Answer: c. –1 and .1587
Step-by-step explanation:
As per given , we have
Null hypothesis : [tex]H_0 : p= 0.35[/tex]
Alternative hypothesis : [tex]H_1 : p< 0.35[/tex]
Since Alternative hypothesis is left-tailed ,so the test must be a left tailed test .
Z -Test statistic for proportion = [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
, where p= population proportion
[tex]\hat{p}[/tex] = sample proportions
n= Sample size.
Let x be the number of successes.
For n= 40 and x= 11
[tex]\hat{p}=\dfrac{x}{n}=\dfrac{11}{40}=0.275[/tex]
Then , [tex]z=\dfrac{0.275-0.35}{\sqrt{\dfrac{0.35(1-0.35)}{40}}}[/tex]
[tex]z=\dfrac{-0.075}{\sqrt{0.0056875}}[/tex]
[tex]z=\dfrac{-0.075}{0.075415515645}[/tex]
[tex]z=-0.99449031619\approx-1[/tex]
By using z-table ,
P-value for left-tailed test = P(z<-1)= 1-P(z<1) [∵ P(Z<-z)= 1-P(Z<z) ]
= 1-0.8413
=0.1587
Hence, the -score and P-value are –1 and 0.1587 .
So the correct option is c. –1 and .1587
