Question

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6969 subjects randomly selected from an online group involved with ears. There were 1326 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

As we are testing the claim that  the return rate is less than​ 20%, therefore the null and alternate hypothesis here are given as:

The sample proportion of surveys returned here is computed as:

p = 1326/6969 = 0.1903

The test statistic is now computed as:

As this is a one tailed test, the p-value here is computed from the standard normal tables as:

p = P( Z < -0.8830) = 0.1886

As the p-value here is 0.1886 > 0.01 which is the level of significance, therefore the test is not significant and we cannot reject the null hypothesis here that the proportion is not less than 0.2.