You wish to test the following claim $\left(H_{a}\right)$ at a significance level of $\alpha=0.01$.
\[
\begin{array}{l}
H_{o}: \mu=79.5 \\
H_{a}: \mu \neq 79.5
\end{array}
\]
You believe the population is normally distributed and you know the standard deviation is $\sigma=20.7$ . You obtain a sample mean of $M=70.2$ for a sample of size $n=26$.
What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic $=$
What is the $p$-value for this sample? (Report answer accurate to four decimal places.) $p$-value $=$
The $p$-value is...
less than (or equal to) $\alpha$
greater than $\alpha$
This test statistic leads to a decision to...
reject the null
accept the null
fail to reject the null

Step 1 ：First, we calculate the test statistic using the formula: \(Z = \frac{M - \mu}{\sigma / \sqrt{n}}\). Here, \(M = 70.2\), \(\mu = 79.5\), \(\sigma = 20.7\), and \(n = 26\).

Step 2 ：Substituting the given values into the formula, we get: \(Z = \frac{70.2 - 79.5}{20.7 / \sqrt{26}}\).

Step 3 ：Calculating the above expression, we get: \(Z = -2.178\) (rounded to three decimal places). So, the test statistic is \(\boxed{-2.178}\).

Step 4 ：Next, we calculate the p-value. The p-value is the probability that a standard normal random variable is less than -2.178 or greater than 2.178.

Step 5 ：Using a standard normal table or a calculator, we find that the probability that a standard normal random variable is less than -2.178 is approximately 0.0148.

Step 6 ：Since the test is two-tailed, we multiply this probability by 2 to get the p-value. So, the p-value is \(2 \times 0.0148 = 0.0296\) (rounded to four decimal places). So, the p-value is \(\boxed{0.0296}\).

Step 7 ：Since the p-value (0.0296) is greater than the significance level (0.01), we fail to reject the null hypothesis \(H_{o}\).