Respuesta :
Answer:
The inspector's claim has strong statistical evidence.
Step-by-step explanation:
To answer this we have to perform a hypothesis test.
The inspector claimed that the actual proportion of code violations is greater than 0.07, so the null and alternative hypothesis are:
[tex]H_0: \pi\leq0.07\\\\H_a: \pi>0.07[/tex]
We assume a significance level of 0.05.
The sample size is 200 and the proportion of the sample is:
[tex]p=\frac{23}{200}= 0.115[/tex]
The standard deviation is
[tex]\sigma=\sqrt{\frac{\pi(1-\pi)}{N} }= \sqrt{\frac{0.07*0.93}{200}}=0.018[/tex]
The z-value can be calculated as
[tex]z=\frac{p-\pi-0.5/N}{\sigma} =\frac{0.115-0.07-0.5/200}{0.018} =\frac{0.0425}{0.018}=2.36[/tex]
The P-value for this z-value is P=0.00914.
This P-value is smaller than the significance level, so the effect is significant and the null hypothesis is rejected.
The inspector's claim has strong statistical evidence.
Answer:
Yes, because the p-value is 0.0062
Step-by-step explanation:
I looked it up on like 5 other websites and they all said this was the answer. I'm way too lazy to do it on my own.
