Question

Quit smoking Nicotine patches are often used to help smokers quit. Does giving medicine to fight depression help? A randomized double-blind experiment assigned $244$smokers to receive nicotine patches and another $245$to receive both a patch and the antidepressant drug bupropion. After a year, $40$subjects in the nicotine patch group had abstained from smoking, as had $87$in the patch-plus-drug group. Construct and interpret a $99\%$confidence interval for the difference in the true proportion of smokers like these who would abstain when using bupropion and a nicotine patch and the proportion who would abstain when using only a patch.

Short answer

Confidence Interval is$(-0.2908,-0.0916)$.

Step-by-step solution

Given Information

It is given that $x_1=40$

$x_2=87$

$n_1=244$

$n_2=245$

$c=99\%=0.99$

Checking the conditions

Random: Samples are independent random samples.

Independent: $224<10\%$of all smokers.

Normal: The success are $40,87$and failures are $204,158$which are greater than ten. So, is satisfied.

Calculations

Sample Proportion: ${\hat{p}}_1=\frac{x_1}{n_1}=\frac{40}{244}=0.1639$

and ${\hat{p}}_2=\frac{x_2}{n_2}=\frac{87}{245}=0.3551$

For $1-\alpha =0.99$, value of role="math" localid="1654639742329" $z_{\alpha /2}=z_{0.005}$using table is

$z_{a/2}=2.575$

Hence, confidence interval is $\left({\hat{p}}_1-{\hat{p}}_2\right)-z_{a/2}\times \sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$=(0.1639-0.3551)-2.575\times \sqrt{\frac{0.1639(1-0.1639)}{244}+\frac{0.3551(1-0.3551)}{245}}$

$\approx -0.2908$

and $\left({\hat{p}}_1-{\hat{p}}_2\right)+z_{a/2}\times \sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$=(0.1639-0.3551)+2.575\times \sqrt{\frac{0.1639(1-0.1639)}{244}+\frac{0.3551(1-0.3551)}{245}}$

$\approx -0.0916$

Hence, confidence interval is$(-0.2908,-0.0916)$