# Homework Assignment 1. Show that SSTOT/a2 N x2-1 by using an argum...

## Transcribed Text

Homework Assignment 1. Show that SSTOT/a2 N x2-1 by using an argument similar to that used in the lectures to find distributions for SSE/a² and SST/a². 2. A test statistic that could be used to test for a significant pairwise differences between the i-th and j-th treatment is Di, = SSE/(n v) The null hypothesis is Ho T; = Tjy Find the distribution of Dis under the null hypothesis. (For your information, R uses the square root of this statistic to get the p-values reported in the lameans output) 3. Recall the soap experiment from Homework 1. Look back at Homework 1 for an explanation of the experiment. The data are the weight lost over 24 hours by different types of soap. Cube Regular Deodorant Moisturizing 1 -0.30 2.63 1.86 2 -0.10 2.61 2.03 3 -0.14 2.41 2.26 4 0.40 3.15 1.82 (a) Construct the ANOVA table for this experiment by hand. Show the calculations needed to construct the quantities in the ANOVA table. (b) Test the null hypothesis that there is no difference in mean weight lost between different soap types. Report the test statistic, the p-value of the statistic under the null hypothesis, and interpret the result of the test. You may use R or an online p-value calculator (like http://graphpad.com/quickeales/ PValuel.cfm) to compute the p-value. Provide the R code used, with output, or clearly describe the online p-value calculator you used. 8

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## Question 1

The variation in the data {$Y_{it}$} around the ground mean $\bar{Y}$ is
given by:

$W^2_i = \frac{1}{n_i-1}\sum^{n_i}_{j=1}(Y_{ij}-\bar{Y})^2$

As, we know that $\frac{(n_i-1)W^2_i}{\sigma^2}$ follows a chi-square
distribution. And it is also known that adding a bunch of independent
chi-square random variable will be a chi-square random variable.

$\sum^v_{i=1}\frac{(n_i-1)W^2_i}{\sigma^2}$ = $\frac{\sum^v_{i=1}\sum^r_{t=1}(Y_{ij}-\bar{Y})^2}{\sigma^2}$
= $\frac{SSTOT}{\sigma^2}$

Since, it is assumed that $Y_{ij}$ are independent observations, so adding those chi-square makes
$\frac{SSTOT}{\sigma^2}$ also a chi-square distribution with n-1 degree of freedom...

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