Answer:
0.0594.
Step-by-step explanation:
The standard error of a proportion p, in a sample of size n, is given by:
[tex]SE_{p} = \sqrt{\frac{p(1-p)}{n}}[/tex]
In a sample of 62 horses with enteroliths, she finds 42 are fed two or more flakes of alfalfa.
This means that [tex]n = 62, p = \frac{42}{62} = 0.6774[/tex]
So
[tex]SE_{p} = \sqrt{\frac{p(1-p)}{n}}[/tex]
[tex]SE_{p} = \sqrt{\frac{0.6774*0.3226}{62}} = 0.0594[/tex]
So the correct answer is:
0.0594.