We are asked to calculate the pooled variance from the given raw data for two groups. Group 1 has the data: 15, 11, 8, 7, 6, 4, 13. Group 2 has the data: 4, 10, 15, 12, 12, 9, 8.

1. Problem Description

We are asked to calculate the pooled variance from the given raw data for two groups. Group 1 has the data: 15, 11, 8, 7, 6, 4,

1

3. Group 2 has the data: 4, 10, 15, 12, 12, 9,

8.

2. Solution Steps

First, we calculate the sample mean for each group.

Let $x_i$ represent the data in Group 1, and $y_i$ represent the data in Group

2.

$\bar{x} = \frac{\sum_{i=1}^{n_1} x_i}{n_1}$

$\bar{y} = \frac{\sum_{i=1}^{n_2} y_i}{n_2}$

where $n_1$ and $n_2$ are the number of data points in Group 1 and Group 2, respectively. Here $n_1 = 7$ and $n_2 = 7$.

$\bar{x} = \frac{15+11+8+7+6+4+13}{7} = \frac{64}{7} \approx 9.1429$

$\bar{y} = \frac{4+10+15+12+12+9+8}{7} = \frac{70}{7} = 10$

Next, we calculate the sample variance for each group.

$s_x^2 = \frac{\sum_{i=1}^{n_1} (x_i - \bar{x})^2}{n_1 - 1}$

$s_y^2 = \frac{\sum_{i=1}^{n_2} (y_i - \bar{y})^2}{n_2 - 1}$

$s_x^2 = \frac{(15-64/7)^2 + (11-64/7)^2 + (8-64/7)^2 + (7-64/7)^2 + (6-64/7)^2 + (4-64/7)^2 + (13-64/7)^2}{7-1}$

$s_x^2 = \frac{(5.8571)^2 + (1.8571)^2 + (-1.1429)^2 + (-2.1429)^2 + (-3.1429)^2 + (-5.1429)^2 + (3.8571)^2}{6}$

$s_x^2 = \frac{34.3061 + 3.4589 + 1.3061 + 4.5918 + 9.8776 + 26.4531 + 14.8776}{6} = \frac{94.8712}{6} \approx 15.8119$

$s_y^2 = \frac{(4-10)^2 + (10-10)^2 + (15-10)^2 + (12-10)^2 + (12-10)^2 + (9-10)^2 + (8-10)^2}{7-1}$

$s_y^2 = \frac{(-6)^2 + (0)^2 + (5)^2 + (2)^2 + (2)^2 + (-1)^2 + (-2)^2}{6}$

$s_y^2 = \frac{36 + 0 + 25 + 4 + 4 + 1 + 4}{6} = \frac{74}{6} \approx 12.3333$

Now, we calculate the pooled variance $s_p^2$.

$s_p^2 = \frac{(n_1-1)s_x^2 + (n_2-1)s_y^2}{n_1+n_2-2}$

$s_p^2 = \frac{(7-1)s_x^2 + (7-1)s_y^2}{7+7-2} = \frac{6s_x^2 + 6s_y^2}{12} = \frac{s_x^2 + s_y^2}{2}$

$s_p^2 = \frac{15.8119 + 12.3333}{2} = \frac{28.1452}{2} \approx 14.0726$

3. Final Answer

14.0726