Respuesta :
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
[tex]H_0:p=0.45\\ H_a:p\neq0.45[/tex]
Now to find z statistic we will use one sample proportion test
[tex]\widehat{p}=\frac{x}{n}[/tex]
[tex]\widehat{p}=\frac{30}{82}[/tex]
[tex]\widehat{p}=0.365[/tex]
Formula of test statistic =[tex]\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
=[tex]\frac{0.365-0.45}{\sqrt{\frac{0.45(1-0.45)}{82}}}[/tex]
=[tex]-1.547[/tex]
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
