In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6977subjects randomly selected from an online group involved with ears. There were 1337surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than 20%. Use the P-value method and use the normal distribution as an approximation to the binomial distribution.Identify the null hypothesis and alternative hypothesis.

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.