Statistics
Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. In 1999, the average LOS for men in the United States was 5.4 days and the average for women was 4.7 days (Statistical Abstract of the United States: 2001). A random sample of 20 hospitals in one state had a mean LOS for women in 2003 of 3.8 days and a standard deviation of 1.2 days. a. Use a 90% confidence interval to estimate the population mean LOS for women for the state's hospitals in 2003. b. Interpret the interval in terms of this application. c. What is meant by the phrase “90% confidence interval”?
Hello Farrah, I notice that you have sent a flurry of questions to Ask NDSL. Perhaps you have homework due tomorrow? This one is very much like the other two that are still available. So if I help you with this one, you ought to be able to do the others. Let's see...a 3.8 day mean with a 1.2 day standard deviation. If we assume a normal distribution, then the 95th percentile is mu + 1.64sigma, while the 5th percentile is mu - 1.64 sigma. With respect to your problem, that translates to a 90% confidence interval of [1.826,5.774]. This interval includes 4.7, so it is quit possible that there has been no actual decrease at all. At the same time, take a closer look at the question. You are comparing the entire United States versus one state. The one state you are looking at may not be representative of the nation as a whole. 90% confidence interval -- you should look it in your own textbook, but it is the best guess on what the true value is based on the available data. Yes, there is a true value known to God with no uncertainty (and you could construct paradoxically overlapping confidence intervals.) Hope this helps, Joe McCollum Information Technology Specialist Forest Inventory and Analysis Knoxville, TN