Medical Malpractice In a study of 1228 randomly selected medic...

Question

Medical Malpractice In a study of 1228 randomly selected medical malpractice lawsuits, it was found that 856 of them were dropped or dismissed (based on data from the Physicians Insurers Association of America). Use the bootstrap method to construct a 95% confidence interval estimate of the proportion of lawsuits that are dropped or dismissed. Use 1000 bootstrap samples. How does the result compare to the confidence interval found in Exercise 16 “Medical Malpractice” from Section 7-1?

Accepted Answer

Welcome back, everyone. In this problem, a call center manager wants to estimate the true proportion of incoming calls that are abandoned. That is when a caller hangs up before reaching an agent. He takes a simple random sample of 500 calls and finds that 150 were abandoned. Using 100 bootstrap replicates, she obtains the 2.5th and 97.5th percentiles of the bootstrap abandonment proportions as 0.2667 and 0.3333 respectively. Compute the standard normal approximation 95% confidence interval and compare it to the bootstrap result. Now, by the percentile method, OK, the 95% confidence interval would be directly. The between the 2.5 and 97.5th percentile, OK? And here we're told that by the bootstrap replicates, then the, those percenters have the bootstrap abandonment proportions as 0.2667 and 0.3333 respectively, OK? So we can say then that by bootstrap. Or by the bootstrap confidence interval. OK. Then We have our proportion between 0.2667. And 0.3333. Now, let's use our normal approximation confidence interval and then compare it to that. So by OK, the normal approximation. Confidence interval. OK, there are a few things that we need. First, we need our point estimate, OK. And a point estimate It's going to be our proportion, OK, which is equal to X divided by N. In this case, it was the 150 abundant from the 500 random calls, OK, which is equal to 0.3. Next, we need the critical value for a 95% confidence interval. So for a 95% confidence, OK. Then the critical value is going to be the Z value or half of alpha, which is, since it's 0. Well, since it's 5%, then half of that would be 2.5% or 0.025. And for 95% confidence, we get that Z value to be approximately equal to 1.96%. Next, we need to calculate our standard error. And the standard error would be equal to the square root of our point estimate multiplied by 1 minus the point estimate divided by n. That's equal to 0.3, multiplied by 1 minus 0.3, 0.7 divided by 500. Which means our standard error is going to be approximately 0.02049. Therefore, the margin of error. It's gonna be equal to or Z value, OK. Multiplied by the standard error, so that's why we needed both of those. So that's gonna be 1.96 multiplied by 0.02 049. That's gonna be approximately 0.040168, OK? Know that we have this margin of error, then this implies that our confidence interval from our normal approximation is going to be the point estimate plus R minus the error. So that's 0.3 plus R minus 0.040168. Which means then that the lower bound is 0.3 minus 0.040168, which is 0.2598, while the upper bound is 0.3 plus the error, which is 0.3402. OK? So for a 95% confidence interval. OK, therefore, for 95% confidence interval, then P is gonna be between 25.98%. And 34.02%. Now if we compare that to our previous interval between 26.67% and 33.33%, then the two intervals are almost identical, OK? So that indicates the normal approximation works well here and the bootstrap provides a useful confirmation. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Bootstrap Method

Confidence Interval

Proportion

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