Q.The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.

The key idea is to find the Combined Standard Deviation of two samples.

1. First, calculate the combined mean ($\bar{x}_c$) of the two samples. Given: $n_1 = 60, \bar{x}_1 = 650, \sigma_1 = 8$ Given: $n_2 = 80, \bar{x}_2 = 660, \sigma_2 = 7$ Total number of bulbs $N = n_1 + n_2 = 60 + 80 = 140$.

2. Next, calculate the deviations of individual sample means from the combined mean:

3. Apply the formula for combined variance ($\sigma_c^2$):

   $\sigma_c^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$

   Substitute the values:

$$ \sigma_c^2 = \frac{60\left(64 + \frac{1600}{49}\right) + 80\left(49 + \frac{900}{49}\right)}{140} $$ …

When combining two or more samples, we cannot simply average their standard deviations or even their variances. This is because standard deviation measures the spread of data points around their own mean. When we combine samples, the data points from each original sample will now be spread around a new, overall mean. The formula for combined variance accounts for both the internal spread within each sample (its own variance) and the spread of each sample's mean relative to the overall mean.

Here is how to find the overall standard deviation:

1. Identify the given data for each sample.

   We have two samples of bulbs:

   • Sample 1: Number of bulbs, $n_1 = 60$ Mean life, $\bar{x}_1 = 650$ hours Standard deviation, $\sigma_1 = 8$ hours
   • Sample 2: Number of bulbs, $n_2 = 80$ Mean life, $\bar{x}_2 = 660$ hours Standard deviation, $\sigma_2 = 7$ hours

2. Calculate the combined mean ($\bar{x}_{combined}$).

   The combined mean is a weighted average of the individual sample means, weighted by their respective sample sizes. This makes intuitive sense: a larger sample contributes more to the overall average.

   $\bar{x}_{combined} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$

   Substituting the given values:

   $\bar{x}_{combined} = \frac{(60 \times 650) + (80 \times 660)}{60 + 80}$

   $\bar{x}_{combined} = \frac{39000 + 52800}{140}$

   $\bar{x}_{combined} = \frac{91800}{140}$

   $\bar{x}_{combined} = \frac{9180}{14}$

   $\bar{x}_{combined} = \frac{4590}{7} \approx 655.714 \text{ hours}$

3. Understand the concept of variance and sum of squares.

   The variance ($\sigma^2$) of a sample is defined as the average of the squared deviations from the mean: $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$. This means that the sum of squared deviations from the mean, $\sum (x_i - \bar{x})^2$, can be expressed as $n\sigma^2$. This quantity, $n\sigma^2$, represents the total "spread" within a sample relative to its own mean.

   When combining samples, we need to find the total sum of squared deviations from the overall combined mean.

4. Use the formula for combined variance ($\sigma_{combined}^2$).

   The formula for the combined variance of two samples is:

   $\sigma_{combined}^2 = \frac{n_1 (\sigma_1^2 + d_1^2) + n_2 (\sigma_2^2 + d_2^2)}{n_1 + n_2}$

   where $d_1 = \bar{x}_1 - \bar{x}_{combined}$ and $d_2 = \bar{x}_2 - \bar{x}_{combined}$.

   This formula essentially states that the total variance is a weighted average of each sample's variance plus a term that accounts for how far each sample's mean is from the overall combined mean. The $d_i^2$ terms are crucial because they capture the additional spread introduced by the difference between individual sample means and the overall mean.

   ›Proof

   Derivation of the Combined Variance Formula

   Let $x_{1i}$ be the $i$-th observation in sample 1 and $x_{2j}$ be the $j$-th observation in sample 2.

   The total sum of squares about the combined mean is $\sum_{k=1}^{n_1+n_2} (x_k - \bar{x}_{combined})^2$.

   This can be split into two parts:

   $\sum_{i=1}^{n_1} (x_{1i} - \bar{x}_{combined})^2 + \sum_{j=1}^{n_2} (x_{2j} - \bar{x}_{combined})^2$.

   Consider the first sum: $\sum_{i=1}^{n_1} (x_{1i} - \bar{x}_{combined})^2$.

   We can rewrite $(x_{1i} - \bar{x}_{combined})$ as $((x_{1i} - \bar{x}_1) + (\bar{x}_1 - \bar{x}_{combined}))$.

   So, $\sum_{i=1}^{n_1} (x_{1i} - \bar{x}_{combined})^2 = \sum_{i=1}^{n_1} ((x_{1i} - \bar{x}_1) + (\bar{x}_1 - \bar{x}_{combined}))^2$

   Expanding the square:

   $= \sum_{i=1}^{n_1} (x_{1i} - \bar{x}_1)^2 + \sum_{i=1}^{n_1} (\bar{x}_1 - \bar{x}_{combined})^2 + 2 \sum_{i=1}^{n_1} (x_{1i} - \bar{x}_1)(\bar{x}_1 - \bar{x}_{combined})$

   We know:

   1. $\sum_{i=1}^{n_1} (x_{1i} - \bar{x}_1)^2 = n_1 \sigma_1^2$ (by definition of variance).
   2. $\sum_{i=1}^{n_1} (\bar{x}_1 - \bar{x}_{combined})^2 = n_1 (\bar{x}_1 - \bar{x}_{combined})^2$ (since $(\bar{x}_1 - \bar{x}_{combined})$ is a constant for all $x_{1i}$).
   3. $2 \sum_{i=1}^{n_1} (x_{1i} - \bar{x}_1)(\bar{x}_1 - \bar{x}_{combined}) = 2 (\bar{x}_1 - \bar{x}_{combined}) \sum_{i=1}^{n_1} (x_{1i} - \bar{x}_1)$. Since $\sum_{i=1}^{n_1} (x_{1i} - \bar{x}_1) = 0$ (the sum of deviations from the mean is always zero), the third term vanishes.

   Thus, $\sum_{i=1}^{n_1} (x_{1i} - \bar{x}_{combined})^2 = n_1 \sigma_1^2 + n_1 (\bar{x}_1 - \bar{x}_{combined})^2$.

   Similarly, for the second sample:

   $\sum_{j=1}^{n_2} (x_{2j} - \bar{x}_{combined})^2 = n_2 \sigma_2^2 + n_2 (\bar{x}_2 - \bar{x}_{combined})^2$.

   Summing these two results gives the total sum of squares about the combined mean:

   $n_1 \sigma_1^2 + n_1 (\bar{x}_1 - \bar{x}_{combined})^2 + n_2 \sigma_2^2 + n_2 (\bar{x}_2 - \bar{x}_{combined})^2$.

   Dividing by the total number of observations $(n_1 + n_2)$ gives the combined variance:

   $\sigma_{combined}^2 = \frac{n_1 \sigma_1^2 + n_1 (\bar{x}_1 - \bar{x}_{combined})^2 + n_2 \sigma_2^2 + n_2 (\bar{x}_2 - \bar{x}_{combined})^2}{n_1 + n_2}$. …