Question

If n₁, n₂ are the sizes, x̅₁, x̅₂ the means, σ₁, σ₂ the standard deviations, the variance of the combined series is (where d₁ = x̅₁ – x̅ and d₂ = x̅₂ – x̅)       
1. \(\left( {\frac{1}{{{n_1}}} + \frac{1}{{{n_2}}}} \right)\left( {{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)} \right)\)
2. \(\frac{1}{{{n_1} + {n_2}}}\left( {{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)} \right)\)
3. \(\left( {\frac{1}{{{n_1}}} - \frac{1}{{{n_2}}}} \right)\left( {{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)} \right)\)
4. \(\frac{1}{{{n_1} - {n_2}}}\left( {{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)} \right)\)

Answer 1

Correct Answer - Option 2 : \(\frac{1}{{{n_1} + {n_2}}}\left( {{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)} \right)\)

Explanation:

Suppose the values in the first group are x1i, i = 1,2,……,n1 and those in the second group are x2i, i = 1,2,……..,n2.

Now since x̅1 and x̅2 are the means so, x̅1=\(\frac{1}{{{n_1}}}\mathop \sum \limits_{i = 1}^{{n_1}} {x_{1i}}\) and  x̅2 =\(\frac{1}{{{n_2}}}\mathop \sum \limits_{i = 1}^{{n_2}} {x_{2i}}\)  

If the mean of the combined series is x̅, then x̅ = \(\frac{{{n_1}\overline {{x_1}} + {n_2}\overline {{x_2}} }}{{{n_1} + {n_2}}}\)

Again σ1 and σ2 are the standard deviations. So, variance \({\rm{\sigma }}_1^2 = \frac{1}{{{n_1}}}\mathop \sum \limits_{i = 1}^{{n_1}} {\left( {{x_{1i}} - {{{\rm{\bar x}}}_1}{\rm{\;}}} \right)^2}\;\) and \({\rm{\sigma }}_2^2 = \frac{1}{{{n_2}}}\mathop \sum \limits_{i = 1}^{{n_2}} {\left( {{x_{2i}} - {{{\rm{\bar x}}}_2}{\rm{\;}}} \right)^2}.\)

Now, the sum of squares of the deviation of the values of the two groups from x̅ is:

\(\Rightarrow\mathop \sum \limits_{i = 1}^{{n_1}} {\left( {{x_{2i}} - \bar x} \right)^2} + \mathop \sum \limits_{i = 1}^{{n_2}} {\left( {{x_{2i}} - \overline {x{\rm{\;}}} } \right)^2}\)

Now,

\(\mathop \sum \limits_{i = 1}^{{n_1}} {\left( {{x_{1i}} - \bar x} \right)^2}\;=\mathop \sum \limits_{i = 1}^{{n_1}} \;{\left\{ {\left( {{x_{1i}} - {\rm{\;}}\overline {{x_1}} } \right) + \left( {\overline {{x_1}} - \bar x{\rm{\;}}} \right)} \right\}^2}\)

\(=\mathop \sum \limits_{i = 1}^{{n_1}} {\left( {{x_{1i}} - {{{\rm{\bar x}}}_1}{\rm{\;}}} \right)^2}\;+2\left( {\overline {{x_i}} - \bar x} \right)\mathop \sum \limits_{i = 1}^{{n_1}} \left( {{x_{1i}} - \overline {{x_1}} } \right) + {n_1}{\left( {\overline {{x_1}} - \bar x} \right)^2}\)

\( = {n_1}\sigma _1^2 + \;{n_1}{\left( {\overline {{x_1}} - \bar x} \right)^2},\;\;\;\mathop \sum \limits_{i = 1}^{{n_1}} \left( {{x_{1i}} - \overline {{x_1}} } \right) = 0\)

Similarly, \(\mathop \sum \limits_{i = 1}^{{n_2}} {\left( {{x_{2i}} - \overline {x{\rm{\;}}} } \right)^2} = {n_2}\sigma _2^2 + {n_2}{\left( {\overline {{x_2}} - \bar x} \right)^2}\)

If the variation of the combined series is σ2,

Then, \({\sigma ^2} = \frac{1}{{{n_1} + {n_2}}}\left\{ {\mathop \sum \limits_{i = 1}^{{n_1}} {{\left( {{x_{2i}} - \bar x} \right)}^2}\; + {\rm{\;}}\mathop \sum \limits_{i = 1}^{{n_2}} {{\left( {{x_{2i}} - \overline {x{\rm{\;}}} } \right)}^2}} \right\}\)

\( = \frac{{{n_1}\sigma _1^2 + {n_2}\sigma _2^2}}{{{n_1} + {n_2}}}+\frac{{{n_1}{{\left( {\overline {{x_1}} - \bar x} \right)}^2} + {n_2}{{\left( {\overline {{x_2}} - \bar x} \right)}^2}}}{{{n_1} + {n_2}}}\)

\(=\frac{{{n_1}\sigma _1^2 + {n_2}\sigma _2^2}}{{{n_1} + {n_2}}}+\frac{{{n_1}d_1^2 + {n_2}d_2^2}}{{{n_1} + {n_2}}}\)

\(=\frac{1}{{{n_1} + {n_2}}}\left( {{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)} \right)\)