Question

27) Read the ullowing and anewer o 27-31 asing A recent sudy examined 384 randomly sclected chilidn d tem aond signs of an abnormality. Des this peovide evidence that the risk p-005, Hap eot equal 005 25) 28) What would the standard ereor be for this problem A) 0.011122 5) 0.0133 D) 10-1-4) C) o.7911 29) What is the resulting test statistic? 4) 0.89 B) 0.623 C) 0.712 D) -089 What is the resulting p-value? 867 B) 0.224 C) 0.712 D) 0.1494

Answer 1

Given that: n = 384 x = 23 hat{p} = x/n = 23/384 = 0.06 To test the hypothesis, H_0: p = 0.05 H_a = p > 0.05 (claim) standard Error, SE = squareroot 0.05 times 0.95/384 = squareroot 0.011122 Test statistics z = hat{p} - p_0/SE = 0.06 - 0.05/0.011122 z = 0.89 p-value = p[z > 0.89] using z-table p-value = 0.1867