Solution20:
given to test for increase in abnormatlities or not
H0:p=0.05
Ha:p>0.05
MARK OPTION A
Solution21:
standard error=sqrt(p*(-p)/n
=(0.05*(1-0.05)/384
=0.011122
MARK OPTION A
Solution22:
z=p^-p/sqrt(p*(1-p)/n
p^=23/384=0.05989583
z=0.05989583-0.05/sqrt(0.05*(1-0.05)/384)
z=0.89
MARK OPTION A
0.89
Solution23:
p=1-NORM.S.DIST(0.89;1)
p=0.1867
ANSWER:
0.1867
MARK OPTION A:
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...