Data Analysis #2 Version 2 (75 points total)

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: +0.7147
## kurtosis: +1.6673
## kurtosis: +4.6673 (3.0 added back)

(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: -0.0939
## kurtosis: +0.5354
## kurtosis: +3.5354 (3.0 added back)

(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:  L_RATIO by mydata$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: L_RATIO exhibits better conformance to a normal distribution with homogeneous variances across age classes. As L_RATIO makes data smoother, which we can tell from the histgram, qq-plot and the calculations, it reduced the skewness to close to 0 as well as reducing kurtosis at the same time.

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.

##               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

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  38.524 < 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 main values of CLASS and SEX are significant in predicting L_RATIO;the interaction term CLASS:SEX is insignificant as we can tell from the 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: 1. We fail to reject the null hypothesis that A1 and A2 are from the same population, i.e. there’s no significant difference between class A1 and A2. 2. For all other class comparison, if we use a P value of 0.05, we can reject NULL hypothesis for all of them. 3. A2 vs A3, A3 ~ A4 have similar P value of .01~.02, it’s statistically significant that they are different with p value .05. 4 all other classes differences are significant even we use p value of .0001. 5. In terms of sex, For M and F we can reject the null hypothesis that M and F has no significant difference. Therefore Males and Females can be combined into an Adult group.

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.

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 appears to be close to normally distributed while the Infant distribution skews to the right. The majority of the adult volume appears to be between 300-500 while the majority of the infant volume is less than 300. This indicates that there’s difficulty separating infants from adults just based on Volume.

(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: (log-transformation greatly reduced the variability presented in the plot. In terms of linear regression, I think adjusted R2 will increase with log-transformation. )

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:

## , ,  = ADULT
## 
##    
##      A1  A2  A3  A4  A5
##   F   5  41 121  82  77
##   I   0   0   0  21  19
##   M  12  62 143  85  79
## 
## , ,  = I
## 
##    
##      A1  A2  A3  A4  A5
##   F   0   0   0   0   0
##   I  91 133  65   0   0
##   M   0   0   0   0   0

(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 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: (We can tell class A2~A5 has negative coefficient, that means when they grow older, their CHUCK (L_CHUCK) grows slower while VOLUME(L_VOLUME) grows, which looks consistent with earlier displays(The display itself is not very obvious);Besides, Class A5 has a bigger weight (.11) compare to other class especially A2, which has a smaller weight (.018), while A3 and A4 falls in between, so class A5 will have a bigger impact compare to other class especially A2; all these been said, L_VOLUME has a much bigger weight overall all compare to other variables and so L_VOLUME has the biggest prediction power)

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 not very important, as the wieght/coefficient is only .021, so it’s contribution to predictions of L_SHUCK for harvesting decision is not that big(although slightly bigger than classes))

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.

## Ratio:
## 
## skewness: -0.0595
## kurtosis: +0.3433
## kurtosis: +3.3433 (3.0 added back)

(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:  model$residuals and mydata$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 residual seem to be roughly normally distributed and averages at 0, so VOLUME can be useful for harvesting decisions)

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] 133.8199

## [1] 384.5138

(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: (If we harvest all volumes greater than 133.82, we’ll harvest 50% infant and majority of adults, if we harvest volumes greater than 384.51, will not harvest most of the infants, but we’ll only get half the adults)

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.

## When max difference cutoff(262.143) is used:
## 
## Adult  harvest proportions(true positive rate): 0.7416
## infant harvest proportion(false positive rate): 0.1765

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.

## When max difference cutoff(206.786) is used:
## 
## Adult  harvest proportions(true positive rate): 0.8260
## infant harvest proportion(false positive rate): 0.2872

(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.

## When max difference cutoff(237.639) is used:
## 
## Adult  harvest proportions(true positive rate): 0.7818
## infant harvest proportion(false positive rate): 0.2180

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.

## AUC(Areas-under-curve) for (1 - prop.infants) vs (1 - prop.adults) is: 0.867

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

##                  Volume   TPR   FPR PropYield
## max.difference  262.143 0.742 0.176     0.584
## zero.A1.infants 206.786 0.826 0.287     0.676
## equal.error     237.639 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 cutoff indicates having the lowest proportional yield. The zero.A1.infants cutoff has the highest true positive rate and proportional yield, with the equal.error in between these two.)

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? 4) What suggestions would you have for planning future abalone studies of this type?

Answer: (1. I’d recommend chosen max.difference option, although it earns us the least adults;it also captures least infant, so that we can have a sustainable way to harvest abalones. 2.I’d prefer lowest FPR for sustainability 3.I chose to implemente the max.difference cutoff. 4. I’d suggest measure more variables to better differentiate between adult and infant so taht FPR can be further reduced)