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

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dfbustos dfbustos · Answer 1 · 2020-06-06T04:38:21+02:00

Answer:

[tex]z=\frac{0.05 -0.06}{\sqrt{\frac{0.06(1-0.06)}{97}}}=-0.415[/tex]

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

[tex]p_v =P(z<-0.415)=0.3409[/tex]

Step-by-step explanation:

Information given

n=97 represent the random sample taken

[tex]\hat p=0.05[/tex] estimated proportion of defective

[tex]p_o=0.06[/tex] is the value to verify

z would represent the statistic

[tex]p_v[/tex] 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:[tex]p\geq 0.06[/tex]

Alternative hypothesis:[tex]p < 0.06[/tex]

The statistic is given by:

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)

Replacing the info given we got:

[tex]z=\frac{0.05 -0.06}{\sqrt{\frac{0.06(1-0.06)}{97}}}=-0.415[/tex]

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

[tex]p_v =P(z<-0.415)=0.3409[/tex]