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?
Hv0 : [?]
Hv1: [?]
right/left or two-tailed test?
Answers
Answer
Compute the value of the test statistic. Round the answer to at least two decimal places.
Answer
No. z= -2.47
Use this to compute the P-Value to at least 4 decimal places
Use this to compute the P-Value to at least 4 decimal places
Answer
at the a=0.05 level, do we reject the null hypothesis Hv0?
Answered by
GPT-4o mini
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.
1. **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:
\[
z = \frac{0.4425 - 0.5}{0.0237} \approx \frac{-0.0575}{0.0237} \approx -2.43
\]
2. **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:
\[
P(Z < -2.43) \approx 0.0075
\]
3. **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.
- **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.
1. **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:
\[
z = \frac{0.4425 - 0.5}{0.0237} \approx \frac{-0.0575}{0.0237} \approx -2.43
\]
2. **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:
\[
P(Z < -2.43) \approx 0.0075
\]
3. **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.
Answered by
GPT-4o mini
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:
\[
\boxed{-2.43}
\]
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:
\[
\boxed{-2.43}
\]
Answered by
GPT-4o mini
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:
\[
\boxed{0.0068}
\]
### 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:
\[
\boxed{0.0068}
\]
Answered by
GPT-4o mini
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.
### 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.
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