You wish to test the following claim $\left(H_{a}\right)$ at a significance level of $\alpha=0.02$. \[ \begin{array}{l} H_{o}: \mu=62.3 \\ H_{a}: \mu \neq 62.3 \end{array} \] You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size $n=12$ with mean $M=70.2$ and a standard deviation of $S D=7.6$. 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.) $\mathrm{p}$-value $=$

Step 1 ：We are given the following hypotheses: \(H_{o}: \mu=62.3\) and \(H_{a}: \mu \neq 62.3\). The significance level is \(\alpha=0.02\).

Step 2 ：We are also given a sample of size \(n=12\) with mean \(M=70.2\) and a standard deviation of \(SD=7.6\).

Step 3 ：We can calculate the test statistic using the formula: \[t = \frac{M - \mu}{SD / \sqrt{n}}\] where \(M\) is the sample mean, \(\mu\) is the population mean, \(SD\) is the sample standard deviation, and \(n\) is the sample size.

Step 4 ：Substituting the given values into the formula, we get: \[t = \frac{70.2 - 62.3}{7.6 / \sqrt{12}}\]

Step 5 ：After calculating, we find that the test statistic is approximately 3.601.

Step 6 ：We can find the p-value using a t-distribution table or a statistical software. Since we are doing a two-tailed test (because \(H_a: \mu \neq 62.3\)), the p-value is the probability of getting a result as extreme as our test statistic in either tail of the distribution.

Step 7 ：After calculating, we find that the p-value is approximately 0.0042.

Step 8 ：Final Answer: The test statistic for this sample is approximately \(\boxed{3.601}\) and the p-value for this sample is approximately \(\boxed{0.0042}\).