Homework Help: Statistics

Posted by Jim on Tuesday, March 25, 2008 at 8:55pm.

Can you please help with this question. The only thing that I have come up thus far is that I am using the z critical value because I am estimating a population proportion. Am I on the right track?

An important issue facing Americans is the large number of medical malpractice lawsuits and the expenses that they generate. In a study of 1228 randomly selected medical malpractice lawsuits, it is found that 856 of them were latter dropped or dismissed (based on data from the Physician Insurers Association of America).

Use the formulas to construct a 99% confidence interval estimate of the proportion of medical malpractice lawsuits that are dropped or dismissed. Round the point estimate and the interval limits to the nearest thousandth. Use the rounded value of the point estimate when computing your interval limits. Make sure you also state the critical value that is used to find the interval limits and interpret your interval with an appropriate statement.

Statistics - MathGuru, Saturday, March 29, 2008 at 8:42am
You can use a proportional confidence interval formula for large samples.
Here's one:
CI99 = p + or - (2.58)[√(pq/n)]
...where p = x/n; q = 1 - p; + or - 2.58 represents the 99% confidence interval using a z-table.

Substituting into the formula, we have this:
CI99 = 856/1228 + or - (2.58)[√(856/1228)(372/1228)/1228]
Convert fractions to decimals. This will make your calculations easier to do.

I hope this will help get you started.

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