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=82 C of I students found 30 with brown eyes.

We test

H0:p=.45

Ha:p≠.45

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

(b) 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

Ha : p 0.45

n =82

x =30

= x / n = 30 / 82 = 0.37

P0 = 0.45

1 - P0 = 1 - 0.45 =0.55

a ) Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

=0.37 - 0.45 / [0.45*0.55 / 80 ]

= -1.438

Test statistic = z = -1.44

b ) P-value = 2 * 0.0749 =0.1498