Question

Xan article in the washington post on march 16, 1993 stated that nearly 45 percent of all americans have brown eyes. a random sample of n=80 c of i students found 30 with brown eyes. we test h0:p=.45 ha:p?.

Answer 1

(a) What is the z-statistic for this test?

• z = -1.35

(b) What is the P-value of the test?

• P-value = 0.0885

Step-by-step explanation:

n = 80

x = 30

H₀:p = 0.45

Hₐ:p ≠ 0.45

Now to find z statistic we will use one sample proportion test

`\widehat{p}\\` = X / n = 30 / 80 = 0.375

z = (
`\widehat{p}\\` - p) / √{[p(1 - p)] / n}

z = (0.375 - 0.45) / √{[0.45(1 - 0.45)] / 80} = -1.3484 or -1.35

P-value for z value for -1.35 (using table) = 0.0885

z = -1.35 and P-value = 0.0885