Question

A manufacturer claims that fewer than 6% of its fax machines are defective. In a random sample of 97 such fax machines, 5% are defective. Find the P-value for a test of the manufacturer's claim. Group of answer choices 0.3264 0.1736 0.3409 0.1591

Answer 1

Answer:

$z=\frac{0.05 -0.06}{\sqrt{\frac{0.06(1-0.06)}{97}}}=-0.415$  

Now we can find the p value with the following probability:

$p_v =P(z$  

Step-by-step explanation:

Information given

n=97 represent the random sample taken

$\hat p=0.05$ estimated proportion of defective

$p_o=0.06$ is the value to verify

z would represent the statistic

$p_v$ represent the p value

Hypothesis to tests

We want to tet if the true proportion is less than 6%, the system of hypothesis are:  

Null hypothesis:$p\geq 0.06$  

Alternative hypothesis:$p < 0.06$  

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

Replacing the info given we got:

$z=\frac{0.05 -0.06}{\sqrt{\frac{0.06(1-0.06)}{97}}}=-0.415$  

Now we can find the p value with the following probability:

$p_v =P(z$