The p-value of the test is of 0.0287 < 0.05(standard significance level), which means that it can be concluded that less than half of HIV-positive smokers have used a nicotine patch.

Step-by-step explanation:

Test if less than half of HIV-positive smokers have used a nicotine patch:

At the null hypothesis, we test if the proportion is of at least half, that is:

$H_0: p \geq 0.5$

At the alternative hypothesis, we test if the proportion is below 0.5, that is:

$H_1: p < 0.5$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.5 is tested at the null hypothesis:

This means that $\mu = 0.5, \sigma = \sqrt{0.5*(1-0.5)} = 0.5$

In a survey of 444 HIV-positive smokers, 202 reported that they had used a nicotine patch to try to quit smoking.

This means that $n = 444, X = \frac{202}{444} = 0.455$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.455 - 0.5}{\frac{0.5}{\sqrt{444}}}$

$z = -1.9$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.455, which is the p-value of z = -1.9.

Looking at the z-table, z = -1.9 has a p-value of 0.0287.

The p-value is of 0.0287 < 0.05(standard significance level), which means that it can be concluded that less than half of HIV-positive smokers have used a nicotine patch.

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