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

walter needs to make at least 5 more cookies but no more than 40. 5 ≥ x ≥ 40;

step-by-step explanation:

walter has to make a minimum of 20 cookies. he already has 15 cookies made, so he must make 5 more to meet the minimum.

the maximum amount of cookies he can make is 55. he already has 15, so he has to make no more than 40 more cookies to meet the maximum.

2nd one in the first row is correct