Question

According to the American Academy of Ophthalmology, 25% of Americans have brown irises. A random survey of students at a local community college found that 58 out of 175 had brown irises. What is the test statistic (ie the Z score) for this sample of students

Answer 1

2.49

Explanation:

Given X= 58

N = 175.

P = 25% = 0.25.

`P' = (X)/(N) = (58)/(175) = 0.3314\\Z = \frac{P'-P}{\sqrt{(pq)/(N) } }`

Now plugging the values we get

`= \frac{0.3314-0.25}{\sqrt{(0.25*0.75)/(175) } }`

=2.49