Respuesta :
Answer:
Option C: fail to reject the null hypothesis; accept the shipment.
Step-by-step explanation:
The question is incomplete. It should say:
"A port inspector working for a trading company needs to decide whether to accept a large shipment of products of a certain foreign country. He will decline the shipment if there is compelling evidence that the proportion of the defective products is more than 5 percent (0.05). Suppose that a random sample of 1000 products from this shipment has yielded p=0.06. At the 5 % level of significance, what decision should the port inspector make?
A) reject the null hypothesis; reject the shipment
B) reject the null hypothesis; accept the shipment
C) fail to reject the null hypothesis; accept the shipment
D) fail to reject the null hypothesis; reject the shipment"
We have to perform an hypothesis test on proportion, with a sginificance level of 0.05.
The null and alternative hypothesis are:
[tex]H_0: \pi\leq0.05\\\\H_1:\pi > 0.05[/tex]
The standard deviation is
[tex]\sigma=\sqrt{\frac{p(1-p)}{n} }= \sqrt{\frac{0.05(1-0.05)}{1000} }=0.007[/tex]
Now we can calculate the z value:
[tex]z=\frac{p-\pi-0.5/n}{\sigma}=\frac{0.06-0.05-0.5/1000}{0.007}=\frac{0.0095}{0.007} =1.357[/tex]
For a one-side test with z=1.357, the P-value is 0.08739.
The P-value is greater than the significance level, so the null hypothesis can not be rejected.
The inspector has no compelling evidence that the proportion of defective products is more than 5%, so he has to accept the shipment.
