Question

Culligan et al. (A-34) compared the long-term results of two treatments: (a) a modified Burch procedure, and (b) a sling procedure for stress incontinence with a low-pressure urethra. Thirty-six women took part in the study with 19 in the Burch treatment group and 17 in the sling procedure treatment group. One of the outcome measures at three months post-surgery was maximum urethral closure pressure (cm H2O). In the Burch group the mean and standard deviation were 16.4 and 8.2 cm, respectively. In the sling group, the mean and standard deviation were 39.8 and 23.0, respectively. Construct the 99 percent confidence interval for the difference in mean maximum urethral closure pressure for the two populations represented by these subjects. State all necessary assumptions.

Answer 1

Here,

$\bar{x_1} = 16.4,s_1 = 8.2,n_1 = 19, \bar{x_2} = 39.8,s_2 = 23,n_2 = 17$

degrees of freedom for t statistic = $\gamma = \frac{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2})}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}} = 19.6207 \approx 19$

a 99% confidence interval for the difference in mean maximum urethral closure pressure for the two populations represented by these subjects

$=((\bar{x_1} - \bar{x_2})-t_{0.005,\gamma }*\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}, (\bar{x_1} - \bar{x_2})+t_{0.005,\gamma }*\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})$

$=((\bar{x_1} - \bar{x_2})-t_{0.005,19 }*\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}, (\bar{x_1} - \bar{x_2})+t_{0.005,19}*\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})$

$=((\bar{x_1} - \bar{x_2})-2.86093*\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}, (\bar{x_1} - \bar{x_2})+2.86093*\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})$

$=(-40.2423, -6.55775)$

Assumption : maximum urethral closure pressure for the two populations follows normal distribution.