Answer to Question #90820 in Statistics and Probability for leonardo barrios

Question

A survey of 100 similar-sized hospitals revealed a mean daily census in the pediatrics service of 27with a standard deviation of 6.5. Do these data provide sufficient evidence to indicate that thepopulation mean is greater than 25? Let a: 0,05

Answer 1

Solution.  We define null and alternative hypothesis:

$H_0: \mu \leq 25,$

$H_a: \mu >25.$

Next we find a critical value. We will use a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

The significance level is $\alpha = 0.05$, and the critical value for a right-tailed test is 

$t_c = t(\alpha, n-1)= t(0.05, 99) = 1.66.$

The rejection region for this right-tailed test is $t > 1.66$ .

The t-statistic is computed as follows:

$t = \frac{\bar{x} - \mu}{s}\cdot \sqrt{n} = \frac{27 - 25}{6.5}\cdot \sqrt{100} = \frac{2}{6.5}\cdot 10 = 3.077.$

Since it is observed that $t = 3.077 > t_c = 1.66$, it is then concluded that the null hypothesis is rejected.

Conclusion:

It is concluded that the null hypothesis $H_0$  is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$  is greater than 25, at the 0.05 significance level.