Question

We have provided data from independent simple random samples from several populations. In each case, determine the following items.

a. SSTR

b. MSTR

c. SSE

d. MSE

e. F

Short answer

The determined values are,

a).$SSTR=36$.

b).$MSTR=9$.

c).$SSE=52$.

d). $MSE=3.47$.

e). $F-\text{statistic}=2.60$.

Step-by-step solution

Part (a) Step 1: Given Information

Given data:

Part (a) Step 2: Explanation

Based on the information provided,

$k=5$

$n_1=4,$

$n_2=3,$

$n_3=5,$

$n_4=5,$

$n_5=3$

$n=4+3+5+5+3$

$n=20$

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

$\overline{x_1}=\frac{7+4+5+4}{4}$

$=5$

$\overline{x_2}=\frac{5+9+4}{3}$

$=6$

$\overline{x_3}=\frac{6+7+5+4+8}{5}$

$=6$

$\overline{x_4}=\frac{3+7+7+4+4}{5}$

$=5$

$\overline{x_5}=\frac{7+9+11}{3}$

$=9$

Part (a) Step 3: Explanation

Calculate the mean:

$\overline{x}=\frac{120}{20}$

$=6$

SSTR (sum of squares treatment)

$SSTR=n_1{\left(\overline{x_1}-\overline{x}\right)}^2+n_2{\left(\overline{x_2}-\overline{x}\right)}^2+n_3{\left(\overline{x_3}-\overline{x}\right)}^2+n_4{\left(\overline{x_4}-\overline{x}\right)}^2+n_5{\left(\overline{x_5}-\overline{x}\right)}^2$

$=4(4-5{)}^2+3(7-6{)}^2+5(3-6{)}^2+5(6-5{)}^2+3(6-9{)}^2$

$=36$

Part (b) Step 1: Given Information

Given data:

Part (b) Step 2: Explanation

We know,

$SSTR=36$

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

$=\frac{36}{5-1}$

$=9$

Part (c) Step 1: Given Information

Given data,

Part (c) Step 2: Explanation

Each sample's variance is

$s_1^2=2,$

$s_2^2=7,$

$s_3^2=2.5,$

$s_4^2=3.5,$

$s_5^2=4$

$SSE=\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2+\left(n_3-1\right)s_3^2$

$=(4-1)\left(2\right)+(3-1)\left(7\right)+(5-1)(2.5)+(5-1)3.5+(3-1)4$

$=52$

Part (d) Step 1: Given Information

Given data,

Part (d) Step 2: Explanation

We know

$n=20,$

$k=5$

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

$=\frac{52}{20-5}$

$=3.47$

Part (e) Step 1: Given Information

Given data:

Part (e) Step 2: Explanation

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

$=\frac{9}{3.47}$

$=2.60$