Answer to Question #213376 in Statistics and Probability for arwa

Question

 The objective of a randomized controlled trial by Adab et al. (A-19) was to determine whether providing

women with additional information on the pros and cons of screening for cervical cancer would increase the willingness to be screened. A treatment group of 138 women received a leaflet on screening that contained more information (average individual risk for cervical cancer, likelihood of positive finding, the possibility of false positive/negative results, etc.) than the standard leaflet developed by the British National Health Service that 136 women in a control group received. In the treatment group, 109 women indicated they wanted to have the screening test for cervical cancer while in the control group, 120 indicated they wanted the screening test. Construct a 95 percent confidence interval for the difference in proportions for the two populations represented by these samples

Answer 1

Given, a = 0.05

and $Z_{0.05}=1.96$ (from standard normal table)

So, according to question:

$p_1= \dfrac{109}{138}=0.79$

and

$p_2=\dfrac{120}{136}=0.88$

We know that , for 95% confidence interval is

$(p_1-p_2)\pm Z\sqrt{\dfrac{p_1(1-p_1)}{n_1}+\dfrac{p_2(1-p_2)}{n_2}}$

So,

CI = $(0.79-0.88)\pm1.96\times\sqrt{\dfrac{0.79(1-0.79)}{138}+\dfrac{0.88(1-0.88)}{136}}$

lower limit of CI

$\implies (-0.09)-1.96\times\sqrt{\dfrac{0.79(1-0.79)}{138}+\dfrac{0.88(1-0.88)}{136}}\\\implies -0.1771846$

upper limit of CI

$\implies (-0.09)+1.96\times\sqrt{\dfrac{0.79(1-0.79)}{138}+\dfrac{0.88(1-0.88)}{136}}\\\implies-0.002815364$

So the confidence interval is (-0.1771846 , -0.002815364)