Question

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.

Answer 1

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

`\widehat{p}=(x)/(n)`

`\widehat{p}=(1334)/(6977)`

`\widehat{p}=0.1911`

We are given that the claim is the return rate is less than​ 20%.

`H_0:p=0.2\\H_a:p<0.2`

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

=
`\frac{0.1911-0.2}{\sqrt{(0.2(1-0.2))/(6977)}}`

=
`−1.858`

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.