# Testing null hypothesis using ANOVA statistic

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?

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 $$F^* the $$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.

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