Question

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6985 subjects randomly selected from an online group involved with ears. There were 1295 surveys 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.

Answer 1

Solution:

Given:

n = Number of  subjects randomly selected from an online group involved with ears = 6985

x = Number of surveys returned = 1295

Significance level = 0.01

Claim: The return rate is less than​ 20%.

Step 1) State H0 and H1:

H0: p = 0.20

Vs

H1: p < 0.20

Step 2) Test statistic:

where

thus

Step 3) Find P-value:

For left tailed test, P-value is given by:

P-value = P( Z < z test statistic)

P-value = P( Z < -3.05)

Look in z table for z = -3.0 and 0.05 and find corresponding area.

P( Z< -3.05) = 0.0011

thus

P-value = P( Z < -3.05)

P-value =0.0011

Step 4) Decision Rule:
Reject H0, if P-value < 0.01 level of significance, otherwise we fail to reject H0.

Since P-value =0.0011 < 0.01 level of significance, we reject null hypothesis.

Step 5) Conclusion:

At  0.01 significance level , we have sufficient evidence to support the claim that the return rate is less than​ 20%.