b. compare your results in part (a) for $\mathrm{SSTR}$and $\mathrm{SSE}$with those you obtained in Exercises $13.24-13.29$, where you employed the defining formulas.

c. construct a one-way $ANOVA$table.

d. decide, at the $5\%$significance level, whether the data provide sufficient evidence to conclude that the means of the populations from which the samples were drawn are not all the same.

The goal is to use the following formulas to compute $\mathrm{SST},\mathrm{SSTR}$, and $\mathrm{SSE}$:

$SST=\sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2$denotes the entire sum of squares.

Where,

$\overline{X}=\frac{{\displaystyle \sum _{i=1}^n}{X}_i}{n}$

Calculate the average:

$$\begin{gathered} \overline{X}=\frac{{\displaystyle \sum _{i=1}^n}{X}_i}{n} \\ =\frac{(4+8+9+...+2+9)}{15} \\ =\frac{75}{15} \\ =5 \end{gathered}$$

Determine the SST,

$$\begin{gathered} SST=\sum _{i=1}^n{(X_i-\overline{X})}^2 \\ ={(4-5)}^2+{(8-5)}^2+{(9-5)}^2+...+{(2-5)}^2+{(9-5)}^2 \\ =112 \end{gathered}$$

Thus,

$SST=112$

b.

Determine the treatment mean square $\left(\mathrm{SSTR}\right)$:

$SSTR=\sum _{i=1}^n{n}_i{({\overline{X}}_i-\overline{X})}^2$

$=n_1{\left({\overline{X}}_1-\overline{X}\right)}^2+n_2{\left({\overline{X}}_2-\overline{X}\right)}^2+n_3{\left({\overline{X}}_3-\overline{X}\right)}^2+n_4{\left({\overline{X}}_4-\overline{X}\right)}^2+n_5\left({\overline{X}}_5-\overline{X}\right)$

localid="1653227695948" $$\begin{gathered} =3{(3-5)}^2+3{(6-5)}^2+3{(8-5)}^2+3{(2-5)}^2+3{(6-5)}^2=3[4+1+9+9+1] \\ =3\times 24 \\ =72 \\ SSTR=72 \end{gathered}$$

$\mathrm{SSE}$Error Mean Square $=\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2+\left(n_3-1\right)s_3^2+\left(n_4-1\right)s_4^2$

The following is the generic formula for variance $S_i^2$

$s_i^2=\frac{1}{n_i-1}\sum _{i=1}^n{\left(X_i-{\overline{X}}_i\right)}^2$

The following is the MINITAB output for variances:

Therefore,

$$\begin{gathered} SSE==\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2+\left(n_3-1\right)s_3^2+\left(n_4-1\right)s_4^2+(n_5-1)s_5^2 \\ =(3-1)1+(3-1)3+(3-1)3+(3-1)4+(3-1)9 \\ =2\times 20 \\ =40 \end{gathered}$$

Hence

$SSE=40$

c.

There is no significant difference in the sample means, according to ${\mathrm{H}}_0$.

There is a significant discrepancy in the sample means ${\mathrm{H}}_1$.

The following is the technique for performing One-Way Analysis of Variance:

1) Open the $\mathrm{MINITAB}$sheet and import the data.

2) Select $\mathrm{Start}\to \mathrm{ANOVA}\to \mathrm{One}-\mathrm{way}$(un-stacked),

3) In the$\mathrm{Response}\mathrm{text}\mathrm{box}$, type $\mathrm{RESPONSE}$.

4) Enter $95$as the confidence level.

5) Finally, press the $\mathrm{OK}$button.

The following is the output:

d.

$F=4.50$is the determined value. $p=0.024$is the value. As a result, $\mathrm{p}$has a value less than$alpha=0.05$. Therefore, the Null hypothesis must be rejected. As a result, draw the conclusion that the sample means differ significantly.