Question

A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of 200 of the over 2500 students at the school. In all, 83 students say that they approve of the new parking arrangement. The principal cites this as evidence that the change was effective. Perform a test of the principal’s claim at the $\alpha$=0.05 significance level.

Short answer

There is not enough evidence to conclude the principal’s claim at$5\%$significance level.

Step-by-step solution

Given Information

Percentage of students approved of the parking that was provided = $37\%$

After the change, the principal surveys an SRS of $200$of the over $2500$students at the school.

The number of students that approve of the new parking arrangement = $80$

Explanation

The number of students is n = $200$

The number of students that approve of the new parking arrangement is x = $83$

Population proportion is p = role="math" localid="1654320587832" $37\%=0.37$

$p_0=1-p=1-0.37=0.63$

Calculating, $\stackrel{-}{p}=\frac{x}{n}=\frac{83}{200}=0.415$

The null and alternative hypotheses,

$\begin{array}{r}{H}_0:p=0.37\\ H_a:p>0.37\end{array}$

Using,

role="math" localid="1654321022197" $z=\frac{\stackrel{-}{p}-p_0}{\sqrt{\frac{P_0\left(1-p_0\right)}{n}}}=\frac{0.415-0.63}{\sqrt{{\displaystyle \frac{0.63(0.37)}{200}}}}=0.0937$

Hence there is not enough evidence to conclude the principal’s claim at $5\%$as the p-value is less than the significance level$\alpha =0.05$