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The Effect of Beech Bark Disease on Radial Growth: A Statistical Investigation

By Julia Worland

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Part 1:

Mathematical Concepts

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The Normal Distribution

In a frequency distribution, data can be distributed in many different ways.

Properties of the Normal Distribution

symmetry about the center
68% of values are within 1 SD of the mean
95% of values are within 2 SD of the mean
99.7% of values are within 3 SD of the mean
mean = median = mode

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The Sampling Distribution of the Mean of Quantitative Data

When a population has a mean of μ and a true standard deviation σ, we predict that sample means taken from the population has a sampling distribution with the same mean μ but whose standard deviation is σ / √n.

inversely proportional

mean = μ

SD = σ

mean = μ

SD = σ / √5

mean = μ

SD = σ / √30

Example:

μ(x̅) = μ

σ(x̅) = σ / √n

There is an explicit relationship between the distribution of data taken from a population and the distribution of means of that data.

When a population has a given mean (μ) and a given standard deviation (σ), we predict that sample means taken from the population has a sampling distribution with the same mean (μ) but a standard deviation of the original standard deviation divided by the square root of the sample size (σ / √n).

As shown by this equation, sample size and the standard deviation of means are inversely proportional. Means, by adding up all values and dividing by the total number of values, cancel out a large amount of variation. Thus, they have less variability than the raw quantitative data they come from. Thus, their standard deviation is σ / √n, which will always be less than σ (assuming n>1).

This example shows that, as sample size increases from 5 to 30, the mean becomes more accurate and has a smaller standard deviation. As n increases, standard deviation of means decreases.

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Null & Alternative Hypothesis

The Null Hypothesis:

states that a population parameter is equal to a hypothesized value
based on previous analyses or specialized knowledge

The Alternative Hypothesis:

states that a population parameter is smaller, greater, or different than the hypothesized value in the null hypothesis

Ho:

parameter = hypothesized value

Ha:

parameter ≠ hypothesized value

hypothesis test for the mean of a single population:

Ho: µo = µ

Ha: µo ≠ µ

hypothesis test for the difference in mean of two populations

Ho: μ1=μ2 , μ1-μ2=0

Ha: μ1≠μ2 , μ1-μ2≠0

One of the main purposes of statistics is to test a hypothesis. Null hypothesis significance testing is a method of statistical inference by which an observation is tested against the null hypothesis

The null hypothesis states that a population parameter (such as the mean, the standard deviation, and so on) is equal to a hypothesized value. The null hypothesis is often an initial claim that is based on previous analyses or specialized knowledge.

We usually write down the null hypothesis in the form...

Ho: parameter = hypothesized value

The alternative hypothesis contains the values of the parameter that we consider plausible if we reject the null hypothesis. It states that a population parameter is smaller, greater, or different than the hypothesized value given by the null hypothesis

We usually write down the alternative hypothesis in the form...

Ha: parameter ≠ hypothesized value

The null hypothesis is the hypothesis that is assumed to be true. We then test to see the probability of the observed data occurring under the null hypothesis, and dependending on our results, either confirm or reject the null. If we reject the null, we conclude the alternative hypothesis.

If we are performing a hypothesis test for the mean of a single population, our null hypothesis is that the initial claim of the mean is equal to its actual mean. Our alternative hypothesis is that the initial claim of the mean of a population is not equal to its actual mean.

Ho: µo = µ

Ha: µo ≠ µ

If we are performing a hypothesis test for the difference in mean of two populations, our null hypothesis is that the means of two populations are equal, in other words, that their difference is zero. Our alternative hypothesis is that the means of two populations are not equal, and thus that their difference is not zero.

Ho: μ1=μ2 , μ1-μ2=0

Ha: μ1≠μ2 , μ1-μ2≠0

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Student’s T-Model

Student’s T-model is used when estimating the mean of a normally-distributed population in situations where the sample size is small and the population's standard deviation is unknown

Characteristics:

unimodal
symmetric
bell-shaped models
change in shape depending on n.

the smaller a sample size, the taller the tails.

= degrees of freedom = n-1

= normal distribution

When estimating the mean of a normally-distributed population in situations where the sample size is small and the population's standard deviation is unknown, we model data using a variation on a normal distribution called student's t distribution.

These models are unimodal, symmetric, bell-shaped, and change in shape depending on n. The larger the sample size, the closer the distribution resembles the normal model. The smaller a sample size, the taller the tails. To find the probability that a mean at least as extreme as a given number occurred, we find the area of the distribution which includes values at least as extreme as the given number. When n is small and the tails are taller, the area and thus the probability that the observed mean occurred under the null is larger. The accounts for the fact that there is a larger probability that an extreme value occurred in a small sample size.

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Assumptions for T-Models

Independence assumption: The sampled values must be independent of and not influence each other.

Randomization condition: The sample population used should be a simple random sample representative of the whole target population

10% condition: the sample size, n, must be no larger than 10% of the entire target population

Nearly normal condition: The distribution of the variable measured should be generally normal in shape

Random Sampling

Population

Sample

Certain assumptions and conditions must be met to ensure t-distribution modeling is appropriate for a dataset.

Independence assumption: the sampled values must be independent of and not influence each other. When working with two sample populations, the values must not influence each other and not influence the other population
Randomization condition: The sample population used should be a simple random sample representative of the whole target population. Subjects should be randomly selected, and if given a treatment, should be randomly assigned one.
10% condition: the sample size, n, must be no larger than 10% of the entire target population.
Nearly normal condition: In order to me modeled with a t-distribution the distribution of the variable measured should be generally normal in shape. The Central Limit Theorem states that a distribution becomes more and more normal and more and more representative of the true distribution of the actual proportion as the sample size increases. Thus, the larger a sample size, the more likely the nearly normal condition will be violated if the distribution is not perfectly normal.

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T-Scores

1-sample t-score:

a measure of how unusual a mean is
how many standard errors a given average is above or below what we assume to be the true average of a population

ȳ - μ

t = --------

SE

2-sample t-score

a measure of how unusual a difference between two means is
how many standard errors a given difference in averages is above what we assume to be the true difference between averages

(ȳ1 - ȳ2) - 0 ← Ho

t = --------------------

SE (ȳ1 - ȳ2)

ȳ = average from a sample population

μ = average under null hypothesis

SE(ȳ) = s/√n

= an estimation of the SD(μ)

(ȳ1 - ȳ2) = difference of averages from 2 sampling populations

0 = difference of averages under null hypothesis

s1^2 s2^2

SE(y1-y2) = √ ------- + -------

n1 n2

= an estimation of the SD(μ1 - μ1)

T scores are a measure of how unusual a mean or a difference of means is under a null hypothesis. Both types, 1-sample t-scores and 2-sample t-scores, when modeled, take on a t-distribution.

1-sample t-score:

a measure of how many standard errors a given average is above or below what we assume to be the true average of a population

ȳ - μ

t = --------

SE

ȳ = average from a sample population

μ (mu) = average under null hypothesis

SE(ȳ) = s/√n

= an estimation of the SD under the null hypothesis

2-sample t-score

how many standard errors a given difference in averages is above what we assume to be the true difference between averages

(ȳ1 - ȳ2) - 0 ← Ho

t = --------------------

SE (ȳ1 - ȳ2)

(ȳ1 - ȳ2) = difference of averages from 2 sampling populations

0 = difference of averages under null hypothesis

s1^2 s2^2

SE(y1-y2) = √ ------- + -------

n1 n2

= an estimation of the true SD under the null hypothesis

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T-Intervals

n-1 degrees of freedom

confidence level % of distribution area

{

t critical value

t-interval for a mean:

ȳ ± ME

ME = t(n-1) * SE(ȳ)
SE(ȳ) = s / √n

t-interval for a difference in means:

(ȳ1-ȳ2) ± ME

ME = t(n-1) * SE(ȳ1-ȳ2)

s1^2 s2^2

SE(ȳ1-ȳ2) = √ ------ + ------

n1 n2

Interpreting a confidence interval:

If we created an interval based around your sample statistic:

“We are confidence-level% confident that the true population parameter is within your interval.

If we created an interval based around your population:

“We are confidence-level% confident that a given sample statistic is within your interval.

A T-interval is a way to estimate likely values of a population parameter given a sample statistic and likely values for a sample statistic given a population parameter. It depends on the statistic given, how confident one wants to be in their estimation, called a confidence level, and the sample size.

Throughout these definitions, we will be assuming we were given a sample statistic and want to estimate a population parameter, but terms can be swapped to estimate the converse.

T interval for a mean:

ȳ ± ME

ME = t(n-1) * SE(ȳ)

SE(ȳ) = s / √n

T interval for a difference in means:

(ȳ1-ȳ2) ± ME

ME = t(n-1) * SE(ȳ1-ȳ2)

s1^2 s2^2

SE(ȳ1-ȳ2) = √ ------ + ------

n1 n2

t critical value = the number of standard deviation which, when subtracted and added to the mean of the t distribution, create a range of values which cover your confidence level% of the t distribution. The t critical value is determined by your degrees of freedom, which determines the shape of the t distribution, and your confidence level. You can determine the t value by

creating a t distribution with n-1 degrees of freedom
finding the distances between the two bounds of the middle 95%
dividing the distance by tw

That gives you the number of standard deviations which, when subtracted and added to the mean of the t distribution, create a range of values which cover your confidence level% of the t distribution.

Interpreting a confidence interval:

After creating an interval based around your sample statistic, you are your confidence level% confident that the true population parameter is within your interval. Similarly, if you create an interval based around your population parameter, are your confidence level% confident that a given sample statistic is within your interval.

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P-Value

degrees of freedom = n-1 OR (n1 + n2) -2

= t-score of sample statistic

= p-value (2-tailed) = 0.0073462754

In null hypothesis significance testing, the p-value is the probability of obtaining test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is correct.

the null hypothesis is rejected if the p-value is less than the probability of a type 1 error = 𝜶 = 0.05, by convention

How to calculate p-value:

calculating a t-score
create a t-distribution with the appropriate degrees of freedom
find the area of the distribution that is at least as extreme as that z score

In null hypothesis significance testing, the p-value is the probability of obtaining test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is correct.

In a formal significance test, the null hypothesis is rejected if the p-value is less than the probability of a type 1 error, which occurs when the null hypothesis is true but is falsely rejected. The probability of a type 1 error is denoted as 𝜶(alpha). By convention, 𝜶(alpha) is commonly set to 0.05 or 5%, though other alpha levels are sometimes used if researchers have reason to believe the probability of a type 1 error is lower or higher.

How to calculate p-value:

calculating a t-score, represented with the orange dot in this diagram
create a t-distribution with the appropriate degrees of freedom
find the area of the distribution that is at least as extreme as that z score. Assuming we are conducting a 2-tailed test, which means we are calculating the probability of obtaining test results at least as extreme in both directions of the mean, the p-value is the sum of the two regions shaded in blue.

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Part 2: Application of Mathematical Concepts

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Background

Neonectria

Cryptococcus fagisuga

Xylococcus betulae

Beech bark disease (BBD):

a disease that causes mortality and defects in beech trees in the eastern United States, Canada and Europe.
involves scale insects (Cryptococcus fagisuga and Xylococcus betulae) and a fungal pathogen (Neonectria ditissima and Neonectria faginata).
occurs when the bark of a tree is attacked by the beech scale insect and is then ultimately destroyed by the fungus
occurs in 3 stages: the advancing front, the killing front, the aftermath zone

(Nbrazee, 2019) (Crandall Park Trees)

effects on tree health

greater levels of damage
poorer tree crown conditions
splitting during windy conditions
drought and cold sensitivity

(Koch, 2010)

uninfected beech tree

infected beech tree

In my project, I applied these statistical concepts to real world situation: beech bark disease

Beech bark disease is a disease that causes mortality and defects in beech trees in the eastern United States, Canada and Europe. Its presence in the US is shown in this maps. Beech bark disease is a disease-insect complex that involves both native and non-native scale insects and two species of a fungal pathogen.

The disease develops in three stages: the advancing front, the killing front, and the aftermath zone.

During the advancing front, previously unaffected beech trees are colonized by the beech scale.
During the killing front, fungi infect trees through the damaged bark caused by the beech scale insect. Once the bark is killed, wood-rotting fungi can invade and further weaken the tree, often leading to premature death of the host.
Finally, the aftermath zone occurs when trees die.

The disease had a wide array of effects on tree health, including greater levels of damage, poorer tree crown conditions, and splitting during harsh weather. Trees suffering from BBD are also more sensitive to drought and cold temperature injury during the winter.

However, some beech trees have been observed to be resilient to this disease, and while others die, they remain healthy and uninfected. These are pictures of two of the trees in our study, one which was infected by the disease, and another which, for over 10 years, remained healthy while other trees around it passed. It is very likely that this tree is resilient to the disease.

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Origin of the Idea

Beech trees in my backyard affected with BBD

The Lewis Deane Nature Preserve, Poultney, Vermont

Growing up surrounded by the forests of Vermont, the presence of beech bark disease has been a constant. As a child, I noticed evidence of the disease: Dead beech bark trees were a common sight in my backyard. Of those which remained alive, I noticed that many had cracked bark and cankering. As I grew older, I gained knowledge of the complexities of forest ecosystems. I began to wonder if there were more effects of beech bark disease beyond exterior manifestations of the disease.

In 2019, I got the opportunity to pursue this investigation. The beginning of my sophomore year, I started an independent study with Natalie Coe, a member of a team of researchers from Green Mountain College who had been collecting data from a sample population of beech trees in the Deane Preserve in Poultney, Vermont. Since 2005, this team had been recording the diameter of trees and ranking their severity of infection.

That same year, I was learning about circle geometry and the relationship between the irrational number pi and circle’s radius, circumference, and area. While perfect circles do not exist in the real world, there are many real world examples of objects which embody their mathematical principles. For instance, if natural variation is removed, trees grow like a circle of increasing radius.

This connection prompted me to wonder if beech bark disease had an effect on the radial growth of beech trees. Through collaborating with my teachers, I was able to investigate this question with statistics.

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Research Question, Variables, and Hypothesis

1. First, I chose to investigate the following question:

Is the mean growth rate of trees infected with beech bark disease different from and those without?

explanatory variable = infection by beech bark disease as my explanatory variable (No Infection or Infected)

dependent variable = average radial growth

Null Hypothesis: The mean growth rate of healthy trees is equal to the mean growth rate of trees infected with beech bark disease.

Ho:

x̅(growth rate healthy trees) = x̅(growth rate infected trees)

Ha:

x̅(growth rate healthy trees) = x̅(growth rate infected trees)

Alternative Hypothesis: The mean growth rate of healthy trees differs from the mean growth rate of trees infected with beech bark disease.

1.

First, I chose to investigate the following question: Is the mean growth rate of trees infected with beech bark disease different frm and those without?

I used infection by beech bark disease as my explanatory variable, categorized into No Infection or Infected, and I used average growth rate as my dependent variable.

I then formulated a null and alternative hypothesis concerning growth rates.

Null Hypothesis: The mean growth rate of healthy trees is equal to the mean growth rate of trees infected with beech bark disease.

x̅(growth rate healthy trees) = x̅(growth rate infected trees) -

Alternative Hypothesis: The mean growth rate of healthy trees differs from the mean growth rate of trees infected with beech bark disease.

x̅(growth rate healthy trees) ≠ x̅(growth rate infected trees)

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Data Collection

Trees were measured using a diameter tape.

Trees were ranked using Jacob M. Griffin 1-5 beech bark severity scale (Griffin, 2003):

1 = no visible sign of infection
2-4 = varying severities of beech bark disease
5 = tree in question is dead due to BBD

Visible Signs of Infection:

white waxy covering

red/brown spots

cankering

foliage reduction

and color change

cracking

At the Deane Preserve, we measured trees using a diameter tape, a tape that converts the circumference of a cylinder into an estimate of its diameter. In an effort to account for the variation in diameter of a tree throughout its length, measurements were taken at the breast height of the data collector. If trees were notably slanted, a data collector’s approximate breast height was measured and then measured along a tree's trunk to find the location to measure the tree.

Trees were then ranked using Jacob M. Griffin 1-5 beech bark severity scale, 1 indicating no visible sign of infection, 2-4 indicating varying severities of beech bark disease, and 5 indicating that a tree in question was dead, presumably due to the disease. A tree’s rank was determined by assessing the presence and severity of visual indicators of infection. This included blotches of a white, waxy covering, red or brownish spots, cankering, and a reduction or color change of foliage (Beech Bark Disease).

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Calculating Growth Rates

Tree ID	healthy range	healthy growth	infected range	infected range	infected growth
1
2
...

Tree ID	healthy growth rate	infected growth rate
1
2
...

growth

growth rate = -------------------

range of years

(final year - initial year)

For each tree, we determined

the range of years the tree was healthy, aka has a ranking of 1
the range of years the tree was sick, aka had a ranking of 2, 3, or 4
the growth in diameter while infected (found by subtracting the first diameter recorded when the tree was healthy from the final diameter recorded when the tree was healthy)
the growth in diameter while infected (found by subtracting the first diameter recorded when the tree was sick from the final diameter recorded when the tree was sick)

(It should be noted that there were many tree that were infected or not infected for the entirety of the study, and so only contributed one type of growth)

From these values we were able to calculate

the average growth rate of the tree when healthy (found by dividing health diameter growth by the number of years healthy)
the average growth rate of the tree when infected (found by dividing infected diameter growth by the

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Distributions

The distributions of healthy and infected growth rates found that healthy

growth rates ranged from 0 to 0.8719 cm in diameter per year, while infected growth rates

ranged from 0 to 0.6873 cm. Considering the skewed distributions of data, the medians of each

were the best measure of center. Healthy trees had a higher median growth rate

than infected trees. Their median growth rate was 0.2358 cm in diameter/year, while the median

The growth rate of infected trees was 0.1551 cm in diameter/year. The middle 50% of healthy growth rates ranged from 0.0782 to 0.3364 cm in diameter per year, and the middle 50% of infected trees ranged from 0.0167 to 0.25 cm in diameter per year. In addition, the standard deviation of healthy trees appeared to be slightly larger than infected trees, indicating that healthy trees had greater variability in yearly growth.

The points surrounding the boxplots represent outliers. If a healthy growth rate was beyond the range of 1.5InterQuartileRanges above and below Q1 and Q3 they were deemed outliers and removed from the study. Similarly, if an infected growth rate was beyond the range of 1.5InterQuartileRanges above and below Q1 and Q3 they were also removed

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Assumptions / Conditions

Independence assumption: While clumping of similar growth rates and years of infection can occur, we can assume that, in general, the growth rates of beech trees are independent of one another.

Randomization condition: We assume that random techniques were employed and a representative sample was selected.

10% condition: The 207 trees used in the study are less than 10% of all American beech trees.

Nearly normal condition: The distributions of both healthy and infected growth rates are unimodal, but both are positively skewed. However, past studies suggested radial tree growth to be a normal variable).

Before conducting an hypothesis testing using the t-distribution, I confirmed that none of the assumptions and conditions for t-modeling were violated.

Independence assumption: Past studies have concluded that, overall, beech bark disease spreads randomly through a forest. We can assume that the growth rates of different trees are independent of each other. However, the two growth rates of an individual tree are dependent on that tree. Thus, this condition is slightly violated and may limit the confidence we can place in later conclusions.
Randomization condition: I do not have in depth knowledge on how trees were selected from the Deane Preserve for this study. However, we assume that random techniques were employed. We assume that our sample of 207 trees is representative of our target population of all American beech trees.
10% condition: The 207 trees used in the study are less than 10% of all American beech trees.
Nearly normal condition: The distributions of both healthy and infected growth rates are unimodal, but both are positively skewed. Both also contain outliers at higher growth rates. Considering the moderately large sample size of both rates, suggesting that both should be relatively normal according to the Central Limit Theorem, the violation of this condition may be a concern and may limit the confidence we can place in later conclusions. (However, past studies suggested radial tree growth to be a normal variable).

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T-Test and P-Value

degrees of freedom = 149.0085797

= t-score of 2.71802195653

= p-value = 0.0073462754 = 0.73462754%

(ȳ1 - ȳ2) - 0 ← Ho

t = ----------------

SE (ȳ1 - y2)

(0.2395 - 0.1654) - 0 ← Ho

t = -----------------------

0.2021^2 0.1614^2

√------- + ---------

81 109

= 2.71802195653

First, I performed a 2-sample t-score for the observed difference in mean growth rates

(ȳ1 - ȳ2) - 0 ← Ho

t = --------------------

SE (ȳ1 - y2)

(0.2395 - 0.1654) - 0 ← Ho

t = -------------------------------------

0.2021^2 0.1614^2

√------------ + -------------

81 109

t = 2.71802195653

Thus ,the observed difference in means was roughly 2.72 standard eros from the differenc under the null hypothesis.

Next, I calculated the p-value, the area of the t-distribution with a t-score at least as extreme as the observed one.

First, I created a t-distribution with the appropriate degrees of freedomDifferent programs have different ways of calculating degrees of freedom during two sample t-tests. One common way is to add the two n values and subtract 2. The program I used on my ti-84 calculator created a similar value to this with about 149 degrees of freedom.
Then, I found the area of the distribution that is at least as extreme as that z score. I did so using a function on my ti-84 called tcdf. This function allows you to find the area of a section of the t distribution with a custom degree of freedom. I found the area from the lower bound of the t-distribution, negative infinity, to a t score of 2.71802195653, which was roughly 0.0036731377. I then multiplied it by two to get the total area of the distribution that was as extreme as our observed difference in z scores. The total area, aka our p-value, was approximately 0.0073462754.

There is only a 0.73% probability that the observed difference in distributions occurred under the null hypothesis. Given that this probability is less than 5%, We reject the null hypothesis and conclude that the mean growth rate of healthy trees differs from the mean growth rate of trees infected with beech bark disease.

However, we do have some concern in these conclusions due to the fact that not all assumptions and conditions for t-distribution modeling was correct. Given to the skewed distributions on both

variables, we are unsure if it was appropriate to model with a d-distribution, which is symmetric and nearly normal.

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T-Interval

formula:

confidence interval = (ȳ1-ȳ2) ± ME

ME = t(df) * SE(ȳ1-ȳ2)

s1^2 s2^2

SE(ȳ1-ȳ2) = √ ------ + ------

n1 n2

my calculation:

ȳ1-ȳ2 = 0.2395 - 0.1654 = 0.0781

ME = t(df) * SE(ȳ1-ȳ2)

t(149.0085797) = 1.976012213.

0.2021^2 0.1614^2

SE(ȳ1-ȳ2) = √ -------- + --------

81 109

= 0.02726247274

ME = 1.976012213 * 0.02726247274

= 0.0538709791

Confidence interval

= 0.0781 ± 0.053870979 = ( 0.024229021 , 0.131970979 )

conclusion:

We are 95% confident that the true difference in mean growth rates between healthy beech trees and those infected by beech bark disease is between 0.0242 and 0.132 cm in diameter per year.
If repeated infinitely, 95% of intervals would capture the true difference in means.
This interval does not include 0, so we conclude that it is unlikely that the true mean growth rates are equal.

I then calculated a t-interval to estimate the true difference in mean growth rates between healthy and beech bark disease infected trees.

formula:

(ȳ1-ȳ2) ± ME

ME = t(df) * SE(ȳ1-ȳ2)

s1^2 s2^2

SE(ȳ1-ȳ2) = √ ------ + ------

n1 n2

my calculation:

(0.2395 - 0.1654) ± ME

0.0781 ± ME

ME = t(df) * SE(ȳ1-ȳ2)

t(149.0085797)

= the number of standard deviation which, when subtracted and added to the mean of the t distribution, create a range of values which cover your confidence level% of the t distribution

I want a 95% confidence interval.

In a t-distribution with 149 degrees of freedom, the cutoff t-score values of the middle 95% is -1.976012213 and 1.976012213.

Therefore, by t critical value is 1.976012213.

invt() = input df and percentage below cutoff point → get cutoff point

0.2021^2 0.1614^2

SE(ȳ1-ȳ2) = √ ----------- + -----------

81 109

= 0.02726247274

ME = 1.976012213 * 0.02726247274

= 0.0538709791

0.0781 ± 0.053870979

( 0.024229021 , 0.131970979 )

We are 95% confident that the true difference in mean growth rates between healthy beech trees and those infected by beech bark disease is between 0.0242 and 0.132 cm in diameter per year. 95% confidence indicates that, if repeated infinitely, 95% of intervals would capture the true difference in means. This interval does not include 0, so we conclude that it is unlikely that the true mean growth rates are equal.

However, we again do have some concern in these conclusions due to the fact that not all assumptions and conditions for t-distribution modeling were correct.

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The Future of Null Hypothesis Significance Testing:

My Ideas

redefine α
make it commonplace to provide justification for α-levels (own experiment or site past experiments)
Depending on the alpha level these prior experiments produce and depending on how exact experimenters need their analysis to be, the experiment may need to be redesigned

�Application to our beech bark disease study:

Perhaps, determining the true alpha level of our experiment and then re-evaluating our data could be the next step in strengthening our statistical analysis.

(On the Past and Future of Null Hypothesis Significance Testing 2001)

Common System

Proposed System

Question

↓

Hypothesis

↓

Design Experiment

↓

Collect Data

↓

Analyze Data

↓

Conclusion

Question

↓

Hypothesis

↓

Design Experiment

↓

Determine alpha

↓

Collect Data

↓

Analyze Data

↓

Conclusion

While working with null hypothesis significance testing, I became intrigued and sceptical of the concept that alpha should always be set to 0.05 or 5% probability. It seemed improbable to me that the great variety of studies employing this statistical technique all had exactly a 5% probability of committing a type 1 error. Upon research, I discovered that the future of null hypothesis testing greatly resides in redefining alpha. To make null hypothesis testing more accurate, I believe we could make it commonplace to justify chosen alpha levels, either by performing an initial experiment to determine the probability of a type 1 error, or at the very least, site past experiments that determined that probability.

Depending on the alpha level these prior experiments produce and depending on how exact experimenters need their analysis to be, the experiment may need to be redesigned. For example, Ronald Fisher, the man who developed the theory behind the p value, always attempted to improve the design when p values were between 0.05 and 0.2.

Perhaps, determining the true alpha level of our experiment and then re-evaluating our data could be the next step in strengthening our statistical analysis. This goes to show that, even once the calculations are done and the conclusions are made, there’s always room for improvement and discovery.

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Sources:

Crandall Park Trees. (n.d.). Retrieved April 03, 2021, from http://mdocs.skidmore.edu/crandallparktrees/invasives/beech-bark-disease/

Griffin, J. M., Lovett, G. M., Arthur, M. A., & Weathers, K. C. (2003). The distribution and severity of beech bark disease in the Catskill Mountains, n.y. Canadian Journal of Forest Research, 33(9), 1754-1760. doi:10.1139/x03-093

Koch, J. L. (2010). Beech bark disease: The OLDEST "NEW" threat to American beech in the United States. Outlooks on Pest Management, 21(2), 64-68. doi:10.1564/21apr03

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