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

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

Formula of test statistic =

                                       =

                                       =

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.

it has the same slope and a different y-intercept

hope this : )