The amount of time single men and women spend on house work is measured for 15 single women and 25 single men. For the women the mean was 7 hours/week with a standard deviation of 1.5. For the men the mean was 4.5 hours/week with a standard deviation of 1.1. What is the value of the pooled standard deviation for the difference in mean time spent on housework between single men and women? A. 1.30 B. 0.45 C. 1.59 D. 1.26

To find the pooled standard deviation of the difference in mean time spent on housework between single men and women, we can use the following formula for the pooled standard deviation when dealing with two independent samples:

\[ s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \]

Where: - \( s_1 \) and \( s_2 \) are the standard deviations of the first and second samples, respectively. - \( n_1 \) and \( n_2 \) are the sample sizes of the first and second samples, respectively. - \( s_p \) is the pooled standard deviation.

Let's plug in the numbers:

For women: - \( n_1 = 15 \) - \( s_1 = 1.5 \)

For men: - \( n_2 = 25 \) - \( s_2 = 1.1 \)

\[ s_p = \sqrt{\frac{(15 - 1)(1.5)^2 + (25 - 1)(1.1)^2}{15 + 25 - 2}} \] \[ s_p = \sqrt{\frac{(14)(2.25) + (24)(1.21)}{38}} \] \[ s_p = \sqrt{\frac{31.5 + 29.04}{38}} \] \[ s_p = \sqrt{\frac{60.54}{38}} \] \[ s_p = \sqrt{1.59} \] \[ s_p ≈ 1.26 \]

Therefore, the value of the pooled standard deviation for the difference in mean time spent on housework between single men and women is approximately 1.26. The correct answer is D. 1.26.