Question

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.

Short answer

Option b is

SSE Error Mean Square $=40$

Option c is

The Output is

Option d is

As a result, draw the conclusion that the sample means differ significantly.

Step-by-step solution

    Given Information 

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,

$SST=112$

Step 2:      Given information Option b

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

    SSE Error Mean Square

$\mathrm{SSE}$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

$SSE=40$

    Given Information Option c

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:

    Given Information of Option d

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.