Question

To study the effect of temperature on yield in a chemical process, five batches were produced at each of three temperature levels.

The results follow.

Construct an analysis of variance table.

Use a $0.05$ level of significance to test whether the temperature level has an effect on the mean yield of the process.

Temperature

Temperature/ Batches	$50°C$	$60°C$	$70°C$
I	$34$	$30$	$23$
II	$24$	$31$	$28$
III	$36$	$34$	$28$
IV	$39$	$23$	$30$
V	$32$	$27$	$31$

Answer 1

Constructing the analysis of variance table:

Step-1:Given data and Assume hypothesis:

Number of groups $k=3$

Number of observation in group $50°C$, $n_1=5$

Number of observation in group $60°C$, $n_2=5$

Number of observation in group $70°C$, $n_3=5$

Total number of observation $N=15$

Null hypothesis $H_0$:

The mean yield for the three temperatures are equal.

Alternative hypothesis $H_1:$The mean yield for the three temperatures are not equal.

Step-2: Sum of the squares of between groups:

SSB$={\underset{i}{\sum n_i\left(\overline{X_i}-\overline{X}\right)}}^2$.

Here, $\overline{X_i}$ is the mean of the j^th group,$\overline{X}$ is the overall mean and $n_i$ is the sample size of the j^th group.

$\overline{X}=\frac{\overline{X_1}+\overline{X_2}+...+\overline{X_i}}{i}$

$X_1=34+24+36+39+32=165$

$X_2=30+31+34+23+27=145$

$X_3=23+28+28+30+31=140$

$\overline{X_i}=\frac{X_i}{n_i}$

$\Rightarrow \overline{X_1}=\frac{165}{5}=33$

$\Rightarrow \overline{X_2}=\frac{145}{5}=29$

$\Rightarrow \overline{X_3}=\frac{140}{5}=28$

$\Rightarrow \overline{X}=\frac{33+29+28}{3}=30$

$$\begin{gathered} \text{SSB}=n_1{\left(\overline{X_1}-\overline{X}\right)}^2+n_2{\left(\overline{X_2}-\overline{X}\right)}^2+n_3{\left(\overline{X_3}-\overline{X}\right)}^2 \\ =5\times {\left(33-30\right)}^2+5\times {\left(29-30\right)}^2+5\times {\left(28-30\right)}^2 \\ =5\times {\left(3\right)}^2+5\times {\left(1\right)}^2+5\times {\left(2\right)}^2 \\ =5\times 9+5\times 1+5\times 4 \\ =45+5+20 \\ =70 \end{gathered}$$

Step-3:Sum of the squares of Errors:

SSE$=\sum _i\sum _j{\left(x_j-\overline{X_i}\right)}^2$

$x_j-\overline{X_1}$	${\left(x_j-\overline{X_1}\right)}^2$	$x_j-\overline{X_2}$	${\left(x_j-\overline{X_2}\right)}^2$	$x_j-\overline{X_3}$	${\left(x_j-\overline{X_3}\right)}^2$
$34-33=1$	$1$	$30-29=1$	$1$	$23-28=-5$	$25$
$24-33=-9$	$81$	$31-29=2$	$4$	$28-28=0$	$0$
$36-33=3$	$9$	$34-29=5$	$25$	$28-28=0$	$0$
$39-33=6$	$36$	$23-29=-6$	$36$	$30-28=2$	$4$
$32-33=-1$	$1$	$27-29=-2$	$4$	$31-28=3$	$9$
	$128$		$70$		$38$

SSE$=128+70+38=236$

Step-4: Anova table:

Source of variation	Sum of squares	Degrees of freedom	Mean square	$F=\frac{MSB}{MSE}$
Between groups	SSB$=70$	$d{f}_1=k-1=2$	MSB$=\frac{SSB}{d{f}_1}=\frac{70}{2}=35$	$F=\frac{35}{19.67}=1.7794$
Error	SSE$=236$	$d{f}_2=N-k=12$	MSE$=\frac{SSE}{d{f}_2}=\frac{236}{2}=19.67$
Total	$306$

Given, $0.05$ level of significance. i.e, $F(0.05,2,12)=3.8853$ Tabulated value.

As $1.7794<3.8853$, tabulated value is higher than calculated value.

So the null hypothesis is accepted.

Hence, The mean yield for the three temperatures are equal.