Answer to Question #350962 in Statistics and Probability for sergu

Question

A medical centre reports that the average cost of rehabilitation for stroke victims is 24672 BD. To see if the average cost of rehabilitation is different at a particular hospital, a researcher selects 35 victims at random and find the average cost is 26343 BD. Standard deviation is 3251 BD. At α=0.01, can it be concluded that the average cost of rehabilitation at a particular hospital is different form 24672 BD

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=24672$

$H_1:\mu\not=24672$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=34$ and the critical value for a two-tailed test is $t_c =2.728394.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.728394\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{26343-24672}{3251/\sqrt{35}}=3.0408$$

Since it is observed that $|t|=3.0408>2.728394=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=34$ degrees of freedom, $t=3.0408$ is $p= 0.004521,$ and since $p=0.004521<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 24672, at the $\alpha = 0.01$ significance level.