. There is noticeable overlap in the middle range (approximately 200-600), which will create challenges in perfectly separating infants from adults based solely on VOLUME. However, the distributions are largely distinct, with smaller volumes reliably identifying infants and larger volumes reliably identifying adults.

(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: he log-transformation significantly reduces the variability in the scatterplots and creates a much more linear relationship between the variables. This linearization is crucial for regression analysis, as it ensures the data better meets the assumptions of constant variance and linearity required for valid statistical inference. The CLASS levels show a clear developmental progression from A1 (concentrated at bottom left) to A5 (concentrated at top right), illustrating the growth trajectory across age classes. For TYPE, the pattern is distinct: infants are heavily concentrated in the lower L_VOLUME range (below approximately 2.2), while adults dominate the upper range (above 1.8), with a zone of overlap between L_VOLUME 1.8 and 2.2 where both types are present. This overlap region represents the critical decision zone where volume alone cannot perfectly discriminate between infants and adults, highlighting the inherent challenge in setting a single harvest cutoff that completely separates the two populations.

#### 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-16

Essay Question: Interpret the trend in CLASS levelcoefficient 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 in CLASS level coefficient estimates shows a consistent decrease from A2 to A5, with values progressing from -0.018 to -0.117. This pattern indicates that older abalones have systematically lower shuck weight even after accounting for their larger volume. This means that as abalones grow through age classes, their bodies become proportionally less composed of meat (shuck) and more composed of shell and other tissues. This aligns with earlier scatterplots showing that while both volume and shuck weight increase with age, the relationship isn’t perfectly proportional - older, larger abalones actually have a lower meat-to-volume ratio than younger, smaller ones.

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 is statistically significant (p = 0.006) but has minimal practical importance for harvesting decisions. The coefficient of -0.021 means infants have only about 2% lower shuck weight than adults of the same volume and age class. In contrast, L_VOLUME’s coefficient of 0.999 indicates that volume is an almost perfect 1:1 predictor of shuck weight. For practical harvesting decisions, knowing an abalone’s volume is overwhelmingly more important than knowing whether it’s an infant or adult. While TYPE provides biologically meaningful information, it adds little predictive value beyond what volume and age class already tell us.

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.

## Residuals - Skewness: -0.05945234 Kurtosis: -2.656692

(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 = 0.4498

Essay 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 demonstrates excellent fit, with residuals showing near-normal distribution and constant variance across all predictors. The near-zero skewness (-0.059) and the non-significant Bartlett’s test (p = 0.450) confirm the model meets key assumptions. More importantly, the extremely strong relationship between L_VOLUME and L_SHUCK (coefficient = 0.999, p < 2e-16) with high R-squared (0.9504) indicates that volume is an exceptionally reliable predictor of meat yield. For harvesting decisions, this means VOLUME can be used with high confidence to predict the shuck weight - the valuable meat portion - making it an ideal metric for optimizing harvest efficiency while allowing managers to estimate yield accurately before harvesting.

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.

## Median infant volume: 133.8214

## [1] 0.4982699

## [1] 0.9330656

## Median adult volume: 384.5584

## [1] 0.02422145

## [1] 0.4993307

(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 median values reveal a critical insight: there is no single cutoff that achieves the ideal of harvesting most adults while protecting most infants. The adult median (384.56) is too conservative - it protects nearly all infants but sacrifices 98% of potential adult harvest. The infant median (133.82) is too aggressive - it captures 93% of adults but also harvests half the infant population. This forces managers to choose between conservation goals (minimizing infant harvest) and economic efficiency (maximizing adult harvest), with the optimal cutoff necessarily falling somewhere in between these extremes, accepting some infant bycatch to achieve reasonable harvest yields.

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.

(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.

(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] 0.7416332

## [1] 0.1764706

## [1] 0.1764706

There 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.

## Zero A1 Infant Cutoff: 206.786

## Infant harvest proportion: 0.2871972

## Adult harvest proportion: 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.

## Area Under ROC Curve (AUC): 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.

##                Cutoff   Volume       TPR         FPR Total_Harvest
## 1 Protect All Infants 526.6383 0.2476573 0.003460208     0.1785714
## 2       Median Infant 133.8214 0.9330656 0.498269896     0.8117761
## 3        Median Adult 384.5584 0.4993307 0.024221453     0.3667954
## 4      Max Difference 262.1430 0.7416332 0.176470588     0.5839768
## 5     Zero A1 Infants 206.7860 0.8259705 0.287197232     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: The five cutoffs represent a spectrum from most conservative to most aggressive harvesting strategies. “Protect All Infants” (526.64) is extremely conservative, sacrificing substantial adult harvest for complete infant protection. “Median Adult” (384.56) is very conservative, while “Max Difference” (262.14) balances adult harvest (74%) with infant protection (93%). “Zero A1 Infants” (206.79) increases harvest efficiency but sacrifices more infants, and “Median Infant” (133.82) is highly aggressive. The “Max Difference” cutoff appears optimal, maximizing the difference between adult and infant harvest rates while maintaining reasonable conservation goals.

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: In presenting this analysis to investigators, I would first outline the spectrum of harvesting choices and their inherent tradeoffs, demonstrating how each cutoff balances conservation goals against harvest efficiency. I would then make a specific recommendation for the “Max Difference” cutoff at 262.14 volume units, as it optimally balances adult harvest (74%) with infant protection (93%) while maximizing the discrimination between populations.I would qualify these findings by acknowledging several limitations: this analysis relies on a single observational dataset that may not represent all abalone populations, excludes potentially important environmental covariates like water temperature and food availability, and provides a static snapshot that cannot account for population dynamics over time. The model also assumes linear relationships that may not hold across all conditions. For implementation based on the current analysis, I would recommend adopting the 262.14 volume cutoff as an initial management rule while establishing a robust monitoring program to track population responses. This should include an adaptive management framework that allows for cutoff adjustments based on observed harvest outcomes and population trends, ensuring the rule remains effective under changing conditions. For future studies, I would recommend implementing longitudinal designs that track individual abalones over time, incorporating environmental and spatial variables across multiple geographic locations, validating harvesting rules through controlled experimental studies, and developing integrated population models that account for growth rates, mortality, and reproductive success to create more sustainable, dynamic management strategies.

Data Analysis Assignment #2 (75 points total)

Deidieva, Nargiza

Test Items starts from here - There are 10 sections - total of 75 points