Statistical Analysis Assignment Solution: Biological Statistics

This assignment solution delves into various statistical techniques used in biological research. It provides a comprehensive analysis of hypothesis testing, including the formulation of null and alternative hypotheses, and the interpretation of p-values to determine statistical significance. The solution extensively utilizes ANOVA (Analysis of Variance) and MANOVA (Multivariate Analysis of Variance) to compare means across different treatment groups and assess the significance of multiple variables simultaneously. The application of TukeyHSD and Bonferroni correction methods are also discussed to determine specific differences between groups and control for Type I errors. Furthermore, the solution incorporates R programming for data analysis, including the generation of matrices, calculation of statistical values, and interpretation of results. The assignment covers several biological scenarios, such as comparing knowledge acquisition, motivation levels, sepal lengths, and amino acid content, making it a valuable resource for students studying biological statistics.

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Biological Statistics
Student’s Name
Institution
Course
Date

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Question 2
a) For the knowledge to be the same in all treatments, the hypothesis will be as follows
H0: T1=T2=T2
H1: T1# T2#T2
From R analysis the results were as follows
The P-value was 0.00283 which was less than 0.05. As a result, the null hypothesis is rejected
and the conclusion is that there are significant differences between knowledge acquired across
the 3 groups of employees. The differences were from T2-T1 and T3-T2 as shown by TukeyHSD
test.
b) The null hypothesis will be as follows
H0: T1=T2=T2
H1: T1# T2 #T2

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The analysis from R program was as follows:
The p-value is less than 0.05 and, therefore, it means that there is a significant difference
between the mean motivation between the three treatments.2 In particular, the difference is from
Treatment 2 and treatment 1 (T2-T1) and Treatment 3 and Treatment 2 (T3-T2) as shown by
TukeyHSD test.
c) MANOVA compares mans between treatments. Therefore, the hypothesis will be set as
follows:
Ho: ᶙT1=ᶙT2=ᶙT3 where U is the mean
Ho: ᶙT1# ᶙT2#ᶙT2
The results from R program were as follows:

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The p value is less than 0.05. As a result, the means between the treatment groups are different.
The null hypothesis is rejected in favor of the alternative hypothesis. Therefore, it is appropriate
to conclude that there is a difference between the means.
Question 3
a) Four one-way ANOVA
Septal.Length~Species ANOVA analysis

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The adjusted p-value form TukeyHSD test are less than 0.05. This shows that sepal lengths are
significantly different amongst the three species of flowers
Petal.Length~Species ANOVA analysis
The adjusted P-value from TukeyHSD test are less than 0.05. As a result, petal lengths are
significantly different between the 3 species.
Petal.Width~Species ANOVA analysis
The adjusted p-value form TukeyHSD test are less than 0.05. Therefore , the petal width were
significantly different between the 3 species of flowers,

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Sepal.Width~Species ANOVA analysis
The adjusted p-value form TukeyHSD test are less than 0.05. Therefore, sepal widths were
significantly different between the 3 species of flowers.
b) The probability of type one error (rejection of null hypothesis when it is indeed correct) is
higher when using One-way ANOVA, to evaluate the accuracy of four independent variables
separately. Bonferroni correction entails testing of hypothesis for each individual variable where
the significance level is set at α/m where α is the significance level and m is the number of
hypothesis being tested1. If the number of hypothesis is 4 and the α=0.05, the significance level
for each variable becomes :0.05/4 = 0.0125. Setting the significance level at 0.0125, in this case,
allows for a precise analysis and increases the accuracy of the conclusion derived from the
analysis.
c) Manova
The R-code is as follows

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Manova analyzes whether the difference between different means is significant by comparing all
the variables in the dataset at a glance3. The P-value calculated provides if greater than 0.05
shows that there is no significant difference between the variable means in a dataset. On the
other hand, if the p-value is less than 0.05, it means that there is a difference in at least one mean
in the dataset.4. In the above analysis, the p-value is less than 0.05, and as a result, there is a
significance difference between the in the means of the observe variables
The difference between the MANOVA and one-way ANOVA is that, Manova analyzes different
variables together and determines if there is any significant different amongst them.4 On the
other hand, one-way ANOVA analysis independent variables separately5. This makes it difficult
to tell whether interaction between different variables is significant or not.

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Question 4
The R code to generate the matrix was as follows
a) The analysis using R

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The P-value is 0.001317 for Pillai trace, Wilks’ Lambda, Hotelling-Lawley and Roy’s statistics
and therefore, it does not matter the type of the test conducted on the data. The p-value is smaller
than 0.05. This means that there is a significant difference between the three amino acids
contains on female and male centipedes.
b)

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The P values for response 1 and response 2 are 0.8492 and 0.0879 respectively. However, the P-
value for response 3 is 0.01006. It therefore means that Alanine and Aspartic Acid are not
significant in both male and female centipedes while Tyrosine is significant between male and
female centipedes
Question 5
Hormone and gender are the two factors of interest.
Matrix for the response variables
Manova test

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The P-values for the main effects (Sex and Hormone) are 0.35844, and 0.00114 for sex and
hormones respectively. The effect of sex in not significant as the p-value is more than 0.35844.
On the other hand, the p-value for Hormone shows that there is as significant effect between the
female and male gender. The interaction effect has a p-value of 0.27435 which means shows that
there is no significance difference in the model occasioned by a combination of gender and
hormones.
The fitted means for different groups are as follows:
sex Hormone Sex: Hormone
Plasma.Ca 14.7408 635.108 7.2075
Water.loss 18.75 102.083 36.75
The table above shows the means at the point of interaction between the variables
Oneway ANOVA for variables
For Plasma, the p value is less than 0.05 hormone-plasma interaction. This shows that there is as
a difference between hormone groupings and interaction with plasma. However, the p-value for
sex and plasma interaction and combined effect of both sex and plasma interaction are more than
0.05, and this means that there is no difference between the variables in their interaction with
Plasma Ca.

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The p-value effect of hormone and water loss has a value of 0.048. this value is closer to 0.05
and as a result, hormone was perceived to have no significant effect on water loss. Sex and
interaction between sex and hormone also disclosed that there was no significant effect
interaction between the variables.