Question

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6989 subjects randomly selected from an online group involved with ears. There were 1328 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 :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.20

H_a : p < 0.20

= x / n = 1328 / 6989 = 0.1900

Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.1900 - 0.20 / [(0.20 * 0.80) / 6989]

= -2.087

P-value = 0.0184

= 0.01

P-value >

Fail to reject the null hypothesis .

There is not sufficient evidence to suggest that the rate is less than​ 20% .