Respuesta :
Answer:
Given :An Internet survey was e-mailed to 6977 subjects randomly selected from an online group involved with ears. There were 1337 surveys returned.
To Find : Use a 0.01 significance level to test the claim that the return rate is less than 20%.
Solution:
n = 6977
x = 1337
We will use one sample proportion test
[tex]\widehat{p}=\frac{x}{n}[/tex]
[tex]\widehat{p}=\frac{1334}{6977}[/tex]
[tex]\widehat{p}=0.1911[/tex]
We are given that the claim is the return rate is less than 20%.
[tex]H_0:p=0.2\\H_a:p<0.2[/tex]
Formula of test statistic = [tex]\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
= [tex]\frac{0.1911-0.2}{\sqrt{\frac{0.2(1-0.2)}{6977}}}[/tex]
= [tex]−1.858[/tex]
Refer the z table
P(z<-1.85)=0.0332
Significance level = 0.01=α
Since p value > α
So, we accept the null hypothesis .
So,the claim that the return rate is less than 20% is false.
