You wish to test the following claim $\left(H_{a}\right)$ at a significance level of $\alpha=0.01$.
\[
\begin{array}{l}
H_{o}: \mu=74.9 \\
H_{a}: \mu< 74.9
\end{array}
\]

You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size $n=17$ with mean $\bar{x}=59.2$ and a standard deviation of $s=20.2$.

What is the p-value for this sample? (Report answer accurate to four decimal places.)
\[
p \text {-value }=
\]

The p-value is...
less than (or equal to) $\alpha$
greater than $\alpha$

This p-value leads to a decision to...
reject the null
accept the null
fail to reject the null

Step 1 ：Given the following hypotheses: \(H_{o}: \mu=74.9\) and \(H_{a}: \mu<74.9\), with a significance level of \(\alpha=0.01\).

Step 2 ：We have a sample of size \(n=17\) with mean \(\bar{x}=59.2\) and a standard deviation of \(s=20.2\).

Step 3 ：First, we calculate the t-score using the formula: \(t = \frac{\bar{x} - \mu}{s / \sqrt{n}}\).

Step 4 ：Substituting the given values into the formula, we get: \(t = \frac{59.2 - 74.9}{20.2 / \sqrt{17}} \approx -3.2046\).

Step 5 ：Next, we calculate the p-value. The p-value is the probability of observing a t-score as extreme as the one calculated (or more extreme) under the null hypothesis.

Step 6 ：Using a t-distribution table or a statistical software, we find that the p-value is approximately 0.0028.

Step 7 ：Since the p-value is less than the significance level \(\alpha=0.01\), we reject the null hypothesis.

Step 8 ：Final Answer: The p-value for this sample is approximately \(\boxed{0.0028}\). Since the p-value is less than the significance level \(\alpha=0.01\), we \(\boxed{reject the null hypothesis}\).