91Ó°ÊÓ

It is given that claim is greater than $37\%$.

$\mathrm{Î\pm }=0.05$

$n=200$

$x=83$

The claim is null or alternate hypothesis.

Null hypothesis: $H_0:p=37\%=0.37$

Alternate Hypothesis: $H_1:p>0.37$

Once null hypothesis is true, type I error rejects the null hypothesis:

Evidence is present that students who approve of new parking is $>0.37$, when students who approve parking is actually $0.37$

Conclusion is change was effective when it was actually not effective and it may lead to wastage of money.

Once null hypothesis is false, type II error fails to reject it:

No evidence is present that students who approve of new parking is really larger than $0.37$.

Change is not effective. When it was, parking is not improved.

The condition of normality is satisfied as:

$n{p}_0=200(0.37)=74$and $n\left(1-p_0\right)=200(1-0.37)=126$. Both are large than $10$.

Conditions are satisfied, we can use hypothesis test.

Sample proportion is $\hat{p}=\frac{x}{n}=\frac{83}{200}=0.415$

Test static: $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}=\frac{0.415-0.37}{\sqrt{\frac{0.37(1-0.37)}{200}}}=1.32$

$P$value is $P=P(z>1.32)$

$=1-P(Z<1.32)=1-0.9066=0.0934$

Hence, $P>0.05⇒\text{Fail to reject}{H}_0$.

No evidence that principal's claim is true.