Step 1 :We are given a sample size of 805 medical malpractice lawsuits, out of which 484 were dropped or dismissed. We are to test the claim that most medical malpractice lawsuits are dropped or dismissed at a 0.01 significance level.
Step 2 :The null hypothesis is that the proportion of dropped or dismissed lawsuits is 0.5 (or 50%), and the alternative hypothesis is that the proportion is greater than 0.5.
Step 3 :We calculate the sample proportion (\(\hat{p}\)) as the number of successes divided by the sample size, which is \(\frac{484}{805} = 0.6012422360248447\).
Step 4 :The hypothesized population proportion (\(p_0\)) is 0.5, and the sample size (\(n\)) is 805.
Step 5 :We use the formula for the z-score: \[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\]
Step 6 :Substituting the values into the formula, we get a z-score of approximately 5.74.
Step 7 :This means that the sample proportion of dropped or dismissed lawsuits is approximately 5.74 standard deviations above the hypothesized population proportion of 0.5.
Step 8 :This is a very high z-score, which suggests that the null hypothesis (that the proportion of dropped or dismissed lawsuits is 0.5) is likely to be rejected.
Step 9 :Final Answer: The test statistic is \(\boxed{5.74}\).