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 here

R 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] 4.5
round(sum(sample.two*successes), digits = 1) # [1] 5.5
## [1] 5.5

Question: 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.

## Registered S3 methods overwritten by 'lme4':
##   method                          from
##   cooks.distance.influence.merMod car 
##   influence.merMod                car 
##   dfbeta.influence.merMod         car 
##   dfbetas.influence.merMod        car
## 
## Attaching package: 'rockchalk'
## The following objects are masked from 'package:moments':
## 
##     kurtosis, skewness

## [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)(c) Test the homogeneity of variance across classes using bartlett.test() (Kabacoff Section 9.2.2, p. 222).

## [[1]]
## [1] "ratio:"
## 
## [[2]]
## 
##  Bartlett test of homogeneity of variances
## 
## data:  RATIO by CLASS
## Bartlett's K-squared = 21.49, df = 4, p-value = 0.0002531
## 
## 
## [[3]]
## [1] "log ratio:"
## 
## [[4]]
## 
##  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: (We see that L_Ratio exhibits better conformance to a normal distribution with homogeneous variances across age classes. For Ratio, we see in the QQ-Plot that the distribution tends to deviate from normality and this is confirmed by the skewness and kurtosis. We also see that once we transform the variable to a log10 form, the kurtosis and skewness resemeble more to a normal distribution. This is also confirmed by the histogram and QQ-Plot. We also run the Bartlett test to test for the homogeneity of variance and fail to reject the null hypothesis. Therefore, we assume homogenity across classes.)

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 interaction term Class:Sex shows to be insignificant as shown by the p-value of 0.87. The individual variables Class and Sex show to be significant with L_Ratio as shown by their p-value.)

(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  0.0167236690 0.9412689
## M-I  0.017959386  0.003340824  0.0325779478 0.0111881

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 only coefficient that does not seem to be significant in the age classes is between A2 and A1. Besides those, all of them are significant. The p-value of Infant and Male or Female leads us to reject the null hypothesis that Infants and Male or Females are the same, indicating that the groups Male and Female can be combined into a single group called “Adults”)

Section 3: (10 points)

(3)(a1) We combine “M” and “F” into a new level, “ADULT”. (While this could be accomplished using combineLevels() from the ‘rockchalk’ package, we use base R code because many students do not have access to the rockchalk package.) This necessitated defining a new variable, TYPE, in mydata which had two levels: “I” and “ADULT”.

## 
## Check on definition of TYPE object (should be an integer):  integer
## 
## mydata$TYPE is treated as a factor:  TRUE
##    
##     ADULT   I
##   F   326   0
##   I     0 329
##   M   381   0

(3)(a2) Present side-by-side histograms of VOLUME. One should display infant volumes and, the other, adult volumes.

## The original levels F I M 
## have been replaced by I ADULT

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 distribution of Infants seems to be skewed to the right, while the distribution of adults seems to be normally distributed. The histograms show that the majority of the Infant’s volume seems to be from 0 to roughly 200, while the majority of the Adult’s volume apprears to be from 300 to 600. It does not seem that there will be any difficulty separting adults from infants.)

(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.

## 
## Attaching package: 'dplyr'
## The following object is masked from 'package:rockchalk':
## 
##     summarize
## The following object is masked from 'package:car':
## 
##     recode
## The following object is masked from 'package:gridExtra':
## 
##     combine
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

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: (There is a lot of overlap between Volume and Shuck initially. It is very difficult to see any difference between classes or types. However, once we transform the data, we are able to see some differences between classes and type. Volume in Adults generally lean to the upper right side.)

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. Reclassify the infants in classes A4 and A5 as ADULTS. This reclassification could have been achieved using combineLevels(), but only on the abalones in classes A4 and A5. We will do this recoding of the TYPE variable using base R functions. We will use this recoded TYPE variable, in which the infants in A4 and A5 are reclassified as ADULTS, for the remainder of this data analysis assignment.

## 
## Check on redefinition of TYPE object (should be an integer):  integer
## 
## mydata$TYPE is treated as a factor:  TRUE
## 
## Three-way contingency table for SEX, CLASS, and TYPE:
## , ,  = I
## 
##    
##      A1  A2  A3  A4  A5
##   F   0   0   0   0   0
##   I  91 133  65   0   0
##   M   0   0   0   0   0
## 
## , ,  = ADULT
## 
##    
##      A1  A2  A3  A4  A5
##   F   5  41 121  82  77
##   I   0   0   0  21  19
##   M  12  62 143  85  79
## The original levels I ADULT 
## have been replaced by ADULT

(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) -0.817512   0.019040 -42.936  < 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 ***
## TYPEADULT    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 coefficients seem to follow a decreasing pattern as the class increases. Paired with the plots seen earlier, this data suggests that L_Shuck increases more as you go lower on the classes. This analysis is consistant with an observation made on the Data Analysis #1 in question 2(b). “This may suggest that weight of the shell (in grams) is greater than the shuck for older classes. In other words, as an abalone gets older, shell growth is faster than shuck growth. Take A3, we can see A3 having a greater proportion of shuck to whole. And if we analyze from younger to older, we can see that older ones (A4, A5) start concentrating downwards as opposed to the younger ones which are closer to the trend line.”)

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: (Despite being statistically significant, Type does not have a strong coefficient in predicting L_Shuck. This means that Type may not help as much predicting the target variable.)

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.

(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 residuals seem to be evenly distributed and close to zero in CLASS and TYPE as we can see by the boxplots. There seems to be large groups distributed to the right formed in L_VOLUME in both TYPE and CLASS as we can see in the scatterplots.)

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 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.

## [1] 4.603851 5.595913 6.587974 7.580036 8.572097 9.564159
## [1] 0.01384083 0.01730104 0.02076125 0.02768166 0.03114187 0.03806228
## [1] 0.000000000 0.000000000 0.001338688 0.001338688 0.001338688 0.001338688

(6)(b) Present a plot showing the infant proportions and the adult proportions versus volume.value. Compute the 50% “split” volume.value for each and show on the plot.

Essay Question: The two 50% “split” values serve a descriptive purpose illustrating the difference between the populations. What do these values suggest regarding possible cutoffs for harvesting?

Answer: (The values suggest good cutoff points for the harvesting. We can see that the populations proportions are different and such cutoff points are good.)

This part will address the determination of a volume.value corresponding to the observed maximum difference in harvest percentages of adults and infants. 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.

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.

(7)(d) What separate harvest proportions for infants and adults would result if this cutoff is used? Show the separate harvest proportions (NOTE: the adult harvest proportion is the “true positive rate” and the infant harvest proportion is the “false positive rate”).

Code for calculating the adult harvest proportion is provided.

## [1] "True positive rate: 0.741633199464525"
## [1] "False positive rate: 0.176470588235294"

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.

## [1] 206.9844
## [1] "cutoff: 206.9843918625"
## [1] "True positive rate: 0.825970548862115"
## [1] "False positive rate: 0.28719723183391"

(8)(b) Another cutoff is one for which the proportion of adults not harvested equals the proportion of infants harvested. This cutoff would equate these rates; effectively, our two errors: ‘missed’ adults and wrongly-harvested infants. This leaves for discussion which is the greater loss: a larger proportion of adults not harvested or infants harvested? This cutoff is 237.7383. Calculate the separate harvest proportions for infants and adults using this cutoff. Show these proportions. Code for determining this cutoff is provided.

## [1] "cutoff: 237.73829751"
## [1] "True positive rate: 0.781793842034806"
## [1] "False positive rate: 0.217993079584775"
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 (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] "Area under ROC curve: 0.85633190200244"
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

## # A tibble: 3 x 5
##   strategy        volume   tpr   fpr prop_yield
##   <chr>            <dbl> <dbl> <dbl>      <dbl>
## 1 max difference    262. 0.742 0.176      0.584
## 2 zero A1 infants   207. 0.826 0.287      0.676
## 3 equal error       238. 0.782 0.218      0.625

Essay Question: Based on the ROC curve, it is evident a wide range of possible “cutoffs” exist. Compare and discuss the three cutoffs determined in this assignment.

Answer: (The max.difference is the best case as the false positive rate is the lowest. The case of zero A1 infants has the lowest cutoff, the highest true positive rate and the highest proportion, however it also has the highest false positive rate. The equal.error is between these other cutoffs.)

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:

  1. Would you make a specific recommendation or outline various choices and tradeoffs?
  2. What qualifications or limitations would you present regarding your analysis?
  3. If it is necessary to proceed based on the current analysis, what suggestions would you have for implementation of a cutoff? 4) What suggestions would you have for planning future abalone studies of this type?

Answer: (If I was presenting these results, 1. I would not recommend anything but i would highlight the choices presented in the table and emphasize on the false positive rate of the options. 2. I would present the distribution of the data, presence of outliers, the difficulty in correctly assessing the classes of abalones and highlight the limited scope of the analysis due to constraints presented by the quality of data collection methods. 3. If the analysis was to continue forward, i would recommend going with the most conservative approach. 4. I would suggest ways to collect better and more data - environmental, location, diet, etc. By collecting more and better data, the FPR could possible be reduced.)