initial import of draft website without any materials

Stats_review_soln.qmd

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+---
+title: "VectorByte Methods Training 2024"
+subtitle: "Probability and Statistics Fundamentals: Solutions to Exercises"
+author: "The VectorByte Team (Leah R. Johnson, Virginia Tech)"
+format:
+  html:
+    toc: true
+    toc-location: left
+    html-math-method: katex
+    css: styles.css
+---
+
+[Main materials](materials.qmd)
+
+[Back to stats review](Stats_review.qmd)
+
+### Question 1: For the six-sided fair die, what is $f_k$ if $k=7$? $k=1.5$?
+
+<br>
+
+Answer: both of these are zero, because the die cannot take these values.
+
+<br> <br> <br> <br> \hfill\break
+
+### Question 2: For the fair 6-sided die, what is $F(3)$? $F(7)$? $F(1.5)$?
+
+Answer: The CDF total probability of having a value less than or equal to its argument. Thus $F(3)= 1/2$, $F(7)=1$, and $F(1.5)=1/6$
+
+<br> <br> <br> <br> \hfill\break
+
+### Question 3: For a normal distribution with mean 0, what is $F(0)$?
+
+<br>
+
+Answer: The normal distribution is symmetric around its mean, with half of its probability on each side. Thus, $F(0)=1/2$
+
+<br> <br> <br> <br> \hfill\break
+
+### Question 4: Summation Notation Practice
+
+| $i$   | 1   | 2    | 3   | 4    |
+|-------|-----|------|-----|------|
+| $Z_i$ | 2.0 | -2.0 | 3.0 | -3.0 |
+
+(i) **Compute** $\sum_{i=1}^{4}{z_i}$ = 0 <br>
+(ii) **Compute** $\sum_{i=1}^4{(z_i - \bar{z})^2}$ = 26 <br>
+(iii) **What is the sample variance? Assume that the** $z_i$ are i.i.d.. *Note that i.i.d.\~stands for "independent and identically distributed".* <br>
+
+Solution: $$
+s^2= \frac{\sum_{i=1}^N(Y_i - \bar{Y})^2}{N-1} = \frac{26}{3}
+= 8\times \frac{2}{3}
+$$ <br>
+
+(iv) **For a general set of** $N$ numbers, $\{X_1, X_2, \dots, X_N \}$ and $\{Y_1, Y_2, \dots, Y_N \}$ show that $$
+     \sum_{i=1}^N{(X_i - \bar{X})(Y_i - \bar{Y})} = \sum_{i=1}^N{(X_i-\bar{X})Y_i}
+     $$
+
+<br> Solution: First, we multiply through and distribute: $$
+\sum_{i=1}^N(X_i-\bar{X})(Y_i-\bar{Y}) = \sum_{i=1}^N(X_i-\bar{X})Y_i
+- \sum_{i=1}^N(X_i-\bar{X})\bar{Y}
+$$ Next note that $\bar{Y}$ (the mean of the $Y_i$s) doesn't depend on $i$ so we can pull it out of the summation: $$
+\sum_{i=1}^N(X_i-\bar{X})(Y_i-\bar{Y}) = \sum_{i=1}^N(X_i-\bar{X})Y_i
+- \bar{Y} \sum_{i=1}^N(X_i-\bar{X}).
+$$ Finally, the last sum must be zero because $$
+\sum_{i=1}^N(X_i-\bar{X}) = \sum_{i=1}^N X_i- \sum_{i=1}^N \bar{X} = N\bar{X} - N\bar{X}=0.
+$$ Thus \begin{align*}
+\sum_{i=1}^N(X_i-\bar{X})(Y_i-\bar{Y}) &= \sum_{i=1}^N(X_i-\bar{X})Y_i - \bar{Y}\times 0\\
+& = \sum_{i=1}^N(X_i-\bar{X})Y_i.
+\end{align*}
+
+<br> <br> <br> <br> \hfill\break
+
+### Question 5: Properites of Expected Values
+
+**Using the definition of an expected value above and with X and Y having the same probability distribution, show that:**
+
+```{=tex}
+\begin{align*}
+ \text{E}[X+Y]  & = \text{E}[X] + \text{E}[Y]\\  
+ & \text{and} \\
+ \text{E}[cX]  & = c\text{E}[X]. \\
+\end{align*}
+```
+**Given these, and the fact that** $\mu=\text{E}[X]$, show that:
+
+```{=tex}
+\begin{align*}
+ \text{E}[(X-\mu)^2]  = \text{E}[X^2] - (\text{E}[X])^2
+\end{align*}
+```
+**This gives a formula for calculating variances (since** $\text{Var}(X)= \text{E}[(X-\mu)^2]$).
+
+Solution: Assuming $X$ and $Y$ are both i.i.d. with distribution $f(x)$. The expectation of $X+Y$ is defined as \begin{align*}
+ \text{E}[X+Y]  & =  \int (X+Y) f(x)dx \\
+              & =  \int (X f(x) +Y f(x))dx  \\
+              & =  \int X f(x)dx  +\int Y f(x)dx  \\
+               & = \text{E}[X] + \text{E}[Y]  
+ \end{align*} Similarly \begin{align*}
+\text{E}[cX]   & =  \int cXf(x)dx \\
+              & =  c \int Xf(x) dx  \\
+              & = c\text{E}[X]. \\
+\end{align*} Thus we can re-write: \begin{align*}
+ \text{E}[(X-\mu)^2]  & = \text{E}[ X^2 - 2X\mu + \mu^2] \\
+                        & = \text{E}[X^2] - 2\mu\text{E}[X] + \mu^2 \\
+                        & = \text{E}[X^2] -2\mu^2 + \mu^2 \\
+                        & = \text{E}[X^2] - \mu^2 \\ 
+ & = \text{E}[X^2] - (\text{E}[X])^2.
+\end{align*}
+
+<br> <br> <br> <br>
+
+\hfill\break
+
+### Question 6: Functions of Random Variables
+
+**Suppose that** $\mathrm{E}[X]=\mathrm{E}[Y]=0$, $\mathrm{var}(X)=\mathrm{var}(Y)=1$, and $\mathrm{corr}(X,Y)=0.5$.
+
+(i) Compute $\mathrm{E}[3X-2Y]$; and
+(ii) $\mathrm{var}(3X-2Y)$.
+(iii) Compute $\mathrm{E}[X^2]$.
+
+Solution:
+
+(i) Using the properties of expectations, we can re-write this as: \begin{align*}
+    \mathrm{E}[3X-2Y] & = \mathrm{E}[3X] + \mathrm{E}[-2Y]\\
+    & = 3 \mathrm{E}[X] -2 \mathrm{E}[Y]\\
+    & = 3 \times 0 -2 \times 0\\
+    &=0
+    \end{align*}
+
+\hfill\break
+
+(ii) Using the properties of variances, we can re-write this as: \begin{align*}
+     \mathrm{var}(3X-2Y) & = 3^2\text{Var}(X) + (-2)^2\text{Var}(Y) + 2(3)(-2)\text{Cov}(XY)\\
+     & =  9 \times 1 + 4 \times 1 -12 \text{Corr}(XY)\sqrt{\text{Var}(X)\text{Var}(Y)}\\
+     & = 9+4 -12 \times 0.5\times1\\
+     &=7
+     \end{align*}
+
+\hfill\break
+
+(iii) Recalling from Question 5 that the variance is $\mathrm{var}(X) = \text{E}[X^2] - (\text{E}[X])^2$, we can re-arrange to obtain: \begin{align*}
+      \mathrm{E}[X^2] & = \mathrm{var}(X) + (\mathrm{E}[X])^2\\
+       & = 1+(0)^2 \\
+       & =1
+      \end{align*}
+
+\hfill\break
+
+### The Sampling Distribution
+
+Suppose we have a random sample $\{Y_i, i=1,\dots,N \}$, where $Y_i \stackrel{\mathrm{i.i.d.}}{\sim}N(\mu,4)$ for $i=1,\ldots,N$.
+
+i.  What is the variance of the *sample mean*?
+
+$$\displaystyle \mathrm{Var}(\bar{Y}) =
+\mathrm{Var}\left(\frac{1}{N}\sum_{i=1}^N Y_i\right) =
+\frac{N}{N^2}\mathrm{Var}(Y) =\frac{4}{N}$$.
+
+This is the derivation for the variance of the *sampling distribution*.
+
+<br> \hfill\break
+
+ii. What is the expectation of the *sample mean*?
+
+$$\displaystyle\mathrm{E}[\bar{Y}] = \frac{N}{N}\mathrm{E}(Y) = \mu.$$ This is the mean of the *sampling distribution*.
+
+<br>
+
+iii. What is the variance for another i.i.d. realization $Y_{ N+1}$?
+
+$\displaystyle \mathrm{Var}(Y) = 4$, because this is a sample directly from the population distribution.
+
+<br> \hfill\break 
+
+iv. What is the standard error of $\bar{Y}$?
+
+Here, again, we are looking at the distribution of the sample mean, so we must consider the sampling distribution, and the standard error (aka the standard distribution) is just the square root of the variance from part i.
+
+$$\displaystyle \mathrm{se}(\bar{Y}) = \sqrt{\mathrm{Var}(\bar{Y})} =\frac{2}{\sqrt{N}}$$.
+
+\hfill\break
+
+### Hypothesis Testing and Confidence Intervals
+
+```{r, echo=FALSE}
+m<-12
+```
+
+Suppose we sample some data $\{Y_i, i=1,\dots,n \}$, where $Y_i \stackrel{\mathrm{i.i.d.}}{\sim}N(\mu,\sigma^2)$ for $i=1,\ldots,n$, and that you want to test the null hypothesis $H_0: ~\mu=`r m`$ *vs.* the alternative $H_a: \mu \neq ` r m`$, at the $0.05$ significance level.
+
+i.  What test statistic would you use? How do you estimate $\sigma$?
+ii. What is the distribution for this test statistic *if* the null is true?
+iii. What is the distribution for the test statistic if the null is true and $n \rightarrow \infty$?
+iv. Define the test rejection region. (I.e., for what values of the test statistic would you reject the null?)
+v.  How would compute the $p$-value associated with a particular sample?
+vi. What is the 95% confidence interval for $\mu$? How should one interpret this interval?
+vii. If $\bar{Y} = 11$, $s_y = 1$, and $n=9$, what is the test result? What is the 95% CI for $\mu$?
+
+<br> <br> \hfill\break
+
+This question is asking you think about the hypothesis that the mean of your distribution is equal to `r m`. I give you the distribution of the data themselves (i.e., that they're normal). To test the hypothesis, you work with the sampling distribution (i.e., the distribution of the sample mean) which is: $\bar{Y}\sim N\left(\mu, \frac{\sigma^2}{n}\right)$.
+
+\hfill\break
+
+i.  If we knew $\sigma$, we could use as our test statistic $z=\displaystyle \frac{\bar{y} - `r m`}{\sigma/\sqrt{n}}$. However, here we need to estimate $\sigma$ so we use $z=\displaystyle \frac{\bar{y} - `r m`}{s_y/\sqrt{n}}$ where $\displaystyle s_{y} = \sqrt{\frac{\sum_{i=1}^n(y_i - \bar{y})^2}{n-1}}$.<br>
+
+\hfill\break
+
+ii. If the null is true, the $z \sim t_{n-1}(0,1)$. Since we estimate the mean frm the data, the degrees of freedom is $n-1$.<br>
+
+\hfill\break
+
+iii. As $n$ approaches infinity, $t_{n-1}(0,1) \rightarrow N(0,1)$.<br>
+
+\hfill\break
+
+iv. You reject the null for $\{z: |z| > t_{n-1,\alpha/2}\}$.<br>
+
+\hfill\break
+
+v.  The p-value is $2\Pr(Z_{n-1} >|z|)$. <br>
+
+\hfill\break
+
+vi. The 95% CI is $\bar{Y} \pm \frac{s_{y}}{\sqrt{n}} t_{n-1,\alpha/2}$.
+
+For 19 out of 20 different samples, an interval constructed in this way will include the true value of the mean, $\mu$. <br>
+
+\hfill\break
+
+vii. $z = (11-12)/(1/3) = -3$ and $2\Pr(Z_{8} >|z|) = .017$, so we do reject the null. <br> The 95% CI for $\mu$ is $11 \pm \frac{1}{3}2.3 = (10.23, 11.77)$.
+
+<br> <br>

_quarto.yml

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+project:
+  type: website
+  output-dir: docs
+website:
+  title: "VectorByte Training 2024"
+  navbar:
+    background:	"#630031"
+    foreground: "light"
+    logo: "graphics/vblogoedit.pdf"
+    logo-href: https://www.vectorbyte.org
+    left:
+      - href: index.qmd
+        text: Home
+      - href: about.qmd
+        text: About
+      - href: schedule2024.qmd
+        text: Schedule
+      - href: materials.qmd
+        text: Materials
+    right:
+      - icon: github
+        href: https://github.com/VectorByteOrg/vectorbyte-training2024
+      - icon: twitter
+        href: https://twitter.com/vectorbite_rcn
+
+format:
+  html:
+    theme: cosmo
+    css: styles.css
+    toc: true
+
+editor: visual
+

about.qmd

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+---
+title: "About"
+---
+
+About this site
+
+```{r}
+1 + 1
+```

index.qmd

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+---
+title: ""
+format:
+  html:
+    toc: false
+    toc-location: left
+    html-math-method: katex
+    css: styles.css
+    grid:
+      margin-width: 1000px
+---
+
+# Welcome to the 2024 VectorByte  Training Workshop!
+
+#### Check out our [about](about.html) and [schedule](schedule2024.html) pages to continue.
+
+<br> <br> <br>
+
+<center>![](graphics/VectorByte-logo_lg.png){width="4in"}</center>

materials.qmd

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+---
+title: "VectorByte Training Materials 2024"
+format:
+  html:
+    toc: true
+    toc-location: left
+    html-math-method: katex
+    css: styles.css
+---
+
+# Overview
+
+## Pre-work and set-up
+
+### Hardware and Software
+
+We will be using [`R`](https://cran.r-project.org/) for all data manipulation and analyses/model fitting. Any operating system (Windows, Mac, Linux) will do, as long as you have `R` (version 3.6 or higher) installed.
+
+You may use any IDE/ GUI for `R` (VScode, RStudio, Emacs, etc). For most people, [`RStudio`](https://www.rstudio.com/) is a good option. Whichever one you decide to use, please make sure it is installed and test it before the workshop. We will have a channel on Slack dedicated to software/hardware issues and troubleshooting.
+
+We will also be using Slack for additional support during the training. Please have these installed in advance.
+
+### Pre-requisites
+
+We are assuming familiarity with R basics. In addition, we recommend that you do the following:
+
+1.  Go to [The Multilingual Quantitative Biologist](https://mhasoba.github.io/TheMulQuaBio/intro.html), and read+work through the [Biological Computing in R Chapter](https://mhasoba.github.io/TheMulQuaBio/notebooks/07-R.html) up to the section on Writing R code. Of course, keep going if you want (although we will cover some similar materials here).
+
+2.  In addition / alternatively to pre-work element (1), here are some resources for brushing up on R [at the end of the Intro R Chapter you can try](https://mhasoba.github.io/TheMulQuaBio/notebooks/07-R.html#readings-and-resources). But there are many more resources online (e.g., [this](https://www.codecademy.com/learn/learn-r) and [this](https://www.dataquest.io/blog/learn-r-for-data-science/) ) -- pick something that suits your learning style.
+
+3.  Review background on [introductory probability and statistics](Stats_review.qmd) ([solutions to exercises](Stats_review_soln.qmd))
+
+4.  Inculcate the coding Jedi inside of you - or the Sith - whatever works.
+
+<br> <br>
+
+## Introduction to the VecTraits database[^1]
+
+[^1]: What is the difference between VectorBiTE and VectorByte? We are glad you asked! [VectorBiTE](http://vectorbite.org/) was an RCN or a research coordination network funded by a 5 year grant from the BBSRC. [VectorByte](https://www.vectorbyte.org/) is hosting this training which is a newly funded NSF grant to establish a global open access data platform to study disease vectors. All the databases have transitioned to VectorByte but the legacy options will still be available on the VectorBiTE website.
+
+-   This component will be delivered live & synchronously. The VecDyn website can be found [here](https://vectorbyte.crc.nd.edu/vecdyn-datasets). It might be an idea to explore this prior to the workshop.
+
+
+
+<br> <br>
+