Question

Suppose we take a sample of 18 men and 17 women to estimate the difference in average height between men and women. The men had an average height of 70 inches with a standard deviation of 3, and the women had an average height of 66 inches with a standard deviation of 2. Calculate the pooled sample variance (Sp^2) for a two-sample confidence interval for the difference in means.

Answer 1

Answer:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

And replacing we got:

$\S^2_p =\frac{(18-1)3^2 +(17 -1)2^2}{18 +17 -2} = 6.576$

Step-by-step explanation:

Data given

For this case we have the following info given:

$n_1 = 18$ represent the random sample size of men

$n_2 = 17$ represent the random sample size of women

$\bar X_1 = 70$ represent the average for men

$\bar X_2 = 6$ represent the average for women

$s_1 = 3$ represent the sample deviation for men

$s_2 = 2$ represent the sample deviation for women

Solution to the problem

The pooled variance is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

And replacing we got:

$\S^2_p =\frac{(18-1)3^2 +(17 -1)2^2}{18 +17 -2} = 6.576$

And the pooled sample deviation would be:

$S_p = 2.564$