In a sample of 88 children selected randomly from one town, it is found that 8 of them suffer from asthma. find the p-value for a test of the claim that the proportion of all children in the town who suffer from asthma is equal to 11%

p-value= 0.5686

Explanation:

11% = 11/100 = 0.11

Null hypothesis is p = 0.11

Alternative hypothesis is p ≠ 0.11

Therefore, this test is a two tailed test.

Given,

Null parameter, P,= 0.11

Sample Statistics, Phat, = 8/88 = 1/11

= 0.09091

n = 88

z= (sample statistics - Null parameter) ÷ SE

z = (phat - p)/√[p(1-p)/n]

z = (0.09091 - 0.11)/ √[0.11(1-0.11)/88]

z = - 0.019091 ÷ √0.0011125

z = - 0.019091 ÷ 0.033354

z = - 0.572375

Z ≈ - 0.57

According to the table

z = -0.57 = 0.2843

Since it's a two tailed test,

Therefore,

p value = 2 x 0.2843 = 0.5686

b is the correct answer