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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

9 answers

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.
Incorrect

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?

No, its Hv0 : pv1 = pv2
Hv1: pv1 > pv2

Its a right tailed test

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!

knowing this is a right-tailed test, find the p-value, round to four decimal places

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!

Do we reject Hv0

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!