Two samples each of size 25 are taken from independent populations assumed to be normally distributed with equal variances. The first sample has a mean of 35.5 and standard deviation of 3.0 while the second sample has a mean of 33.0 and standard deviation of 4.0.
The pooled (i.e., combined) variance is ________.
12.5
To find the pooled variance, we first need to calculate the variances of the individual samples.
For the first sample, we know the mean (35.5) and standard deviation (3.0). The variance is calculated as the square of the standard deviation. Therefore, we have:
Variance1 = (Standard deviation1)^2 = 3.0^2 = 9.0.
For the second sample, we know the mean (33.0) and standard deviation (4.0). Again, we calculate the variance as the square of the standard deviation:
Variance2 = (Standard deviation2)^2 = 4.0^2 = 16.0.
Now that we have the variances for both samples, we can find the pooled variance by taking their weighted average. Since both samples have the same size (25) and are assumed to have equal variances, we can use the following formula:
Pooled Variance = [(Size of Sample1 * Variance1) + (Size of Sample2 * Variance2)] / (Total Size of Samples)
Plugging in the values, we get:
Pooled Variance = [(25 * 9.0) + (25 * 16.0)] / (25 + 25)
= (225 + 400) / 50
= 625 / 50
= 12.5.
Therefore, the pooled variance is 12.5.
To find the pooled variance, we can use the formula:
Pooled Variance = [(n1 - 1) * s1^2 + (n2 - 1) * s2^2] / (n1 + n2 - 2)
Given that:
n1 = 25 (size of the first sample)
s1 = 3.0 (standard deviation of the first sample)
n2 = 25 (size of the second sample)
s2 = 4.0 (standard deviation of the second sample)
Substituting these values into the formula:
Pooled Variance = [(25 - 1) * 3.0^2 + (25 - 1) * 4.0^2] / (25 + 25 - 2)
Pooled Variance = [24 * 9.0 + 24 * 16.0] / 48
Pooled Variance = [216 + 384] / 48
Pooled Variance = 600 / 48
Pooled Variance = 12.5
Therefore, the pooled variance is 12.5.