Question

An extensive study of the cost of health care in the U.S. showed that the mean annual spending per Medicare enrollee was $6,883. To investigate the differences across the country, a researcher took a sample of 40 Medicare enrollees in Indianapolis. For the Indianapolis sample, the mean annual Medicare spending was $5,980 and the standard deviation was $2,518. Using a level of significance of 3%, test to see if the mean annual Medicare spending in Indianapolis is lower than the national mean. Interpret your result.

Answer 1

Solution :

Given :

$\mu = 6883$

$s=2518$

n = 40

$\overline x = 5980$

Therefore, determining the null hypothesis and alternate hypothesis is :

$H_0: \mu = 6883$

$H_a: \mu < 6883$

The value of the test statistics is :

$t=\frac{\overline x - \mu}{s/\sqrt n}$

$t=\frac{5980 - 6883}{2518/\sqrt 40}$

$t \approx -2.27$

So, df =  n - 1

          = 40 - 1

          = 39

∴  0.01 < P < 0.025

If the P value is less than α = 0.03, we reject the null hypothesis. Therefore, we conclude that mean annual Medicare spending in Indianapolis is lower than the national mean.

Using the value of $\alpha = 0.03$, the critical value of for the test statistics is 2.17

There we reject the null hypothesis. And thus we conclude that the mean annual Medicare spending in Indianapolis is lower than the national mean.