Consider the following hypothetical samples

a. Obtain the sample mean and sample variance of each of the three samples.

b. Obtain SST, SSTR and SSE by using the defining formulas and verify that the one-way ANOVA identity holds.

c. Obtain SST, SSTR and SSE by using the computing formulas.

d. Construct the one-way ANOVA table.

Given,

From the given data

Number of samples \(k=3\)

Number of values in each sample \(n_{1}=3, n_{2}=5, n_{3}=4\)

Sum of values of each sample \(T_{1}=9, T_{2}=15, T_{3}=24\)

Find the mean of each sample

\(\bar{x_{A}}=\frac{T_{1}}{n_{1}}=\frac{9}{3}=3\)

\(\bar{x_{B}}=\frac{T_{2}}{n_{2}}=\frac{15}{5}=3\)

\(\bar{x_{C}}=\frac{T_{3}}{n_{3}}=\frac{24}{4}=6\)

Find the sample variance

\(S_{A}^{2}=\sum \frac{(x_{i}-\bar{x})}{n-1}=2\)

\(S_{B}^{2}=\sum \frac{(x_{i}-\bar{x})}{n-1}=2.449\)

\(S_{C}^{2}=\sum \frac{(x_{i}-\bar{x})}{n-1}=4.243\)

Hence,

For the three samples, the sample means are \(3,3,6\) and sample variance of the are \(2,2.45, 4.24\).

From the given data

The overall mean is

\(\bar{x}=\frac{\sum x_{i}}{n}\)

\(=\frac{9+15++24}{12}\)

\(=4\)

Then, the total sum of squares is

\(SST=\sum (\bar{x_{i}-\bar{x}})^{2}\)

\(=(1-4)^{2}+(3-4)^{2}+..+(3-4)^{2}\)

\(=110\)

The treatment sum of squares is

\(SSTR=\sum n_{i}(\bar{x_{i}-\bar{x}})^{2}\)

\(=3(3-4)^{2}+5(3-4)^{2}+4(6-4)^{2}\)

\(24\)

The error sum of squares is

\(SSE=\sum (n_{i}-1)(s_{i})^{2}\)

\(=(3-1)(2)^{2}+(5-1)(2.4)^{2}+(4-1)(4.2)^{2}\)

\(=86\)

\(SST=SSTR+SSE\)

\(=24+86\)

\(=110\)

Hence,

\(SST=110, SSTR=24, SSE=86\)

From the given data

The overall mean is

\(\bar{x}=\frac{\sum x_{i}}{n}\)

\(=\frac{9+15++24}{12}\)

\(=4\)

Then, the total sum of squares is

\(SST=\sum (\bar{x_{i}-\bar{x}})^{2}\)

\(=(1-4)^{2}+(3-4)^{2}+..+(3-4)^{2}\)

\(=110\)

The treatment sum of squares is

\(SSTR=\sum n_{i}(\bar{x_{i}-\bar{x}})^{2}\)

\(=3(3-4)^{2}+5(3-4)^{2}+4(6-4)^{2}\)

\(24\)

The error sum of squares is

\(SSE=\sum (n_{i}-1)(s_{i})^{2}\)

\(=(3-1)(2)^{2}+(5-1)(2.4)^{2}+(4-1)(4.2)^{2}\)

\(=86\)

Hence,

\(SST=110, SSTR=24, SSE=86\)

From the given data

The mean treatment of sum of squares is

\(MSTR=\frac{SSTR}{k-1}\)

\(=\frac{24}{3-1}\)

\(=12\)

Then, the mean error of sum of squares is

\(MSE=\frac{SSE}{n-k}\)

\(=\frac{86}{12-3}\)

\(=9.556\)

The \(F-\)static is

\(F-static=\frac{MSTR}{MSE}\)

\(=\frac{12}{9.556}\)

\(=1.26\)

Hence, the ANOVA table is