Question

The data from independent simple random samples from several populations are given.

a. Compute SST, SSTR, and SSE by using the computing formulas.

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 values of $SST=184$

$SSTR=40$

$SSE=144$

(b) Both outcomes are the same.

(c)

(d) The statistics do not give enough information to indicate that the populations from which the samples were collected have different means.

Step-by-step solution

Part (a) Step 1: Given information

The given values are

Part (a) Step 2: Explanation

From the given,

$k=3$

$n=\sum n_j$

$=2+5+3$

$=10$

$\sum x_i=\sum T_j$

$=10+30+30$

$=70$

The sum of the squares of all the data in the table above is

$\sum x_i^2=(1{)}^2+(9{)}^2+\dots \dots ..+(16{)}^2+(10{)}^2=674$

Then,

$SST=\sum x_i^2-\frac{{\left(\sum x_i\right)}^2}{n}$

$=674-\frac{(70{)}^2}{10}$

$=674-490$

$SST=184.$

$SSTR=\sum \frac{T_j^2}{n_j}-\frac{{\left(\sum x_j\right)}^2}{n}$

$=\frac{(10{)}^2}{2}+\frac{(30{)}^2}{5}+\frac{(30{)}^2}{3}-\frac{(70{)}^2}{10}$

$=530-490$

$SSTR=40$

$SSE=SST-SSTR$

$=184-40$

$SSE=144$

Part (b) Step 1: Given information

The given values are

Part (b) Step 2: Explanation

Both outcomes are the same.

Despite using multiple versions of computations, we get the same results.

Part (c) Step 1: Given information

The given values are

Part (c) Step 2: Explanation

Using the given values

Find the treatment mean square

$MSTR=\frac{SSTR}{k-1}$

$=\frac{40}{3-1}$

$MSTR=20$

The error mean square is

$MSE=\frac{SSE}{n-k}$

$=\frac{144}{10-3}$

$MSE=20.57$

The value of the F-statistic is

$F=\frac{MSTR}{MSE}$

$=\frac{20}{20.57}$

$F=0.97$

Therefore the one-way ANOVA table is

Part (d) Step 1: Given information

The given values are

Part (d) Step 2: Explanation

The null and alternate hypotheses are

$H_0:{\mu }_1={\mu }_2={\mu }_3$

$H_1:$Not all the means are equal

We'll run the test at a 5% significance level, so . We'll look at three populations, or $k=3$, and ten observations, or $n=10$.

As a result, the F-degrees statistics of freedom are

$df=(k-1,n-K)$

$=(3-1,10-3)$

$df=2,7$

Table VIII shows that the crucial value at the $5\%$level of significance is $F_{0.05}=4.74$.

$P>0.10$is found in table VIII with $df=(2,7)$.

We do not reject $H_0$because the P-value is bigger than the significance level.

The statistics do not give enough information to indicate that the populations from which the samples were collected have different means.