The article “HIV-positive Smokers Considering Quitting: Differences by Race/Ethnicity” (E. Lloyd- Richardson, C. Stanton, et al., Am J Health Behav, 2008:3-15) surveyed 444 HIV-positive smokers. Of these, 170 reported that they had used a nicotine patch. Consider this to be a simple random sample.

a. Find a 95% confidence interval for the proportion of HIV-positive smokers who have used a nicotine patch.

b. Find a 99% confidence interval for the proportion of HIV-positive smokers who have used a nicotine patch.

c. Someone claims that the proportion is less than 0.40. With what level of confidence can this statement be made?

d. Find the sample size needed for a 95% confidence interval to specify the proportion to within ±0.03.

e. Find the sample size needed for a 99% confidence interval to specify the proportion to within ±0.03.

Step 1 of 4:

An article says that 444 people survived from HIV. Of these,170 reported that they had used a nicotine patch.

We have to find

95% CI for the proportion of HIV-positive smokers who have used a nicotine patch. 99% CI for the proportion of HIV- positive smokers who have used a nicotine patch.With what level of confidence the the claim that the proportion is less than 0.40 made. Find the sample size needed for 95% CI to specifies the proportion within .Find the sample size needed for 99% CI to specifies the proportion within .Step 2 of 4 :

95% CI for the proportion of HIV- positive smokers who have used nicotine patch.Let X be the number of success in an independent bernoulli trials with success probability P, so that X~B(n,P).

Lets define = n+4.

And =

Then a level 100(1-confidence interval for P is

Let X denote the number of people who had used nicotine. And n be the total number of HIV-positive smokers.

X = 170

n = 444

444+4

= 448

And

= 0.3839.

Since here the Z-score is =

We can find from the standard normal table,

So the 95% confidence interval for the proportion of HIV-positive smokers they have used nicotine.

( 0.3839 )

( 0.3388 , 0.4289 )

Therefore the 95% confidence interval for the proportion of HIV-positive smokers they had used nicotine (0.3388, 0.4289).

(b) The 99% CI for the proportion of HIV- positive smokers they had used nicotine.

Since the Z- score is =

We can find the value from the standard normal table, .

So the 99% CI for the proportion of HIV-positive smokers they had used nicotine.

( 0.3839 )

( 0.3246 , 0.4431)

Therefore the 99% CI for the proportion of HIV- positive smokers they had used nicotine (0.3246, 0.4431).