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Question
The mean and standard deviation of a set of n1 observations are `barx_1` and s1, respectively while the mean and standard deviation of another set of n2 observations are `barx_2` and s2, respectively. Show that the standard deviation of the combined set of (n1 + n2) observations is given by
S.D. = `sqrt((n_1(s_1)^2 + n_2(s_2)^2)/(n_1 + n_2) + (n_1n_2 (barx_1 - barx_2)^2)/(n_1 + n_2)^2)`
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Solution
Let xi' = 1, 2, 3, 4, ..., n1
And yj' = 1, 2, 3, 4, ..., n2
∴ `barx_1 = 1/n_1 sum_(i = 1)^n x_i`
And `barx_2 = 1/n_2 sum_(j = 1)^n y_j`
⇒ `sigma_1^2 = 1/n_1 sum_(i = 1)^(n_1) (x_i - barx_1)^2`
And `sigma_2^2 = 1/n_2 sum_(j = 1)^(n_2) (y_i - barx_2)^2`
Now mean of the combined series is given by
`barx = 1/(n_1 + n_2) [sum_(i = 1)^(n_1) + sum_(j = 1)^(n_2) y_j]`
= `(n_1 barx_1 + n_2 x_2)/(n_1 + n_2)`
Therefore, `sigma^2` of the combined series is
`sigma^2 = 1/(n_1 + n_2) [sum_(i = 1)^(n_1) (x_i - barx)^2 + sum_(j = 1)^(n_2) (y_j - barx)^2]`
Now, `sum_(i = 1)^(n_1) (x_i - barx)^2 = sum_(i = 1)^(n_1) (x_i - barx_j + bar_j - barx)^2`
= `sum_(i = 1)^(n_1) (x_i - x_j)^2 + n_1 (barx_j - barx)^2 + 2(barx_j - barx) sum_(i = 1)^(n_1) (x_i - barx_j)^2`
But `sum_(i = 1)^n (x_i - barx_i)` = 0
∵ The algebraic sum of the deviation of values of first series from their mean is zero
Also `sum_(i = 1)^(n_1) (x_i - barx)^2 = n_1s_1^2 + n_1(barx_1 - barx)^2`
= `n_1s_1^2 + n_1d_1^2`
Where `d_1 = (barx_1 - barx)`
Similarly, we have
`sum_(j = 1)^(n_2) (y_j - barx)^2 = sum_(j = 1)^(n_2) (y_j - barx_i + barx_i - barx)^2`
= `n_2s_2^2 + n_2d_2^2`
Where `d_2 = (barx_2 - barx)`
Now combined Standard Deviation (S.D.)
`sigma = sqrt((n_1(s_1^2 + d_1^2) + n_2(s_2^2 + d_2^2))/(n_1 + n_2))`
Where `d_1 = barx_1 - barx`
= `barx_1 - ((n_1barx_1 + n_2 barx_2)/(n_1 + n_2))`
= `(n_2(barx_1 - barx_2))/(n_1 + n_2)`
And `d_2 = barx_2 - barx`
= `barx_2 - ((n_1barx_1 + b_2barx_2)/(n_1 + n_2))`
= `(n_1(barx_2 - barx_1))/(n_1 + n_2)`
∴ `sigma^2 = 1/(n_1 + n_2)[n_1s_1^2 + n_2s_2^2 + (n_1n_2^2(barx_1 - barx_2)^2)/(n_1 + n_2)^2 + (n_2n_1^2(barx_2 - barx_1)^2)/(n_1 + n_2)^2]`
So, `sigma = sqrt((n_1s_1^2 + n_2s_2^2)/(n_1 + n_2) + (n_1n_2(barx_1 - barx_2)^2)/(n_1 + n_2)^2`
Hence proved.
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