Question

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6955 subjects randomly selected from an online group involved with ears. There were 1307 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.

The test statistic is

z = ____

​(Round to two decimal places as​ needed.)

What is the​ P-value?

​P-value =____

​(Round to three decimal places as​ needed.)

Answer 1

Solution :

This is the left tailed test .

The null and alternative hypothesis is

H₀ : p = 0.20

H_a : p < 0.20

= x / n = 1307 / 6955 = 0.1879

P₀ = 0.20

1 - P₀ = 0.80

Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.1879 - 0.20 / [(0.20 * 0.80) / 6955]

= -2.52

P(z < -2.52) = 0.0059

P-value = 0.006