Final answer:
To construct a 99% confidence interval for the percentage of Ziljac users who experienced dizziness, we use the sample proportion, the z-score for the confidence level, and the sample size in the confidence interval formula for a proportion.
Step-by-step explanation:
The subject of this question is mathematics, specifically statistics as it relates to constructing a confidence interval. The student is asked to construct a 99% confidence interval estimate for the percentage of all Ziljac users who experienced dizziness based on the results of a clinical test. Ziljac is used to treat high blood pressure, a cardiovascular condition. In the provided clinical test, 5.3% of 210 users experienced dizziness.
To calculate the confidence interval, we need to use the formula for a proportion, which is:
p ± z*sqrt[p(1-p)/n], where p is the sample proportion, z is the z-score corresponding to the confidence level, and n is the sample size.
Here, p = 0.053 (5.3%), n = 210, and the z-score for 99% confidence is approximately 2.576. Plugging in the values:
- Calculate the standard error (SE): SE = sqrt[0.053(1-0.053)/210]
- Find the margin of error (ME): ME = 2.576 * SE
- Construct the confidence interval: 0.053 ± ME
Through these steps, we will determine the 99% confidence interval for the percentage of all Ziljac users who may experience dizziness.
The complete question is: The drug Ziljac is used to treat hypertension, In a clinical test, 5.3% of 210 Zijac users experienced dizziness. Construct a 99% confidence interval estimate of the percentage of all Ziljac user who experienced dizziness is: