Chapter 10: Q 12. (page 640)

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

Expert verified

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

$z_{a / 2} = 2.575$

Hence, confidence interval is $({\hat{p}}_{1} - {\hat{p}}_{2}) - z_{a / 2} \times \sqrt{\frac{{\hat{p}}_{1} (1 - {\hat{p}}_{1})}{n_{1}} + \frac{{\hat{p}}_{2} (1 - {\hat{p}}_{2})}{n_{2}}}$

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

$\approx - 0.2908$

and $({\hat{p}}_{1} - {\hat{p}}_{2}) + z_{a / 2} \times \sqrt{\frac{{\hat{p}}_{1} (1 - {\hat{p}}_{1})}{n_{1}} + \frac{{\hat{p}}_{2} (1 - {\hat{p}}_{2})}{n_{2}}}$

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

$\approx - 0.0916$

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

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Short Answer

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