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

Answer:

2.49

Step-by-step explanation:

Given  X= 58

N = 175.

P = 25% = 0.25.

$P' = \frac{X}{N} = \frac{58}{175} = 0.3314\\Z = \frac{P'-P}{\sqrt{\frac{pq}{N} } }$

Now plugging the values we get

$= \frac{0.3314-0.25}{\sqrt{\frac{0.25\times0.75}{175} } }$

=2.49