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

Answer:

The z-statistic for this test is -1.547 and P-value of the test is  0.0618

Step-by-step explanation:

An article in the Washington Post on March 16, 1993 stated that nearly 45 percent of all Americans have brown eyes.

We are given that  A random sample of n=82 C of I students found 30 with brown eyes.

n = 82

x = 30

$H_0:p=0.45\\ H_a:p\neq0.45$

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

$\widehat{p}=\frac{x}{n}$

$\widehat{p}=\frac{30}{82}$

$\widehat{p}=0.365$

Formula of test statistic =$\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

                                      =$\frac{0.365-0.45}{\sqrt{\frac{0.45(1-0.45)}{82}}}$

                                      =$-1.547$

So,  the z-statistic for this test is -1.547

Now refer the z table for p value .

p value = 0.0618

The P-value of the test is  0.0618

Hence the z-statistic for this test is -1.547 and P-value of the test is  0.0618