You're conducting a significance test for h0 : p = .35, ha : p < .35. in a sample of size 40, you identify a count of 11 successes. the computed z-score and p-value are:
a. –1.06 and .1446b. –1 and .3174c. –1 and .1587d. 1 and .3174e. 1 and .8413

c. –1 and .1587

Step-by-step explanation:

As per given , we have

Null hypothesis :

Alternative hypothesis :  

Since Alternative hypothesis is left-tailed ,so the test must be a left tailed test .

Z -Test statistic for proportion =

, where p= population proportion

= sample proportions

n= Sample size.

Let x be the number of successes.

For n= 40 and x= 11

Then ,  

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