Question

$\begin{array}{ccc}\text{Sample 1}& \text{Sample 2}& \text{Sample 3}\\ 1& 10& 4\\ 9& 4& 16\\ & 8& 10\\ & 6& \\ & 2& \end{array}$

we provide data from independent simple random samples from several populations. In each case,

a. compute SST, SSTR, and SSE by using the computing formulas given in Formula 13.l on page 535

b. compare your results in part (a) for SSTR and 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

(a) The value of

$SSC=210$

$SSTR=138$

$SSE=72$

(b) On both portions, the values of SST, SSTR, and SSE are the same

(c) $\begin{array}{|lllll|}\hline \text{Source}& \text{df}& \text{SS}& \text{MS}& F\\ \text{Treatment}& 3& 138& 46& 5.11\\ \text{Error}& 8& 72& 9& \\ \text{Total}& 11& 210& & \\ \hline\end{array}$

(d)At the $5\%$level, the statistics on the number of $4$population samples provide sufficient information to determine that a difference exists in the mean number of $4$population samples.

Step-by-step solution

Part (a)Step 1: Given information

Given in the question that, data from independent simple random samples from several populations.

We need to compute SST, SSTR, and SSE by using the computing formulas given in Formula 13.l on page 535.

Part(a) Step 2: Explanation

There are four different samples with a total of 12 distinct terms. So $k=4$and $n=12$in this data.

$\mathrm{SST}=\sum {\mathrm{x}}_{\mathrm{i}}^2-\left(\sum {\mathrm{x}}_{\mathrm{i}}\right)2/\mathrm{n}$

$SSTR=\sum T_j^2/n_j-\left(\sum x_i\right)2/n$

$SSE=SST-SSTR$

SST

$\sum {\mathrm{x}}_{\mathrm{i}}^2=798$

$\left(\sum x_{\mathrm{i}}\right)2/n=588$

$798-588$

$210$

$SSTR$

$\sum T_j^2/n_j=(192+75+432+27)=726$

$\left(\sum x_i\right)2/n=588$

$726-588$

$138$

$SSE=SST-SSTR$

$210-138$

$72$

Part(b) Step 1: Given information

Given in the question that, we provide data from independent simple random samples from several populations.

We need to compare your results in part (a) for SSTR and SSE with those you obtained in Exercises 13.24-13.29,

Part (b) Step 2: Explanation

There are four separate samples with a total of 12 different terms. So $k=4$and $n=12$in this data. SST, SSTR, and SSE values from component a.

When comparing the result to the exercise $13.24-13.29$, the result is $13.24-13.29$.

On both portions, the values of SST, SSTR, and SSE are the same.

Part(c) Step 1: Given information

Given in the question that, we provide data from independent simple random samples from several populations.

We need to construct a one-way ANOVA table.

Part(c) Step 2: Explanation

There are four separate samples with a total of 12 different terms. So$k=4$ and $n=12$in this data.

$\mathrm{SST}=\sum {\mathrm{x}}_{\mathrm{i}}^2-\left(\sum {\mathrm{x}}_{\mathrm{i}}\right)2/\mathrm{n}$

$SSTR=\sum T_j^2/n_j-\left(\sum x_i\right)2/n$

$SSE=SST-SSTR$

$MSTR=SSTR/k-1$

$MSE=SSE/n-k$

Let's find the SST$SST$

$\sum x_i^2=798$

$\left(\sum x_{\mathrm{i}}\right)2/n=588$

$798-588$

$210$

Then we have to find$SSTR$

$\sum T_j^2/n_j=(192+75+432+27)=726$

$\left(\sum x_i\right)2/n=588$

$726-588$

138

$\mathrm{SSE}=\mathrm{SST}-SSTR$

$210-138$

$72$

$MSTR=SSTR/k-1$

$138/3$

$46$

$MSE=SSE/n-k$

• $72/8$
• $9$

$F=MSTR/MSE$

• $46/9$
• $5.11$

Part(d) Step 1: Given information

Given in the question that, we provide data from independent simple random samples from several populations.

We need to find that 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 at the$5\%$significance level.

Part(d) Step 2: Explanation

${\mathrm{H}}_{\mathrm{o}}:{\mu }_1={\mu }_2={\mu }_3={\mu }_4$

${\mathrm{H}}_{\mathrm{a}}:$Not all the mean are equal

The worth of$T1=24,T2=15,T3=36$and$T4=9$

$\sum x^2{}_i=7056$

$F=5.11$

$Criticalvalue=4.07$

$\mathrm{F}>$Critical value

In contrast, the$0.025<\mathrm{P},0.05$

Reject${\mathrm{H}}_{\mathrm{o}}$

The data on the number of $4$population samples provides adequate information to draw the conclusion that there is a difference in the mean number of$4$population samples.