Answer:
The null and alternative hypotheses are given below:
[tex]H_{0}:p=0.6[/tex]
[tex]H_{a}:p \neq 0.6[/tex]
Under the null hypothesis, the test statistic is:
[tex]z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p}{n} } }[/tex]
[tex]=\frac{0.52-0.60}{\sqrt{\frac{0.6(1-0.6)}{51} } }[/tex]
[tex]=-1.17[/tex]
Now, we have to find the critical values at 0.05 significance level. Using the standard normal table, the critical values are:
[tex]z_{critical} = \pm 1.96[/tex]
Conclusion:
Since the test statistic does not lie beyond the critical region, therefore, we fail to reject the null hypothesis and conclude that there is not sufficient evidence to support the claim that the hospital's success rate different from the national average.
