Question

An 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=78 C of I students found 30 with brown eyes.
We test

H0: p=.45
Ha: p≠.45

What is the P-value of the test?  

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.45

H_a : p 0.45

= x / n = 30 / 78 = 0.38

P₀ = 0.45

1 - P₀ = 1 - 0.45 = 0.55

Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.38 - 0.45/ [(0.45 * 0.55) / 78]

= -1.16

P(z < -1.16 ) = 0.123

P-value = 2 * P(z < -1.16 )

P-value = 2 * 0.1230

P-value = 0.2460