91Ó°ÊÓ

Sample one: $1,9$.

Sample two: $10,4,8,6,2$.

Sample three: $4,16,10$.

Based on the information provided,

$k=3$

$n_1=2,$

$n_2=5,$

$n_3=3$

$n=2+5+3$

$n=10$

Samples one, two, and three have the following mean values:

$\stackrel{¯}{x_1}=\frac{1+9}{2}$

$=5$

$\stackrel{¯}{x_2}=\frac{10+4+8+6+2}{5}$

$=5$

$\stackrel{¯}{x_3}=\frac{4+16+10}{3}$

$=10$

Calculate the mean

$\stackrel{¯}{x}=\frac{1+9+10+4+8+6+2+4+16+10}{10}$

$=7$

SSTR (sum of squares treatment)$SSTR=n_1{\left(\stackrel{¯}{x_1}-\stackrel{¯}{x}\right)}^2+n_2{\left(\stackrel{¯}{x_2}-\stackrel{¯}{x}\right)}^2+n_3{\left(\stackrel{¯}{x_3}-\stackrel{¯}{x}\right)}^2$

$=40$

Sample one: $1,9$.

Sample two: $10,4,8,6,2$,

Sample three: $4,16,10$.

We've $SSTR=40$

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

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

$=20$

Sample one: $1,9$.

Sample two: $10,4,8,6,2$.

Sample three: $4,16,10$.

The variance of each sample is $s_1^2=32,$

$s_2^2=10,$

$s_3^2=36$

$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$$=(2-1)\left(32\right)+(5-1)\left(10\right)+(3-1)\left(36\right)$

$=144$

Sample one: $1,9$.

Sample two: $10,4,8,6,2$.

Sample three: $4,16,10$.

We've $n=10,$

$k=3$

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

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

$=20.57$

Sample one: $1,9$.

Sample two: $10,4,8,6,2$.

Sample three: $4,16,10$.

We've $n=10,$

$k=3$

$F-\text{statistic}\frac{MSTR}{MSE}$

$=\frac{20}{20.57}$

$=0.97$