Problem 1

The insulating life of protective fluids at an accelerated load is being studied. The experiment has been performed for four types of fluids, with $n = 5$ trials per fluid type. Suppose fluid types 1 and 2 are from manufacturer A, and that fluid types 3 and 4 are from manufacturer B. The data is available on Blackboard as an Excel File.

(a)

(i)

\fbox{\begin{minipage}{\textwidth} Perform a test of the global null hypothesis $H_0 : \mu_1 = \mu_2 = \mu_3 = \mu_4$. Compute the $F_0$ statistic, and the $p$-value. \end{minipage}}

This is a one-factor experiment with $a=4$ levels of the factor (fluid level) and $n=6$ replicates for a total of $N = a n = 24$ observations.

The test statistic

under the null hypothesis $\mu_1 = \ldots = \mu_a$ (or $\tau_1 = \ldots = \tau_a = 0$ in the effects model) has a reference distribution

where $N = n a$.

We compute the observed test statistic and its $p$-value with:

library("readxl")
h4.data = read_excel("handout2data.xlsx")
fluid = as.factor(na.omit(h4.data$fluid))
life = na.omit(h4.data$life)
#head(data.frame(fluid=fluid,life=life))

aov.mod = aov(life ~ fluid)
summary(aov.mod)
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##             Df Sum Sq Mean Sq F value Pr(>F)  
## fluid        3  30.16   10.05   3.047 0.0525 .
## Residuals   20  65.99    3.30                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
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We see that $F_0 = 3.047$ which has a $p$-value $.0525$.

(ii)

\fbox{\begin{minipage}{\textwidth} Comment on the additional information provided by the $p$-value, beyond a determination of statistical significance alone. \end{minipage}}

The $p$-value quantifies a measure of evidence beyond a determination of statistical significance.

Earlier, we obtained a $p$-value of $.0525$. If we were to perform a dichotomous hypothesis test which specifies a significance level $\alpha = 0.05$, we would decide $H_0$. However, had we obtained a $p$-value of $.049$, while the strength of the evidence is nearly the same, we would decide $H_A$. Presenting the $p$-value provides more information.

(b)

Consider the orthogonal contrasts \begin{align*} \Gamma_1 &= \mu_1 - \mu_2\ \Gamma_2 &= \mu_3 - \mu_4\ \Gamma_3 &= (\mu_1 + \mu_2) - (\mu_3 + \mu_4). \end{align*}

Preliminary analysis

A \emph{constrast} $\Gamma$ is a linear combination of parameters

such that $c_1 + \cdots + c_a = 0$. Assuming a balanced design,

and thus

under $H_0$. The $F$-test may also be used, where $F_0 = t_0^2 = \frac{\msc}{\mse} = \frac{\ssc/1}{\mse}$ where

We have $a=4$ fluid types, $1$, $2$, $3$, and $4$. Fluid types $1$ and $2$ are from manufacturer $A$ and fluid types $3$ and $4$ are from manufacturer $B$.

$\Gamma_1$ compares the average effect (on lifetime) of fluid $1$ with the average effect of fluid $2$, $\Gamma_2$ compares the average effect of fluid $3$ with the average effect of fluid $4$, and $\Gamma_3$ compares the average effect of manufacturer $A$ with the average effect of manufacturer $B$.

(i)

\fbox{\begin{minipage}{\textwidth} Compute SSC for each contrast. Describe a general property for the sums of squares of orthogonal contrasts. Why is this property desirable? \end{minipage}}

Orthogonal contrasts

Orthogonality is desirable since the treatment sum of squares can be decomposed into specific effects.

Additional remarks:

Two contrasts (assuming a balanced design) with coefficients ${c_i}$ and ${d_i}$ are orthogonal if $\sum_{i=1}^{a} c_i d_i = 0$.

For $a$ factor levels (or treatments), a set of $(a-1)$ orthogonal contrasts $\Gamma_1,\ldots,\Gamma_{a-1}$ are independent with $\operatorname{df} = 1$ and thus tests performed on them are independent.

We see that $\Gamma_1$, $\Gamma_2$, and $\Gamma_3$ are orthogonal contrasts.

Computing $\ssc$