Question

A building inspector believes that the percentage of new construction with serious code violations may be even greater than the previously claimed 7%. She conducts a hypothesis test on 200 new homes and finds 23 with serious code violations. Is this strong evidence against the .07 claim?

Answer 1

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:

$H_0: \pi\leq0.07\\\\H_a: \pi>0.07$

We assume a significance level of 0.05.

The sample size is 200 and the proportion of the sample is:

$p=\frac{23}{200}= 0.115$

The standard deviation is

$\sigma=\sqrt{\frac{\pi(1-\pi)}{N} }= \sqrt{\frac{0.07*0.93}{200}}=0.018$

The z-value can be calculated as

$z=\frac{p-\pi-0.5/N}{\sigma} =\frac{0.115-0.07-0.5/200}{0.018} =\frac{0.0425}{0.018}=2.36$

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 2

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.