Question

example
the medical rehabilitation education foundation reports that the average cost of rehabilitation for stroke victims is $24,672. to see if the average cost of rehabilitation is different at a particular hospital, a researcher selects a random sample of 35 stroke victims at the hospital and finds that the average cost of their rehabilitation is $26,343. the standard deviation of the population is $3251. at α = 0.01, can it be concluded that the average cost of stroke rehabilitation at a particular hospital is different from $24,672?

Explanation:

Step1: State the hypotheses

$H_0:\mu = 24672$ (null hypothesis, where $\mu$ is the population mean cost of rehabilitation)
$H_1:\mu
eq24672$ (alternative hypothesis)

Step2: Calculate the test - statistic

The formula for the $z$ - test statistic when the population standard deviation $\sigma$ is known is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean under the null hypothesis, $\sigma$ is the population standard deviation, and $n$ is the sample size.

Given $\bar{x} = 26343$, $\mu = 24672$, $\sigma=3251$, and $n = 35$

First, calculate $\frac{\sigma}{\sqrt{n}}=\frac{3251}{\sqrt{35}}\approx\frac{3251}{5.916}\approx550.9$

Then, $z=\frac{26343 - 24672}{550.9}=\frac{1671}{550.9}\approx3.03$

Step3: Determine the critical values

For a two - tailed test with $\alpha=0.01$, the critical values are $z_{\alpha/2}=\pm z_{0.005}$. From the standard normal table, $z_{0.005}=\pm 2.576$

Step4: Make a decision

Since the calculated $z$ - value ($z = 3.03$) is greater than the positive critical value ($z_{0.005}=2.576$), we reject the null hypothesis.

Answer:

Yes, it can be concluded that the average cost of stroke rehabilitation at the particular hospital is different from $\$24,672$.