Question

What is P-value?

It the​ 1980s, it was generally believed that congenital abnormalities affected about

7

​%

of a large​ nation's children. Some people believe that the increase in the number of chemicals in the environment has led to an increase in the incidence of abnormalities. A recent study examined

416

randomly selected children and found that

31

of them showed signs of an abnormality. Is this strong evidence that the risk has​ increased? (We consider a​ P-value of around

5

​%

to represent reasonable​ evidence.) Complete parts a through f. Assume the independence assumption is met.

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.07

H_a : p > 0.07

n = 416

x = 31

= x / n = 31 / 416 = 0.0745

P₀ = 0.07

1 - P₀ = 1 - 0.07 = 0.93

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.0745 - 0.07 / [(0.07 * 0.93) / 416]

= 0.36

This is the right tailed test .

P(z > 0.36) = 1 - P(z < 0.36) = 1 - 0.6406 = 0.3594

P-value = 0.3594

= 0.05

P-value >

Fil to reject the null hypothesis .

This is not strong evidence that the risk has​ increased

There is no evidence to suggest that