Find the $P$-value for the following values of the test statistic $t$, sample size $n$, and alternate hypothesis $H_{1}$. Use the Critical Values for the Student's t Distribution Table and specify that $P$ is between two values. Part: $0 / 2$ Part 1 of 2 (a) $t=-2.635, n=22$, and $H_{1}: \mu \neq \mu_{0}$ \[ \square

Step 1 ：Given values are test statistic \(t = -2.635\) and sample size \(n = 22\).

Step 2 ：Calculate the degrees of freedom as \(df = n - 1 = 22 - 1 = 21\).

Step 3 ：Calculate the cumulative distribution function (CDF) for the given t-statistic. The CDF at the negative t-statistic is approximately 0.00774 and at the positive t-statistic is also approximately 0.00774.

Step 4 ：The P-value for a two-tailed test is twice the CDF at the negative t-statistic, which is approximately 0.0155.

Step 5 ：This means that assuming the null hypothesis is true, there is a 1.55% chance of observing a test statistic as extreme as \(t=-2.635\) or \(t=2.635\). This is a relatively small P-value, suggesting that if the null hypothesis is true, our observed data is quite unlikely.

Step 6 ：\(\boxed{\text{Final Answer: The P-value for the given test statistic } t=-2.635, \text{ sample size } n=22, \text{ and alternate hypothesis } H_{1}: \mu \neq \mu_{0} \text{ is between 0 and 0.02. In other words, } 0