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

H_{0} : p = 0.07

H_{a} : p > 0.07

n = 416

x = 31

= x / n = 31 / 416 = 0.0745

P_{0} = 0.07

1 - P_{0} = 1 - 0.07 = 0.93

z = - P_{0} / [P_{0 *} (1 -
P_{0} ) / 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

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