Question

Previously, 6​% of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less than 6​% today. She randomly selects 150 pregnant mothers and finds that 7 of them smoked 21 or more cigarettes during pregnancy. Test the​ researcher's statement at the alpha equals 0.1 level of significance.

Answer 1

p= 6​% =0.06​, n=150, x=7, = 0.1

Ho: P = 0.06

H1: P < 0.06

formula for test statistics is

Z= -0.6876

Calculate P-Value

P-Value = P(Z < -0.6876 )

find P(Z < -0.6876 ) using normal z table we get

P(Z < -0.6876 ) = 0.2458

P-Value = 0.2458

Now decision rule is

if(P-Value) () then Reject Ho

here

(P-Value=0.2458) > (=0.1)

hence Do not Reject Ho

Therefore there is not suficient evidence to claim that  the percentage of mothers who smoke 21 cigarettes or more is less than 6​% today.