Question

1. the athletic director believes the average vertical jump height of students is 20 inches. a coach thinks students have been training harder and the true average is greater than 20 inches. \\( h_0: mu = 20 \\) \\( h_a: mu > 20 \\) evidence: a random sample of 36 athletes has a sample mean of 22.4 inches with a standard deviation of 6 inches. \\( z = square \\) unusualness (z - score) = \\( square \\) (three decimal places) chances: p - value = \\( square \\) (three decimal places)

Explanation:

Step1: Calculate z-score

Given $\mu=20$, $\bar{x}=22.4$, $\sigma=6$, $n=36$.
Standard error: $\frac{\sigma}{\sqrt{n}}=\frac{6}{\sqrt{36}}=1$.
$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{22.4-20}{1}=2.4$.

Step2: Find unusualness (z-score)

Unusualness is the absolute z-score, here $2.4$.

Step3: Calculate P-value

Right-tailed test: P-value = $P(Z>2.4)$.
From z-table, $P(Z\leq2.4)=0.9918$, so $P(Z>2.4)=1-0.9918=0.0082\approx0.008$.

Answer:

$z = 2.4$
Unusualness (z) = 2.4
Chances (P-value) = 0.008