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The pooled estimatoris an estimate obtained by combining information from two or more independent samples taken from populations believed to have the same mean. The pooled variance is a way to estimate common variance.

The formula to find pooled estimator of the variance of two samples is:

${{\mathbf{\text{s}}}_{\mathbf{\text{p}}}}^{\mathbf{\text{2}}}\mathbf{\text{=}}\frac{\mathbf{(}{\mathbf{\text{n}}}_{\mathbf{\text{1}}}\mathbf{鈭�}\mathbf{\text{1}}\mathbf{)}{{\mathbf{\text{s}}}_{\mathbf{\text{1}}}}^{\mathbf{\text{2}}}\mathbf{\text{+}}\mathbf{(}{\mathbf{\text{n}}}_{\mathbf{\text{2}}}\mathbf{鈭�}\mathbf{\text{1}}\mathbf{)}{{\mathbf{\text{s}}}_{\mathbf{\text{2}}}}^{\mathbf{\text{2}}}}{{\mathbf{\text{n}}}_{\mathbf{\text{1}}}{\mathbf{\text{+n}}}_{\mathbf{\text{2}}}\mathbf{鈭�}\mathbf{\text{2}}}$

It is given that ${s_1}^2=120,{s_2}^2=100,\text{and}{n}_1=n_2=25$

$\begin{array}{l}{s_p}^2=\frac{(25鈭�1)120+(25鈭�1)100}{25+25鈭�2}\\ =\frac{2880+2400}{48}\\ =110\end{array}$

Therefore, the pooled estimate of variance lies between ${s_1}^2=120\text{and}{s_2}^2=100$.

It is given that ${s_1}^2=12,{s_2}^2=20,n_1=20,\text{and}{n}_2=10$

$\begin{array}{l}{s_p}^2=\frac{(20鈭�1)12+(10鈭�1)20}{20+10鈭�2}\\ =\frac{228+180}{28}\\ =14.57\end{array}$

Therefore, the pooled estimate of variance lies between${s_1}^2=12\text{and}{s_2}^2=20$ .

It is given that${s_1}^2=0.15,{s_2}^2=\text{0}.20,n_1=6,\text{and}{n}_2=10$

$\begin{array}{l}{s_p}^2=\frac{(6鈭�1)0.15+(10鈭�1)0.20}{6+10鈭�2}\\ =\frac{0.75+1.8}{14}\\ =0.18\end{array}$

Therefore, the pooled estimate of variance lies between${s_1}^2=0.15\text{and}{s_2}^2=\text{0}.20$.

It is given that${s_1}^2=3000,{s_2}^2=2500,n_1=16,\text{and}{n}_2=17$

$\begin{array}{l}{s_p}^2=\frac{(16鈭�1)3000+(17鈭�1)2500}{16+17鈭�2}\\ =\frac{45000+40000}{31}\\ =2742\end{array}$

Therefore, the pooled estimate of variance lies between${s_1}^2=3000\text{and}{s_2}^2=2500$