Answer:
The standard error is 0.02849
Step-by-step explanation:
Explanation:-
given data is 42% of primary case doctors think their patients receive un-necessary medical care.
That is The proportion 'p' = 42% = 0.42
Given sample size is n =300
The standard error of the sampling distribution of the sample proportion is
[tex]se (p) = \frac{\sqrt{p(1-p)} }{\sqrt{n} }[/tex]
[tex]se (p) = \frac{\sqrt{0.42(1-0.42))} }{\sqrt{300} }[/tex]
use calculator on simplification , we get
standard error = 0.02849