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Q. The temperature in degree Celsius at which a particular chemical reaction takes place was recorded for 5 different catalysts. The refining company had the test repeated at each of its 3 refineries to be sure that the reaction took place at all of them. Use a $10 \%$ significance level to determine if

(a) The average temperature is the same for all catalysts

(b) The location causes a refinery effect to be present.

\begin{array}{|cccccc|} \hline \text{Refinary} &&& \text{Catalysts}&\\ \hline & 1 & 2 & 3 & 4 & 5 \\ \hline \text{US} & 66 & 58 & 70 & 64 & 68 \\ \hline \text{Canada} & 71 & 74 & 75 & 69 & 69 \\ \hline \text{Mexico} & 54 & 60 & 62 & 59 & 67 \\ \hline \end{array}

I hope that the statement in part (a) can be tested using ANOVA test statistic $F=\frac{(N-r)\sum n_j(\bar x_j- \bar x)^2}{(r-1)(\sum_{i=1}^r \sum_{j=1}^{n_j}(x_{ij}- \bar x_j)^2},$ where $N$ is the total sample size and $r$ is the number of groups (Catalysts). But, how the part (b) is working?

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Here is the standard technique for performing one factor ANOVA analysis. I'll demonstrate (a). Please double check my calculations.

We shall build the following ANOVA table.

\begin{array}{|cccccc|} \hline \hline \text{Catalyst} & 1 & 2 & 3 & 4 & 5 \\ \hline \text{Sample Mean} & 191/3 & 64 & 69 & 64 & 68 \\ \hline \text{Sample Variance} & 229/3 & 76 & 43 & 25 & 1 \\ \hline \end{array}

Identify the following preliminary information:

$$J=\text{# of groups}=5 \\n=\text{# of data points in each group} = 3 \\\text{df}_{\text{b/w}}=J-1=4 \\ \text{df}_{\text{w/in}}=J(n-1)=10 \\ \bar{x}_{\bullet}=\text{grand mean}=\frac{986}{15}$$ Let's first calculate the sum of squares between groups using our table.

$$\begin{eqnarray*}\text{SS}_{\text{b/w}}&=&\sum_{j=1}^Jn(\bar{x}_j-\bar{x}_{\bullet})^2 \\&=&3\Bigg[(191/3-229/3)^2+\dots +(68-229/3)^2\Bigg] \\ &=& \frac{1174}{15}\end{eqnarray*}$$

Now let's calculate the sum of squares within groups using our table. $$\begin{eqnarray*}\text{SS}_{\text{w/in}}&=&\sum_{j=1}^J(n-1)S_j^2 \\ &=& 2\Bigg[229/3+\dots +1\Bigg] \\ &=& \frac{1328}{3}\end{eqnarray*}$$ Your test statistic is $F^*$ is $\frac{\text{SS}_{\text{b/w}}/\text{df}_{\text{b/w}}}{\text{SS}_{\text{w/in}}/\text{df}_{\text{w/in}}}=\frac{587}{1328}\approx 0.442$ while the critical value for this hypothesis test $F_{(4,10)}$ is approximately $2.61$. Since $F^*<F_{(4,10)}$ the $p-$value for $(a)$ exceeds $0.10$ so we don't reject the hypothesis that the average temperature is the same for all the catalysts.

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  • $\begingroup$ @ Mathew Pilling Thanks for your very clear demonstration of part (a), and what does (b) mean exaclty? $\endgroup$
    – Riaz
    Commented Nov 17, 2021 at 4:11
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    $\begingroup$ Perform the same exact produce on the $J=3$ groups for US, Mexico, and Canada. $\endgroup$
    – user801306
    Commented Nov 17, 2021 at 4:16
  • $\begingroup$ @ Mathew Pilling Could you pls explain how $J=3$ is connected with the part (b)? $\endgroup$
    – Riaz
    Commented Nov 17, 2021 at 7:02

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