Homework Assignment 2: Due Tuesday April 1st Homework Assignment 2 INSTRUCTIONS: Data sets for each study are in the accompanying Excel data file (Homework 2 data 2022.xls); one worksheet per data...

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Homework Assignment 2: Due Tuesday April 1st Homework Assignment 2 INSTRUCTIONS: Data sets for each study are in the accompanying Excel data file (Homework 2 data 2022.xls); one worksheet per data set. You should use R to complete this assignment, include pertinent sections from the computer printouts in your answer, and highlight the relevant values on the printouts. Show all hand calculations. · This assignment is due on Thursday, March 31th. · Please submit your homework through Canvas. · Please follow the standard homework answer format (see document on Canvas for details). · UNLESS THE QUESTION INSTRUCTS OTHERWISE, you must check parametric test assumptions and, if violated, determine the best approach for dealing with those violations. Question 1: Animals often possess phenomenal navigational capabilities - even without smartphones! The cues they use to navigate vary, among them are: magnetic fields, sun compasses, celestial maps, sound, and olfactory cues. A researcher was interested in whether fiddler crabs that foraged along the water's edge at low tide used a sun compass, magnetism, or both to help them relocate their burrows high on the beach. So she conducted an experiment in which she collected 48 crabs at random from an area and randomly assigned 12 crabs to each of four conditions: (a) eye-cups were placed over their eyes so that they were blindfolded to disrupt orientation using the sun, (b) a small magnet was glued to their back to disrupt magnetic navigation, (c) both blindfolded and outfitted with a magnet, and (d) a control where the crabs were handled but no eye-cups or magnets applied to them. Each crab was then released near the water's edge 10m from their burrow and the time it took each of them to find their burrow (in seconds) was recorded. Does the navigational capability of crabs differ among the experimental treatments? CHECK ASSUMPTIONS. Question 2: The pathogenic fungus Cryphonectria parasitica was introduced into North American forests in the early 1900’s from Asia and over the next few decades killed over 4 billion Chestnut trees (Castanea dentata), dramatically changing the ecology of American forests. One study investigated whether point mutations in an autoimmune gene within Chestnut trees might influence their susceptibility to infection by the fungus, as compared to a non-mutated gene. To do so, researchers took tissue samples from 40 trees total that varied in 3 point mutations plus non-mutated trees (10 S-type, C-type, 10 E-type, 10 non-mutated control) and measured the number of fugus infected tree cells per mm of tissue sample. Do any of the gene mutation types (S, C, or E) confer more resistance to the fungus compared to the non-mutant control? CHECK ASSUMPTIONS, Question 3: A study was conducted to determine if temperature tolerance in an endangered species of coral varied among individual corals sampled from randomly selected locations on reefs in the Florida Keys. If so, one might suspect that this physiological response varied due to genetic isolation of subpopulations exposed to different natural temperature regimes. Therefore, researchers randomly selected six locations (labeled A - F) along the Florida Keys reef tract from which they collected seven live coral samples per location. The coral samples were brought to a marine laboratory where each coral was subjected to a high seawater temperature (30oC) and their physiological response (measured as % of maximum photosynthetic output) determined using pulsed amplitude flourometry. Does % photosynthetic efficiency in corals vary more among locations or among individual corals within a location? What percentage of the variation in photosynthetic efficiency is attributable to differences among locations versus individual corals? Question 4: An experiment was conducted to determine if the growth of three different species of tree snails common in Florida (manatee treesnail, Drymaeus dormani; West Indian Bulimulus, Bulimulus guadalupensis; lined forest snail, Drymaeus multilineatus) changes when exposed to different concentrations (0 = control and 10, 15, 20, 25, and 30ppm) of an herbicide suspected of being a growth inhibitor. Individuals of each snail species were randomly assigned to aquaria containing natural tree bark and then each aquarium was randomly assigned a concentration of the herbicide to be sprayed on the tree bark in each aquarium. After 60 days, the mean change in snail growth (expressed as % change) in each aquarium was determined. Does exposure to any of the concentrations of the herbicide effect the growth of any of the snail species? Question 5: It is hypothesized that the length of the tibia (leg bone that affects jumping ability) differs among four species of mice (Florida mouse, cotton mouse, oldfield mouse, eastern harvest mouse) that dwell in Florida, but may also be influenced by their location habitat. So researchers collected 10 mice of each species from each of five different randomly selected locations in Florida. The average length of the tibia of the 10 mice from each location was then calculated yielding a single mean tibia length for each species per location. Does tibia length differ among mice species? Does location influence that result? PAGE 1 Question 1 TrtTime Control3.2 Control4.4 Control4.7 Control2.9 Control5 Control4.2 Control5.7 Control5.7 Control6.2 Control3.9 Control3.8 Control3 BLINDFOLD4.1 BLINDFOLD6.7 BLINDFOLD6.6 BLINDFOLD5.8 BLINDFOLD7.5 BLINDFOLD6.1 BLINDFOLD7.2 BLINDFOLD8.5 BLINDFOLD9.9 BLINDFOLD5.9 BLINDFOLD6.3 BLINDFOLD9.1 MAGNET7.6 MAGNET4.6 MAGNET3.1 MAGNET7 MAGNET5.7 MAGNET2.5 MAGNET2.8 MAGNET6.7 MAGNET4.3 MAGNET5.7 MAGNET4.1 MAGNET2.9 BOTH6.9 BOTH7.3 BOTH4.7 BOTH5.7 BOTH4.9 BOTH9.9 BOTH10.3 BOTH3.9 BOTH8.1 BOTH8.8 BOTH4.5 BOTH3.8 Question 2 TrtCells S-type1 S-type0 S-type0 S-type3 S-type7 S-type6 S-type9 S-type7 S-type0 S-type2 C-type11 C-type18 C-type17 C-type11 C-type3 C-type9 C-type18 C-type20 C-type15 C-type20 E-type3 E-type15 E-type23 E-type13 E-type21 E-type25 E-type15 E-type31 E-type125 E-type28 Control50 Control25 Control61 Control36 Control22 Control19 Control29 Control40 Control41 Control39 Question 3 SitePhotosynthetic Efficiency A45 A47 A56 A32 A37 A40 A49 B56 B59 B71 B34 B28 B65 B69 C33 C33 C32 C29 C30 C31 C35 D28 D18 D15 D30 D28 D27 D30 E88 E90 E82 E73 E63 E81 E77 F9 F13 F16 F32 F29 F33 F31 Question 4 Snail SpeciesHerbicide Concentration% Change in Growth 1034 1041 1033 1030 11040 11046 11045 11040 11542 11549 11552 11548 12037 12045 12030 12032 12528 12541 12520 12518 1303 13050 1304 1303 2036 2040 2030 2045 21047 21042 21040 21042 21553 21543 21549 21543 22040 22039 22044 22039 22537 22540 22536 22540 23044 23043 2308 23014 3025 3022 3022 3024 31020 31019 31022 31019 31515 31513 31517 31514 32010 32011 32015 32011 3255 3255 32511 3259 3305 3302 3306 3305 Question 5 LocationMouse speciesmean tibia length (cm) 110.7 210.99 310.85 410.51 511.03 120.53 220.57 320.47 420.35 520.77 130.49 230.76 330.55 430.28 530.84 140.88 240.89 340.81 440.33 540.91 Format to Use for Homework Answers BSC 5935/PCB 4934 Spring 2021 Format to Use for Homework Answers 1. Put your Panther ID at the top of the page, NOT YOUR NAME. 2. Homework answers should be brief and to the point. DO NOT RESTATE THE QUESTION, THE ORIGINAL DATA, OR MAKE LONG-WINDED EXPLANATIONS – IT WON’T HELP! 3. Include RELEVANT portions of your R output and highlight those sections of the output that directly pertain to the answer. 4. If any hand calculations are required, include them in your answer. 5. Below is the format for typical homework answer that you should use. On the following page is an example of a homework answer for a t-test. BSC 5935/PCB 4934 Spring 2021
Answered 1 days AfterMar 31, 2022

Answer To: Homework Assignment 2: Due Tuesday April 1st Homework Assignment 2 INSTRUCTIONS: Data sets for each...

Suraj answered on Mar 31 2022
103 Votes
Solution 1:
Test Required: One Way ANOVA
Independent variable: 4 Conditions
4 levels, Control, BLINDFOLD, MAGNET, BOTH

Dependent Variable: Time it took each of them to find their burrow

Number & Description of replicates: 12 Crabs in each of the 4 conditions
Statistical hypotheses:
At least two of the population means are different.
Assumptions:

(1) Assume iid assumption is met.

(2) I checked normality condition using Shapiro wilk test. Normality assumption is satisfied with W = 0.95832, p-value = 0.08638.
(3) The equal variance assumption is check using the Levane test. The assumption is not satisfied with p-value of 0.0435 <0.05.
Result:
The test-statistic value for the test is 6.509 and the p-value is 0.00097. The p-value is less than 0.05. Hence, the null hypothesis is rejected.
Conclusion:
The null hypothesis is rejected and it is concluded that the navigational capability of crabs differs among the experimental treatments.
R-Code:
#1
> df<-read_excel("Q1.xlsx")
> shapiro.test(df$Time)
    Shapiro-Wilk normality test
data: df$Time
W = 0.95832, p-value = 0.08638
> leveneTest(Time~Trt,df)
Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 3 2.939 0.04349 *
44
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> ANOVA<-aov(Time~Trt,df)
> summary(ANOVA)
Df Sum Sq Mean Sq F value Pr(>F)
Trt 3 59.85 19.950 6.509 0.000968 ***
Residuals 44 134.87 3.065
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Solution 2:
Test Required: One Way ANOVA
Independent variable: gene mutation types
4 levels, 10 S-type, C-type, 10 E-type, 10 non-mutated control

Dependent Variable: number of fugus infected tree cells per mm of tissue
Number & Description of replicates: 10 trees in each of the 4 gene type mutations
Statistical hypotheses:
At least two of the population means are different.
Assumptions:

(1) Assume iid assumption is met.

(2) I checked normality condition using Shapiro wilk test. Normality assumption is not satisfied W = 0.73878, p-value = 4.484e-07 < 0.05.
(3) The equal variance assumption is check using the Levane test. The assumption is satisfied with p-value of 0.235 > 0.05.
Result:
The test-statistic value for the test is 6.331 and the p-value is 0.0015. The p-value is less than 0.05. Hence, the null hypothesis is rejected.
Conclusion:
The null hypothesis is rejected and it is concluded that the gene mutation types (S, C, or E) confer more resistance to the fungus compared to the non-mutant control.
R-Code:
#2
> df2<-read_excel("Q2.xlsx")
> shapiro.test(df2$Cells)
    Shapiro-Wilk normality test
data: df2$Cells
W = 0.73878, p-value = 4.484e-07
>...
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