Assume that ${{\mathbf{\text{σ}}}_{\mathbf{\text{1}}}}^{\mathbf{\text{2}}}{{\mathbf{\text{=σ}}}_{\mathbf{\text{2}}}}^{\mathbf{\text{2}}}{\mathbf{\text{=σ}}}^{\mathbf{\text{2}}}$. Calculate the pooled estimator ${\mathbf{\text{σ}}}^{\mathbf{\text{2}}}$ for each of the following cases:

a.${{\mathbf{\text{s}}}_{\mathbf{\text{1}}}}^{\mathbf{\text{2}}}{{\mathbf{\text{=120,s}}}_{\mathbf{\text{2}}}}^{\mathbf{\text{2}}}{\mathbf{\text{=100,n}}}_{\mathbf{\text{1}}}{\mathbf{\text{=n}}}_{\mathbf{\text{2}}}\mathbf{\text{=25}}$

b.${{\mathbf{\text{s}}}_{\mathbf{\text{1}}}}^{\mathbf{\text{2}}}{{\mathbf{\text{=12,s}}}_{\mathbf{\text{2}}}}^{\mathbf{\text{2}}}{\mathbf{\text{=20,n}}}_{\mathbf{\text{1}}}{\mathbf{\text{=20,n}}}_{\mathbf{\text{2}}}\mathbf{\text{=10}}$

c.${{\mathbf{\text{s}}}_{\mathbf{\text{1}}}}^{\mathbf{\text{2}}}{{\mathbf{\text{=.15,s}}}_{\mathbf{\text{2}}}}^{\mathbf{\text{2}}}{\mathbf{\text{=.20,n}}}_{\mathbf{\text{1}}}{\mathbf{\text{=6,n}}}_{\mathbf{\text{2}}}\mathbf{\text{=10}}$

d.${{\mathbf{\text{s}}}_{\mathbf{\text{1}}}}^{\mathbf{\text{2}}}{{\mathbf{\text{=3000,s}}}_{\mathbf{\text{2}}}}^{\mathbf{\text{2}}}{\mathbf{\text{=2500,n}}}_{\mathbf{\text{1}}}{\mathbf{\text{=16,n}}}_{\mathbf{\text{2}}}\mathbf{\text{=17}}$

Note that the pooled estimate is a weighted average of the sample variances. To which of the variances does the pooled estimate fall nearer in each of the above cases?

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:

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

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

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

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

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

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

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

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

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

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

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

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