An extensive study of the cost of health care int e United States presented data

Question

An extensive study of the cost of health care int e United States presented data showing that the mean spending per Medicare enrolled in 2003 was $6883. To investigate differences across the country, a researcher took a sample of 40 Medicare enrollees in Indianapolis. For the Indianapolis sample, the mean 2003 Medicare spending was $5980 and the standard deviation was $2518.

- State the hypotheses that should be used if we would like to determine whether the mean annual Medicare spending in Indianapolis is lower that the national mean.

- use the preceding sample results to compute the test statistic and the p-value.

- Use a=.05 : What is your conclusion ?

- Repeat the hypthesis test using the critical value approach.

Explanation / Answer

This question asks us to test a claim and see if a different claim is more accurate(for a certain area). When testing claims, one develops two hypotheses: the null hypothesis and alternative hypothesis.
A).

The null hypothesis is what is already assumed to be true. In this case, that 5980>=6883. In words, that the Indianapolis mean spending is greater than or equal to the nationwide spending.

The alternative hypothesis is what we're trying to prove. Here, 5980<6883. In other words, spending in Indianapolis is less than nationwide spending.

B). To test this claim, we'll use the TTest(since is unknown.) The TTest uses values (,x-bar,sx,n, <>)

We know that the population mean()=6883 Sample mean(x-bar)=5980 Sample Standard Deviation(sx)=2518

N(number sampled)=40 and we're testing if X-bar<. This gives us a test statistic of -2.268 and a p-value of .01447.

C). We're asked what our conclusion is if =.05. If the p-value is higher than , we cannot prove the alternative hypothesis. If it is below , we can prove the alternative hypothesis. .01447<.05, therefore we reject the null hypothesis and say that we can prove that average spending in Indianapolis is less than average nationwide spending.

D). To use the critical value approach, we first find the critical value by finding Critical T. This t is based off 1- and may be given to you or in a table. Critical T=-1.96 for .95 confidence level. Since, t is lower than the critical, we reject the null hypothesis and say that we can prove that average spending in Indianapolis is less than average nationwide spending.

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