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)

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

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

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

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

If n1, n2 are the sizes, x̅1, x̅2 the means, σ1, σ2 the standard d

1.	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)
A.	\(\left( {\frac{1}{{{n_1}}} + \frac{1}{{{n_2}}}} \right)\left( {{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)} \right)\)
B.	\(\frac{1}{{{n_1} + {n_2}}}\left( {{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)} \right)\)
C.	\(\left( {\frac{1}{{{n_1}}} - \frac{1}{{{n_2}}}} \right)\left( {{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)} \right)\)
D.	\(\frac{1}{{{n_1} - {n_2}}}\left( {{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)} \right)\)
Answer» C. \(\left( {\frac{1}{{{n_1}}} - \frac{1}{{{n_2}}}} \right)\left( {{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)} \right)\)

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