Question

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

Answer 1

Answer: c. –1 and .1587

Step-by-step explanation:

As per given , we have

Null hypothesis : $H_0 : p= 0.35$

Alternative hypothesis :  $H_1 : p< 0.35$

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

Z -Test statistic for proportion = $z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

, where p= population proportion

$\hat{p}$ = sample proportions

n= Sample size.

Let x be the number of successes.

For n= 40 and x= 11

$\hat{p}=\dfrac{x}{n}=\dfrac{11}{40}=0.275$

Then ,  $z=\dfrac{0.275-0.35}{\sqrt{\dfrac{0.35(1-0.35)}{40}}}$

$z=\dfrac{-0.075}{\sqrt{0.0056875}}$

$z=\dfrac{-0.075}{0.075415515645}$

$z=-0.99449031619\approx-1$

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