Question

sample of 40 electric batteries gives a mean life span of 600 hrs with a standard deviation of 20 hours. Another sample of 50 electric batteries gives a mean lifespan of 520 hours with a standard deviation of 30 hours. If these two samples were combined and used in a given project simultaneously, determine the combined new mean for the larger sample and hence determine the combined or pulled standard deviation.

Answer 1

Answer on Question #82458 – Math – Statistics and Probability

Question

Sample of 40 electric batteries gives a mean life span of 600 hours with a standard deviation of 20 hours. Another sample of 50 electric batteries gives a mean lifespan of 520 hours with a standard deviation of 30 hours. If these two samples were combined and used in a given project simultaneously, determine the combined new mean for the larger sample and hence determine the combined or pulled standard deviation.

Solution

Let $m$ be the combined mean.

Let $x_1$ be the mean of first sample.

Let $x_2$ be the mean of the second sample.

Let $n_1$ be the size of the first sample.

Let $n_2$ be the size of the second sample.

Let $s_1$ be the standard deviation of the first sample.

Let $s_2$ be the standard deviation of the second sample.

$$\text{combined mean} = m = \frac{n_1 x_1 + n_2 x_2}{n_1 + n_2}$$ $$\text{combined standard deviation} = \sqrt{\frac{n_1 s_1^2 + n_1 (m - x_1)^2 + n_2 s_2^2 + n_2 (m - x_2)^2}{n_1 + n_2}} =$$ $$s^2 = \frac{n_1 s_1^2 + n_1 (m - x_1)^2 + n_2 s_2^2 + n_2 (m - x_2)^2}{n_1 + n_2} =$$ $$= \frac{\sum_{i=1}^{n_1} (x_i - x_1)^2 + n_1 (m - x_1)^2 + \sum_{i=1}^{n_2} (y_i - x_2)^2 + n_2 (m - x_2)^2}{n_1 + n_2} =$$ $$= \frac{\sum_{i=1}^{n_1} \left((x_i - x_1)^2 + (x_1 - m)^2\right) + \sum_{i=1}^{n_2} \left((y_i - x_2)^2 + (x_2 - m)^2\right)}{n_1 + n_2} =$$ $$= \frac{\sum_{i=1}^{n_1} \left(x_i^2 - 2 x_i x_1 + 2 x_1^2 - 2 m x_1 + m^2\right)}{n_1 + n_2} +$$ $$+ \frac{\sum_{i=1}^{n_2} \left(y_i^2 - 2 y_i x_2 + 2 x_2^2 - 2 m x_2 + m^2\right)}{n_1 + n_2} =$$ $$= \frac{\sum_{i=1}^{n_1} \left(x_i^2 + m^2 - 2 m \sum_{j=1}^{n_1} \frac{x_j}{n_1}\right) + 2 n_1 x_1^2 - 2 x_1 \sum_{i=1}^{n_1} x_i}{n_1 + n_2} +$$ $$+ \frac{\sum_{i=1}^{n_2} \left(y_i^2 + m^2 - 2 m \sum_{j=1}^{n_2} \frac{y_j}{n_2}\right) + 2 n_2 x_2^2 - 2 x_2 \sum_{i=1}^{n_2} y_i}{n_1 + n_2} =$$ $$\begin{array}{l} = \frac {\sum_ {i = 1} ^ {n _ {1}} \left(x _ {i} ^ {2} + m ^ {2} - 2 m \sum_ {j = 1} ^ {n _ {1}} \frac {x _ {j}}{n _ {1}}\right) + 2 n _ {1} x _ {1} ^ {2} - 2 x _ {1} n _ {1} x _ {1}}{n _ {1} + n _ {2}} + \\ + \frac {\sum_ {i = 1} ^ {n _ {2}} \left(y _ {i} ^ {2} + m ^ {2} - 2 m \sum_ {j = 1} ^ {n _ {2}} \frac {y _ {j}}{n _ {2}}\right) + 2 n _ {2} x _ {2} ^ {2} - 2 x _ {2} n _ {2} x _ {2}}{n _ {1} + n _ {2}} = \\ = \frac {\sum_ {i = 1} ^ {n _ {1}} \left(x _ {i} ^ {2} + m ^ {2} - 2 m \sum_ {j = 1} ^ {n _ {1}} \frac {x _ {j}}{n _ {1}}\right)}{n _ {1} + n _ {2}} + \frac {\sum_ {i = 1} ^ {n _ {2}} \left(y _ {i} ^ {2} + m ^ {2} - 2 m \sum_ {j = 1} ^ {n _ {2}} \frac {y _ {j}}{n _ {2}}\right)}{n _ {1} + n _ {2}} \\ \end{array}$$

Since each $\left(-2m\sum_{j = 1}^{n_1}\frac{x_j}{n_1}\right)$ term appears $n_1$ times, then we can reorder the sums

$$\sum_ {i = 1} ^ {n _ {1}} \left(x _ {i} ^ {2} + m ^ {2} - 2 m \sum_ {j = 1} ^ {n _ {1}} \frac {x _ {j}}{n _ {1}}\right) = \sum_ {i = 1} ^ {n _ {1}} \left(x _ {i} ^ {2} + m ^ {2} - 2 m x _ {i}\right)$$

Hence,

$$\begin{array}{l} s ^ {2} = \frac {n _ {1} s _ {1} ^ {2} + n _ {1} (m - x _ {1}) ^ {2} + n _ {2} s _ {2} ^ {2} + n _ {2} (m - x _ {2}) ^ {2}}{n _ {1} + n _ {2}} = \\ = \frac {\sum_ {i = 1} ^ {n _ {1}} \left(x _ {i} ^ {2} + m ^ {2} - 2 m x _ {i}\right)}{n _ {1} + n _ {2}} + \frac {\sum_ {i = 1} ^ {n _ {2}} \left(y _ {i} ^ {2} + m ^ {2} - 2 m y _ {i}\right)}{n _ {1} + n _ {2}} = \\ = \frac {\sum_ {i = 1} ^ {n _ {1} + n _ {2}} \left(z _ {i} ^ {2} + m ^ {2} - 2 m z _ {i}\right)}{n _ {1} + n _ {2}} = \\ = \frac {\sum_ {i = 1} ^ {n _ {1} + n _ {2}} (z _ {i} - m) ^ {2}}{n _ {1} + n _ {2}} \stackrel {\text {def}} {=} s ^ {2} \\ \end{array}$$ $$\begin{array}{l} n _ {1} = 4 0, x _ {1} = 6 0 0 h, s _ {1} = 2 0 h \\ n _ {2} = 5 0, x _ {2} = 5 2 0 h, s _ {2} = 3 0 h \\ \end{array}$$ $$\text{combined mean} = m = \frac {4 0 (6 0 0) + 5 0 (5 2 0)}{4 0 + 5 0} = \frac {5 0 0 0}{9} \approx 5 5 5. 5 6 (h)$$ $$\begin{array}{l} \text{combined standard deviation} = \\ = \sqrt {\frac {4 0 (2 0) ^ {2} + 4 0 \left(\frac {5 0 0 0}{9} - 6 0 0\right) ^ {2} + 5 0 (3 0) ^ {2} + 5 0 \left(\frac {5 0 0 0}{9} - 5 2 0\right) ^ {2}}{4 0 + 5 0}} \approx 4 7. 5 2 (h) \\ \end{array}$$

combined mean = 555.56 hours

combined standard deviation = 47.52 hours

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