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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...

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.
Â· This assignment is due on Thursday, March 31th.
Â· 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 bu
ows 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 bu
ow and the time it took each of them to find their bu
ow (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 1900â€™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
ought 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,Â Drymaeus dormani; West Indian Bulimulus,Â Bulimulus guadalupensis; lined forest snail,Â Drymaeus multilineatus) changes when exposed to different concentrations (0 = control and 10, 15, 20, 25, and 30ppm) of an he
icide 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 he
icide 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 he
icide 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?
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1

Question 1
Trt    Time
Control    3.2
Control    4.4
Control    4.7
Control    2.9
Control    5
Control    4.2
Control    5.7
Control    5.7
Control    6.2
Control    3.9
Control    3.8
Control    3
BLINDFOLD    4.1
BLINDFOLD    6.7
BLINDFOLD    6.6
BLINDFOLD    5.8
BLINDFOLD    7.5
BLINDFOLD    6.1
BLINDFOLD    7.2
BLINDFOLD    8.5
BLINDFOLD    9.9
BLINDFOLD    5.9
BLINDFOLD    6.3
BLINDFOLD    9.1
MAGNET    7.6
MAGNET    4.6
MAGNET    3.1
MAGNET    7
MAGNET    5.7
MAGNET    2.5
MAGNET    2.8
MAGNET    6.7
MAGNET    4.3
MAGNET    5.7
MAGNET    4.1
MAGNET    2.9
BOTH    6.9
BOTH    7.3
BOTH    4.7
BOTH    5.7
BOTH    4.9
BOTH    9.9
BOTH    10.3
BOTH    3.9
BOTH    8.1
BOTH    8.8
BOTH    4.5
BOTH    3.8
Question 2
Trt    Cells
S-type    1
S-type    0
S-type    0
S-type    3
S-type    7
S-type    6
S-type    9
S-type    7
S-type    0
S-type    2
C-type    11
C-type    18
C-type    17
C-type    11
C-type    3
C-type    9
C-type    18
C-type    20
C-type    15
C-type    20
E-type    3
E-type    15
E-type    23
E-type    13
E-type    21
E-type    25
E-type    15
E-type    31
E-type    125
E-type    28
Control    50
Control    25
Control    61
Control    36
Control    22
Control    19
Control    29
Control    40
Control    41
Control    39
Question 3
Site    Photosynthetic Efficiency
A    45
A    47
A    56
A    32
A    37
A    40
A    49
B    56
B    59
B    71
B    34
B    28
B    65
B    69
C    33
C    33
C    32
C    29
C    30
C    31
C    35
D    28
D    18
D    15
D    30
D    28
D    27
D    30
E    88
E    90
E    82
E    73
E    63
E    81
E    77
F    9
F    13
F    16
F    32
F    29
F    33
F    31
Question 4
Snail Species    He
icide Concentration    % Change in Growth
1    0    34
1    0    41
1    0    33
1    0    30
1    10    40
1    10    46
1    10    45
1    10    40
1    15    42
1    15    49
1    15    52
1    15    48
1    20    37
1    20    45
1    20    30
1    20    32
1    25    28
1    25    41
1    25    20
1    25    18
1    30    3
1    30    50
1    30    4
1    30    3
2    0    36
2    0    40
2    0    30
2    0    45
2    10    47
2    10    42
2    10    40
2    10    42
2    15    53
2    15    43
2    15    49
2    15    43
2    20    40
2    20    39
2    20    44
2    20    39
2    25    37
2    25    40
2    25    36
2    25    40
2    30    44
2    30    43
2    30    8
2    30    14
3    0    25
3    0    22
3    0    22
3    0    24
3    10    20
3    10    19
3    10    22
3    10    19
3    15    15
3    15    13
3    15    17
3    15    14
3    20    10
3    20    11
3    20    15
3    20    11
3    25    5
3    25    5
3    25    11
3    25    9
3    30    5
3    30    2
3    30    6
3    30    5
Question 5
Location    Mouse species    mean tibia length (cm)
1    1    0.7
2    1    0.99
3    1    0.85
4    1    0.51
5    1    1.03
1    2    0.53
2    2    0.57
3    2    0.47
4    2    0.35
5    2    0.77
1    3    0.49
2    3    0.76
3    3    0.55
4    3    0.28
5    3    0.84
1    4    0.88
2    4    0.89
3    4    0.81
4    4    0.33
5    4    0.91

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.
ief and to the point. DO NOT RESTATE THE
QUESTION, THE ORIGINAL DATA, OR MAKE LONG-WINDED EXPLANATIONS
â€“ IT WONâ€™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 After Mar 31, 2022

## Solution

Suraj answered on Mar 31 2022
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 bu
ow
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
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 â€˜***â€™ 0.001 â€˜**â€™ 0.01 â€˜*â€™ 0.05 â€˜.â€™ 0.1 â€˜ â€™ 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 â€˜***â€™ 0.001 â€˜**â€™ 0.01 â€˜*â€™ 0.05 â€˜.â€™ 0.1 â€˜ â€™ 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
shapiro.test(df2\$Cells)
Shapiro-Wilk normality test
data: df2\$Cells
W = 0.73878, p-value = 4.484e-07
...
SOLUTION.PDF

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