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Answer 1

Ans. The given problem is studied under the Analysis of variance(ANOVA), i.e., we need to calculate the variation between and within the groups.

$A$ $(x_{1})$	$\mathbf{A}$ $x_{1}^{2}$	$\mathbf{B}$ $(x_{2})$	$x_{2}^{2}$	$\mathbf{C}$ $(x_{3})$	$x_{3}^{2}$	$\mathbf{D}$ $(x_{4})$	$x_{4}^{2}$
3	9	4	16	4	16	4	16
1	1	1	1	4	16	2	4
4	16	3	9	0	0	4	16
2	4	2	4	5	25	4	16
0	0	5	25	6	36	5	25
5	25	0	0	4	16	3	9
3	9	0	0	3	9	3	9
5	25	4	16	0	0	2	4
1	1	1	1	5	25	2	4
$\Sigma x_{1}=24$	$\mathbf{\Sigma x_{1}^{2}=90}$	$\Sigma x_{2}=20$	$\mathbf{\Sigma x_{2}^{2}=72}$	$\Sigma x_{3}=31$	$\mathbf{\Sigma x_{3}^{2}=143}$	$\Sigma x_{4}=29$	$\mathbf{\Sigma x_{4}^{2}=103}$
$\mathbf{\bar{x_{1}}=2.6667}$		$\mathbf{\bar{x_{2}}=2.2222}$		$\mathbf{\bar{x_{3}}=3.4444}$		$\mathbf{\bar{x_{4}}=3.2222}$
$\mathbf{\bar{x_{t}}=2.8889}$

$\bar{x_{t}}: Total \:mean$

$\bar{x_{1}}, \bar{x_{2}},\bar{x_{3}},\bar{x_{4}}: Means \:of \:different \:groups$

As we know that ANOVA is used to study the variation between and within the groups.

$F=\frac{Difference\:between\:groups}{Difference\:within\:groups}$

$F=\frac{MS_{b}}{MS_{w}}$

where, $MS_{w}=\frac{SS_{w}}{df_{w}},\:\:\:MS_{b}=\frac{SS_{b}}{df_{b}}$

$MS_{b}:$ mean square between groups

$MS_{w}:$ mean square within groups.

$df_{w}:$ degree of freedom for within group

$df_{b}:$degree of freedom for between group.

Calculation for $\mathbf{SS_{b}:}$

$\mathbf{SS_{b}=\Sigma N_{k}(\bar{x_{k}}-\bar{x_{t}})^{2}}$

$SS_{b}=N_{1}(\bar{x_{1}}-\bar{x_{t}})^{2}+N_{2}(\bar{x_{2}}-\bar{x_{t}})^{2}+N_{3}(\bar{x_{3}}-\bar{x_{t}})^{2}+N_{4}(\bar{x_{4}}-\bar{x_{t}})^{2}$

$SS_{b}=9(2.6667-2.8889)^{2}+9(2.2222-2.8889)^{2}+9(3.4444-2.8889)^{2}+9(3.2222-2.8889)^{2}$

$SS_{b}=0.4444+4.0004+2.7772+0.9998$

$\mathbf{SS_{b}=8.2218}$

Calculation for $\mathbf{SS_{T}}\::$

$\mathbf{SS_{T}=(\Sigma x_{k}^{2})-N\bar{x}_{t}^{2}}$

$SS_{T=\Sigma x_{1}^{2}+\Sigma x_{2}^{2}+\Sigma x_{3}^{2}+\Sigma x_{4}^{2}-N\bar{x}_{t}^{2}$                    

N=4*9(no. of groups*no of dataset in each group)=36

$SS_{T}=90+72+143+103-36(2.8889)^{2}$

$SS_{T}=408-300.4468$

$SS_{T}=107.5532$

$SS_{w}=SS_{T}-SS_{b}$

we have,

$SS_{T}=107.5532,\:\:SS_{b}=8.2218$

$SS_{w}=107.5532-8.2218$

$SS_{w}=99.3314$

Calculation for degree of freedom $\mathbf{(df):}$

$df_{w}=N-k=36-4=32$

$df_{b}=k-1=4-1=3$

$N:$ number of cases in dataset(i.e., 36=4*9 in our case)

$k:$ number of groups(i.e., 4 in our case)

$MS_{w}=\frac{SS_{w}}{df_{w}}$ ,   $MS_{b}=\frac{SS_{b}}{df_{b}}$

since we have calculated all the desired value:

$MS_{w}=\frac{SS_{w}}{df_{w}}=\frac{99.3314}{32}=3.10410625\approx 3.1041$

$MS_{w}=3.1041\approx 3.10$

$MS_{b}=\frac{SS_{b}}{df_{b}}=\frac{8.2218}{3}=2.7406$

$MS_{b}=2.7406\approx 2.74$

So now we can calculate the value of $\mathbf{F:}$

$F=\frac{MS_{b}}{MS_{w}}$

$F=\frac{MS_{b}}{MS_{w}}=\frac{2.74}{3.10}$

$F=0.8839\approx 0.88$

$F=0.88$

The table for the calculated values:

Sources of variation	SS	df	MS	F
Between group	8.2218 $\approx$ 8.22	3	2.74	$0.8839\approx 0.88$
Within group(error)	99.3314 $\approx$ 99.33	32	3.10
Total	107.5532 $\approx$107.55	35