Problem
In clinical trials of a medication used to treat arthritis, 67% of users reported dizziness as a side effect out of the 100 trial participants. a. The z-score would be: b. The margin of error at a 90% confidence level is: c. Construct a 90%confidence interval for the proportions of users that experience dizziness on this medication. d. The lowerbound of the confidence interval is: and the upperbound of the confidence interval is e. Interpret the confidence interval from part d. f. (Extra Credit) Based on the results of the confidence interval, can you conclude more than $75.71 % of users would experience dizziness? Why or why not?
Solution
Step 1 ：Given that the proportion (p) of users who reported dizziness as a side effect is 0.67 and the sample size (n) is 100.
Step 2 ：The standard deviation of a proportion is calculated using the formula \(\sqrt{p(1-p)/n}\). Substituting the given values, we get \(\sqrt{0.67(1-0.67)/100} = 0.04702127178203499\).
Step 3 ：The z-score is calculated using the formula \((x - μ) / σ\), where x is the value of the element, μ is the mean, and σ is the standard deviation. In this case, since we are dealing with proportions, the mean is the proportion itself, so the z-score would be \((0.67 - 0.67) / 0.04702127178203499 = 0\).
Step 4 ：Final Answer: The z-score is \(\boxed{0}\).
