In a survey of 444 HIV-positive smokers, 196 reported that they had used a nicotine patch to try to quit smoking. Can you conclude that less than half of HIV-positive smokers have used a nicotine patch? Use the a=0.05 level of significance and the P-value method.

Question

In a survey of 444 HIV-positive smokers, 196 reported that they had used a nicotine patch to try to quit smoking. Can you conclude that less than half of HIV-positive smokers have used a nicotine patch? Use the a=0.05 level of significance and the P-value method.
Hv0 : [?]
Hv1: [?]

right/left or two-tailed test?

GPT-4o mini · Answer

To conduct the hypothesis test regarding the proportion of HIV-positive smokers who have used a nicotine patch, we need to first define our null and alternative hypotheses:

Null Hypothesis (H0): The proportion of HIV-positive smokers who have used a nicotine patch is 0.5 or more.

$ H_0: p \geq 0.5 $
Alternative Hypothesis (H1): The proportion of HIV-positive smokers who have used a nicotine patch is less than 0.5.

$ H_1: p < 0.5 $

Since we are interested in testing if the proportion is less than 0.5, this is a left-tailed test.

Next, let's summarize the information given:

Sample size ($ n $) = 444
Number of smokers who used a nicotine patch ($ x $) = 196

We can calculate the sample proportion ($ \hat{p} $):

\[ \hat{p} = \frac{x}{n} = \frac{196}{444} \approx 0.4425 \]

Now, we can conduct the hypothesis test using the P-value method.

Calculate the test statistic:

The test statistic for testing proportions is calculated using the following formula:

\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]

where $ p_0 = 0.5 $ (the hypothesized proportion).

Substituting in the values:

\[ z = \frac{0.4425 - 0.5}{\sqrt{\frac{0.5(1 - 0.5)}{444}}} \]

Calculating the standard error:

\[ \sqrt{\frac{0.5 \times 0.5}{444}} = \sqrt{\frac{0.25}{444}} \approx \sqrt{0.000563} \approx 0.0237 \]

Now, calculate the z-value:

\[ z = \frac{0.4425 - 0.5}{0.0237} \approx \frac{-0.0575}{0.0237} \approx -2.43 \]

Find the P-value:

Next, we will find the P-value corresponding to the calculated z-value of -2.43. Using a standard normal distribution table (or a calculator), we can find the P-value:

\[ P(Z < -2.43) \approx 0.0075 \]

Compare the P-value to the significance level ($ \alpha $):

The significance level ($ \alpha $) = 0.05. Since the P-value (0.0075) is less than $ \alpha $ (0.05), we can reject the null hypothesis.

Conclusion:

Based on the analysis, we can conclude that there is sufficient evidence to support the claim that less than half of HIV-positive smokers have used a nicotine patch, at the 0.05 significance level.

Answer

Compute the value of the test statistic. Round the answer to at least two decimal places.

GPT-4o mini · Answer

Let's calculate the test statistic step-by-step again, rounding to at least two decimal places.

Given:

Sample size ($ n $) = 444
Number of smokers who used a nicotine patch ($ x $) = 196

First, calculate the sample proportion ($ \hat{p} $): \[ \hat{p} = \frac{x}{n} = \frac{196}{444} \approx 0.4425 \]

Next, the hypothesized proportion ($ p_0 $) is 0.5.

Now, we need to calculate the standard error of the proportion:

\[ \text{Standard Error (SE)} = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.5(1 - 0.5)}{444}} = \sqrt{\frac{0.5 \times 0.5}{444}} = \sqrt{\frac{0.25}{444}} \]

Calculating:

\[ \sqrt{\frac{0.25}{444}} \approx \sqrt{0.000563} \approx 0.0237 \]

Now, compute the z-test statistic using the following formula:

\[ z = \frac{\hat{p} - p_0}{\text{Standard Error}} = \frac{0.4425 - 0.5}{0.0237} \] Calculating the numerator:

\[ 0.4425 - 0.5 = -0.0575 \]

Now divide by the standard error:

\[ z \approx \frac{-0.0575}{0.0237} \approx -2.43 \]

Rounding to at least two decimal places, the value of the test statistic $ z $ is:

\[ \boxed{-2.43} \]

Answer

No. z= -2.47 Use this to compute the P-Value to at least 4 decimal places

GPT-4o mini · Answer

Thank you for your clarification. Let's recalculate the P-value using the corrected z-value of $-2.47$.

Finding the P-value

Since we are conducting a left-tailed test, we will find the P-value associated with the z-value of $-2.47$.

Using a standard normal distribution table or a z-score calculator, we look up the z-value of $-2.47$.

The P-value for $ z = -2.47 $ is approximately:

\[ P(Z < -2.47) \]

Using a standard normal distribution table or calculator, we find this value:

\[ P(Z < -2.47) \approx 0.0068 \]

Result

Thus, the P-value, rounded to at least four decimal places, is:

\[ \boxed{0.0068} \]

Answer

at the a=0.05 level, do we reject the null hypothesis Hv0?

GPT-4o mini · Answer

Yes, at the $ \alpha = 0.05 $ level of significance, we will reject the null hypothesis $ H_0 $. ### Reasoning: 1. **Null Hypothesis**: - $ H_0: p \geq 0.5 $ (The proportion of HIV-positive smokers who have used a nicotine patch is 0.5 or more). 2. **Alternative Hypothesis**: - $ H_1: p < 0.5 $ (The proportion of HIV-positive smokers who have used a nicotine patch is less than 0.5). 3. **Test Statistic**: - We calculated the z-value to be approximately $-2.47$. 4. **P-value**: - The calculated P-value is approximately $ 0.0068 $. 5. **Decision Rule**: - Since the P-value ($ 0.0068 $) is less than the significance level ($ \alpha = 0.05 $), we reject the null hypothesis. ### Conclusion: At the $ \alpha = 0.05 $ level of significance, there is sufficient evidence to reject the null hypothesis. Therefore, we conclude that less than half of HIV-positive smokers have used a nicotine patch.