Question

(2 points) 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=76n=76 Americans found 28 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

Answer: a) z=1.43

b) 0.1528

Step-by-step explanation:

The given set of hypothesis :

$H_0:p=0.45$

$H_a:p\neq0.45$

Since the alternative hypothesis $H_a$ is two-tailed , so we perform two-tailed test.

Also, it is given that : A random sample of n=76 Americans found 28 with brown eyes.

Sample proportion: $\hat{p}=\dfrac{28}{76}=0.3684$

a) The z-statistic would be :-

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

$\Rightarrow\ z=\dfrac{0.3684-0.45}{\sqrt{\dfrac{0.45(1-0.45)}{76}}}=-1.4299121397\approx-1.43$

b) P-value for two-tailed test = 2P(Z>|z|)= 2P(z>|-1.43|)

=2P(z>1.43)

=2(1-P(z≤1.43)

=2-2P(z≤1.43)

= 2-2(0.9236)[Using standard z-table]

= 2-1.8472=0.1528

Hence,  the P-value of the test= 0.1528