Question
Medical researchers performed a comparison of two drugs, clopidogrel and ticagrelor, which are designed to reduce the risk of heart attack or stroke in coronary patients. A total of 6680 patients were given clopidogrel, and 6723 were given ticagrelor. Of the clopidogrel patients, 663 suffered a heart attack or stroke within one year, and of the ticagrelor patients, 577 suffered a heart attack or stroke. Can you conclude that the proportion of patients suffering a heart attack or stroke is less for ticagrelor? Let pv1 denote the proportion of patients suffering a heart attack or stroke with clopidogrel and pv2 denote the proportion of patients suffering a heart attack or stroke with ticagrelor. Use the a=0.01 level of significance and the P-value method.
Hv0: [?]
Hv1: [?]
This is a [right/left/ or two tailed?] test
Hv0: [?]
Hv1: [?]
This is a [right/left/ or two tailed?] test
Answers
GPT-4o mini
To test whether the proportion of patients suffering a heart attack or stroke is less for ticagrelor compared to clopidogrel, we can set up our hypotheses as follows:
### Hypotheses
- \( H_0 \) (Null Hypothesis): \( p_1 \geq p_2 \) (the proportion of patients suffering a heart attack or stroke with clopidogrel is greater than or equal to the proportion with ticagrelor)
- \( H_1 \) (Alternative Hypothesis): \( p_1 < p_2 \) (the proportion of patients suffering a heart attack or stroke with clopidogrel is less than the proportion with ticagrelor)
### Test Type
This is a **left-tailed** test since we are testing if the proportion for clopidogrel is less than that for ticagrelor.
### Sample Proportions
To analyze the proportions, we can calculate:
- The proportion of patients suffering a heart attack or stroke for clopidogrel (\( p_1 \)):
\[
p_1 = \frac{663}{6680} \approx 0.0993
\]
- The proportion of patients suffering a heart attack or stroke for ticagrelor (\( p_2 \)):
\[
p_2 = \frac{577}{6723} \approx 0.0858
\]
### Test Statistic
To perform the test, we can use the z-test for the difference between two proportions. The pooled proportion (\( p \)) can be computed as:
\[
p = \frac{x_1 + x_2}{n_1 + n_2} = \frac{663 + 577}{6680 + 6723} = \frac{1240}{13403} \approx 0.0925
\]
Where:
- \( x_1 = 663 \) (number of successes in group 1)
- \( x_2 = 577 \) (number of successes in group 2)
- \( n_1 = 6680 \) (total in group 1)
- \( n_2 = 6723 \) (total in group 2)
The standard error (SE) for the difference in proportions is calculated as:
\[
SE = \sqrt{p(1 - p) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.0925(1 - 0.0925) \left(\frac{1}{6680} + \frac{1}{6723}\right)}
\]
### Z-Score Calculation
We then calculate the z-score:
\[
z = \frac{p_1 - p_2}{SE}
\]
You would then compare the z-score to the critical value from the z-table for the left-tailed test at \( \alpha = 0.01 \) level of significance or calculate the P-value associated with the z-score to make a decision regarding the null hypothesis.
### Conclusion
If the z-score is less than the critical value (or if the P-value is less than 0.01), you would reject the null hypothesis in favor of the alternative hypothesis, concluding that the proportion of patients suffering a heart attack or stroke is indeed less for ticagrelor compared to clopidogrel.
### Summary
- \( H_0: p_1 \geq p_2 \)
- \( H_1: p_1 < p_2 \)
- This is a left-tailed test.
### Hypotheses
- \( H_0 \) (Null Hypothesis): \( p_1 \geq p_2 \) (the proportion of patients suffering a heart attack or stroke with clopidogrel is greater than or equal to the proportion with ticagrelor)
- \( H_1 \) (Alternative Hypothesis): \( p_1 < p_2 \) (the proportion of patients suffering a heart attack or stroke with clopidogrel is less than the proportion with ticagrelor)
### Test Type
This is a **left-tailed** test since we are testing if the proportion for clopidogrel is less than that for ticagrelor.
### Sample Proportions
To analyze the proportions, we can calculate:
- The proportion of patients suffering a heart attack or stroke for clopidogrel (\( p_1 \)):
\[
p_1 = \frac{663}{6680} \approx 0.0993
\]
- The proportion of patients suffering a heart attack or stroke for ticagrelor (\( p_2 \)):
\[
p_2 = \frac{577}{6723} \approx 0.0858
\]
### Test Statistic
To perform the test, we can use the z-test for the difference between two proportions. The pooled proportion (\( p \)) can be computed as:
\[
p = \frac{x_1 + x_2}{n_1 + n_2} = \frac{663 + 577}{6680 + 6723} = \frac{1240}{13403} \approx 0.0925
\]
Where:
- \( x_1 = 663 \) (number of successes in group 1)
- \( x_2 = 577 \) (number of successes in group 2)
- \( n_1 = 6680 \) (total in group 1)
- \( n_2 = 6723 \) (total in group 2)
The standard error (SE) for the difference in proportions is calculated as:
\[
SE = \sqrt{p(1 - p) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.0925(1 - 0.0925) \left(\frac{1}{6680} + \frac{1}{6723}\right)}
\]
### Z-Score Calculation
We then calculate the z-score:
\[
z = \frac{p_1 - p_2}{SE}
\]
You would then compare the z-score to the critical value from the z-table for the left-tailed test at \( \alpha = 0.01 \) level of significance or calculate the P-value associated with the z-score to make a decision regarding the null hypothesis.
### Conclusion
If the z-score is less than the critical value (or if the P-value is less than 0.01), you would reject the null hypothesis in favor of the alternative hypothesis, concluding that the proportion of patients suffering a heart attack or stroke is indeed less for ticagrelor compared to clopidogrel.
### Summary
- \( H_0: p_1 \geq p_2 \)
- \( H_1: p_1 < p_2 \)
- This is a left-tailed test.
Incorrect
GPT-4o mini
I apologize for any confusion in my previous response. Let's break it down step-by-step accurately.
### Given Data
- Number of patients taking clopidogrel (\( n_1 \)): 6680
- Number of patients taking ticagrelor (\( n_2 \)): 6723
- Number of events (heart attacks or strokes) in the clopidogrel group (\( x_1 \)): 663
- Number of events in the ticagrelor group (\( x_2 \)): 577
### Step 1: Define Hypotheses
- **Null Hypothesis \( H_0 \)**: \( p_1 \geq p_2 \)
(The proportion of patients suffering a heart attack or stroke with clopidogrel is greater than or equal to that with ticagrelor.)
- **Alternative Hypothesis \( H_1 \)**: \( p_1 < p_2 \)
(The proportion of patients suffering a heart attack or stroke with clopidogrel is less than that with ticagrelor.)
### Step 2: This is a left-tailed test
Since we are interested in determining if the proportion of patients suffering events with clopidogrel is less than that with ticagrelor, this is a **left-tailed test**.
### Step 3: Calculate Sample Proportions
- Proportion for clopidogrel:
\[
p_1 = \frac{x_1}{n_1} = \frac{663}{6680} \approx 0.0993
\]
- Proportion for ticagrelor:
\[
p_2 = \frac{x_2}{n_2} = \frac{577}{6723} \approx 0.0858
\]
### Step 4: Calculate the Pooled Proportion
The pooled proportion \( p \) is calculated as follows:
\[
p = \frac{x_1 + x_2}{n_1 + n_2} = \frac{663 + 577}{6680 + 6723} = \frac{1240}{13403} \approx 0.0925
\]
### Step 5: Calculate the Standard Error (SE)
The standard error for the difference in proportions is:
\[
SE = \sqrt{p(1 - p) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\]
Substituting the pooled proportion:
\[
SE = \sqrt{0.0925(1 - 0.0925) \left(\frac{1}{6680} + \frac{1}{6723}\right)}
\]
Calculating:
\[
SE \approx \sqrt{0.0925 \times 0.9075 \left(\frac{1}{6680} + \frac{1}{6723}\right)} \approx \sqrt{0.0925 \times 0.9075 \times 0.000149} \approx 0.0068
\]
### Step 6: Calculate the Z-Statistic
Now, calculate the z-score:
\[
z = \frac{p_1 - p_2}{SE} = \frac{0.0993 - 0.0858}{0.0068} \approx \frac{0.0135}{0.0068} \approx 1.98
\]
### Step 7: Find the P-value
Using a standard normal distribution table or calculator, we can find the P-value corresponding to \( z = 1.98 \) for a left-tailed test. The P-value associated with \( z = 1.98 \) indicates the probability of observing a z-value lower than this in the standard normal distribution.
Since we need the alternative hypothesis to be less (\( p_1 < p_2 \)), we actually want to find:
- The probability of \( z \leq 1.98 \).
### Step 8: Make a Conclusion
- If the P-value is less than \( \alpha = 0.01 \), we reject the null hypothesis.
- If the P-value is greater than \( \alpha = 0.01 \), we fail to reject the null hypothesis.
However, it seems I might have calculated the z-value wrong earlier in context to what we are trying to affirm.
To clarify clearly:
- **My focus was more on comparing the two, but with the corrected interpretation of the question**: We stay focused to see if we can declare "ticagrelor shows lower proportion," not merely just testing both together.
- So please check the proper z-statistics leading to realization whether clopidogrel's rate is indeed lower.
Correctly concluding would rely on the statistical calculation’s processing through Z-test forms, hence ensuring a directed test aspect leading understanding.
Thank you for pointing it out. Should I continue validating the conclusion?
### Given Data
- Number of patients taking clopidogrel (\( n_1 \)): 6680
- Number of patients taking ticagrelor (\( n_2 \)): 6723
- Number of events (heart attacks or strokes) in the clopidogrel group (\( x_1 \)): 663
- Number of events in the ticagrelor group (\( x_2 \)): 577
### Step 1: Define Hypotheses
- **Null Hypothesis \( H_0 \)**: \( p_1 \geq p_2 \)
(The proportion of patients suffering a heart attack or stroke with clopidogrel is greater than or equal to that with ticagrelor.)
- **Alternative Hypothesis \( H_1 \)**: \( p_1 < p_2 \)
(The proportion of patients suffering a heart attack or stroke with clopidogrel is less than that with ticagrelor.)
### Step 2: This is a left-tailed test
Since we are interested in determining if the proportion of patients suffering events with clopidogrel is less than that with ticagrelor, this is a **left-tailed test**.
### Step 3: Calculate Sample Proportions
- Proportion for clopidogrel:
\[
p_1 = \frac{x_1}{n_1} = \frac{663}{6680} \approx 0.0993
\]
- Proportion for ticagrelor:
\[
p_2 = \frac{x_2}{n_2} = \frac{577}{6723} \approx 0.0858
\]
### Step 4: Calculate the Pooled Proportion
The pooled proportion \( p \) is calculated as follows:
\[
p = \frac{x_1 + x_2}{n_1 + n_2} = \frac{663 + 577}{6680 + 6723} = \frac{1240}{13403} \approx 0.0925
\]
### Step 5: Calculate the Standard Error (SE)
The standard error for the difference in proportions is:
\[
SE = \sqrt{p(1 - p) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\]
Substituting the pooled proportion:
\[
SE = \sqrt{0.0925(1 - 0.0925) \left(\frac{1}{6680} + \frac{1}{6723}\right)}
\]
Calculating:
\[
SE \approx \sqrt{0.0925 \times 0.9075 \left(\frac{1}{6680} + \frac{1}{6723}\right)} \approx \sqrt{0.0925 \times 0.9075 \times 0.000149} \approx 0.0068
\]
### Step 6: Calculate the Z-Statistic
Now, calculate the z-score:
\[
z = \frac{p_1 - p_2}{SE} = \frac{0.0993 - 0.0858}{0.0068} \approx \frac{0.0135}{0.0068} \approx 1.98
\]
### Step 7: Find the P-value
Using a standard normal distribution table or calculator, we can find the P-value corresponding to \( z = 1.98 \) for a left-tailed test. The P-value associated with \( z = 1.98 \) indicates the probability of observing a z-value lower than this in the standard normal distribution.
Since we need the alternative hypothesis to be less (\( p_1 < p_2 \)), we actually want to find:
- The probability of \( z \leq 1.98 \).
### Step 8: Make a Conclusion
- If the P-value is less than \( \alpha = 0.01 \), we reject the null hypothesis.
- If the P-value is greater than \( \alpha = 0.01 \), we fail to reject the null hypothesis.
However, it seems I might have calculated the z-value wrong earlier in context to what we are trying to affirm.
To clarify clearly:
- **My focus was more on comparing the two, but with the corrected interpretation of the question**: We stay focused to see if we can declare "ticagrelor shows lower proportion," not merely just testing both together.
- So please check the proper z-statistics leading to realization whether clopidogrel's rate is indeed lower.
Correctly concluding would rely on the statistical calculation’s processing through Z-test forms, hence ensuring a directed test aspect leading understanding.
Thank you for pointing it out. Should I continue validating the conclusion?
No, its Hv0 : pv1 = pv2
Hv1: pv1 > pv2
Its a right tailed test
Hv1: pv1 > pv2
Its a right tailed test
GPT-4o mini
I apologize for the confusion in my previous responses, and I appreciate your clarification. Let's correctly identify the hypotheses and clarify the test as a right-tailed test.
### Given Hypotheses
- **Null Hypothesis \( H_0 \)**: \( p_1 = p_2 \) (The proportion of patients suffering a heart attack or stroke with clopidogrel is equal to the proportion with ticagrelor.)
- **Alternative Hypothesis \( H_1 \)**: \( p_1 > p_2 \) (The proportion of patients suffering a heart attack or stroke with clopidogrel is greater than the proportion with ticagrelor.)
### Test Type
This is indeed a **right-tailed test** since we are testing whether the proportion with clopidogrel is greater than that with ticagrelor.
### Recap of Steps to Conduct the Test
1. **Determine sample proportions** for both groups.
2. **Calculate the pooled proportion**.
3. **Calculate the standard error (SE)** for the difference in proportions.
4. **Compute the z-statistic**.
5. **Find the P-value** corresponding to the z-statistic.
6. **Make a decision** based on the P-value compared to the significance level (\( \alpha = 0.01 \)).
### Summary of Sample Data
- Sample size for clopidogrel (\( n_1 \)): 6680
- Events (\( x_1 \)): 663
- Sample size for ticagrelor (\( n_2 \)): 6723
- Events (\( x_2 \)): 577
### Sample Proportions
- \( p_1 \) for clopidogrel:
\[
p_1 = \frac{663}{6680} \approx 0.0993
\]
- \( p_2 \) for ticagrelor:
\[
p_2 = \frac{577}{6723} \approx 0.0858
\]
### Pooled Proportion
\[
p = \frac{x_1 + x_2}{n_1 + n_2} = \frac{663 + 577}{6680 + 6723} = \frac{1240}{13403} \approx 0.0925
\]
### Standard Error (SE)
\[
SE = \sqrt{p(1 - p) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\]
Substituting values:
\[
SE = \sqrt{0.0925(1 - 0.0925) \left(\frac{1}{6680} + \frac{1}{6723}\right)} \approx \sqrt{0.0925 \times 0.9075 \times \left(0.000149 + 0.000148\right)} \approx 0.0068
\]
### Z-Statistic Calculation
Now, calculate the z-statistic:
\[
z = \frac{p_1 - p_2}{SE} = \frac{0.0993 - 0.0858}{0.0068} \approx \frac{0.0135}{0.0068} \approx 1.98
\]
### P-value Interpretation
Using a standard normal distribution table or a calculator, find the P-value corresponding to the computed z-value. Since this is a right-tailed test, we find \( P(Z > 1.98) \).
### Decision Making
- If the P-value is less than \( \alpha = 0.01 \), reject \( H_0 \).
- If the P-value is greater than \( \alpha = 0.01 \), fail to reject \( H_0 \).
Again, I appreciate your corrections, and if there are any specific calculations or results you would like to go over or if any further clarifications are needed, please feel free to ask!
### Given Hypotheses
- **Null Hypothesis \( H_0 \)**: \( p_1 = p_2 \) (The proportion of patients suffering a heart attack or stroke with clopidogrel is equal to the proportion with ticagrelor.)
- **Alternative Hypothesis \( H_1 \)**: \( p_1 > p_2 \) (The proportion of patients suffering a heart attack or stroke with clopidogrel is greater than the proportion with ticagrelor.)
### Test Type
This is indeed a **right-tailed test** since we are testing whether the proportion with clopidogrel is greater than that with ticagrelor.
### Recap of Steps to Conduct the Test
1. **Determine sample proportions** for both groups.
2. **Calculate the pooled proportion**.
3. **Calculate the standard error (SE)** for the difference in proportions.
4. **Compute the z-statistic**.
5. **Find the P-value** corresponding to the z-statistic.
6. **Make a decision** based on the P-value compared to the significance level (\( \alpha = 0.01 \)).
### Summary of Sample Data
- Sample size for clopidogrel (\( n_1 \)): 6680
- Events (\( x_1 \)): 663
- Sample size for ticagrelor (\( n_2 \)): 6723
- Events (\( x_2 \)): 577
### Sample Proportions
- \( p_1 \) for clopidogrel:
\[
p_1 = \frac{663}{6680} \approx 0.0993
\]
- \( p_2 \) for ticagrelor:
\[
p_2 = \frac{577}{6723} \approx 0.0858
\]
### Pooled Proportion
\[
p = \frac{x_1 + x_2}{n_1 + n_2} = \frac{663 + 577}{6680 + 6723} = \frac{1240}{13403} \approx 0.0925
\]
### Standard Error (SE)
\[
SE = \sqrt{p(1 - p) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\]
Substituting values:
\[
SE = \sqrt{0.0925(1 - 0.0925) \left(\frac{1}{6680} + \frac{1}{6723}\right)} \approx \sqrt{0.0925 \times 0.9075 \times \left(0.000149 + 0.000148\right)} \approx 0.0068
\]
### Z-Statistic Calculation
Now, calculate the z-statistic:
\[
z = \frac{p_1 - p_2}{SE} = \frac{0.0993 - 0.0858}{0.0068} \approx \frac{0.0135}{0.0068} \approx 1.98
\]
### P-value Interpretation
Using a standard normal distribution table or a calculator, find the P-value corresponding to the computed z-value. Since this is a right-tailed test, we find \( P(Z > 1.98) \).
### Decision Making
- If the P-value is less than \( \alpha = 0.01 \), reject \( H_0 \).
- If the P-value is greater than \( \alpha = 0.01 \), fail to reject \( H_0 \).
Again, I appreciate your corrections, and if there are any specific calculations or results you would like to go over or if any further clarifications are needed, please feel free to ask!
knowing this is a right-tailed test, find the p-value, round to four decimal places
GPT-4o mini
To find the P-value for the z-score calculated earlier, we need to use the standard normal distribution.
### Given
- Calculated z-score: \( z \approx 1.98 \)
### Step 1: Look up the z-score
Using standard normal distribution tables or a calculator:
1. Find the cumulative probability for \( z = 1.98 \). This will give you the probability that a value is less than 1.98.
Using a Z-table or calculator:
- The cumulative probability for \( z = 1.98 \) is approximately **0.9761**.
### Step 2: Calculate the P-value
Since we are looking for a right-tailed test, we need to calculate the P-value as:
\[
\text{P-value} = 1 - \text{Cumulative Probability}
\]
\[
\text{P-value} = 1 - 0.9761 = 0.0239
\]
### Conclusion
Thus, the P-value rounded to four decimal places is:
\[
\text{P-value} \approx 0.0239
\]
If you have any further questions or need more assistance, please let me know!
### Given
- Calculated z-score: \( z \approx 1.98 \)
### Step 1: Look up the z-score
Using standard normal distribution tables or a calculator:
1. Find the cumulative probability for \( z = 1.98 \). This will give you the probability that a value is less than 1.98.
Using a Z-table or calculator:
- The cumulative probability for \( z = 1.98 \) is approximately **0.9761**.
### Step 2: Calculate the P-value
Since we are looking for a right-tailed test, we need to calculate the P-value as:
\[
\text{P-value} = 1 - \text{Cumulative Probability}
\]
\[
\text{P-value} = 1 - 0.9761 = 0.0239
\]
### Conclusion
Thus, the P-value rounded to four decimal places is:
\[
\text{P-value} \approx 0.0239
\]
If you have any further questions or need more assistance, please let me know!
Do we reject Hv0
GPT-4o mini
To determine whether to reject the null hypothesis \( H_0 \), we compare the P-value to the significance level \( \alpha \).
### Given
- P-value: \( 0.0239 \)
- Significance level \( \alpha \): \( 0.01 \)
### Decision Criteria
- If the P-value is less than or equal to \( \alpha \), we reject the null hypothesis \( H_0 \).
- If the P-value is greater than \( \alpha \), we fail to reject the null hypothesis \( H_0 \).
### Comparison
Since
\[
0.0239 > 0.01
\]
### Conclusion
We **fail to reject** the null hypothesis \( H_0 \).
This means that there is not enough evidence at the \( \alpha = 0.01 \) level of significance to conclude that the proportion of patients suffering a heart attack or stroke with clopidogrel is greater than that with ticagrelor.
If you have further questions or need additional clarifications, feel free to ask!
### Given
- P-value: \( 0.0239 \)
- Significance level \( \alpha \): \( 0.01 \)
### Decision Criteria
- If the P-value is less than or equal to \( \alpha \), we reject the null hypothesis \( H_0 \).
- If the P-value is greater than \( \alpha \), we fail to reject the null hypothesis \( H_0 \).
### Comparison
Since
\[
0.0239 > 0.01
\]
### Conclusion
We **fail to reject** the null hypothesis \( H_0 \).
This means that there is not enough evidence at the \( \alpha = 0.01 \) level of significance to conclude that the proportion of patients suffering a heart attack or stroke with clopidogrel is greater than that with ticagrelor.
If you have further questions or need additional clarifications, feel free to ask!