Suppose that we are testing the claims $H_{o}: \mu_{1}=\mu_{2}, H_{1}: \mu_{1} \neq \mu_{2}$ and our sample produces $n=26, \bar{d}=7.832, s_{d}=35.822$. Calculate the P-Value for the test. Round your answer using 4 decimal places.

Step 1 ：We are testing the claims \(H_{o}: \mu_{1}=\mu_{2}\), \(H_{1}: \mu_{1} \neq \mu_{2}\) and our sample produces \(n=26\), \(\bar{d}=7.832\), \(s_{d}=35.822\). We need to calculate the P-Value for the test.

Step 2 ：The P-value is the probability that, if the null hypothesis were true, we would observe a more extreme test statistic in the direction of the alternative hypothesis than we did. So, we need to calculate the test statistic first, which is a t-score in this case because we are dealing with means and we know the standard deviation.

Step 3 ：The formula for the t-score is: \[t = \frac{\bar{d} - \mu_{0}}{s_{d}/\sqrt{n}}\] where \(\bar{d}\) is the sample mean, \(\mu_{0}\) is the hypothesized population mean under the null hypothesis, \(s_{d}\) is the sample standard deviation, and \(n\) is the sample size. In this case, \(\mu_{0}\) is 0 because we are testing the hypothesis that the means are equal, so the difference would be 0.

Step 4 ：After calculating the t-score, we can find the P-value using the t-distribution. The P-value is the probability of observing a t-score as extreme as, or more extreme than, the observed t-score, under the null hypothesis. Because the alternative hypothesis is two-sided (\(\mu_{1} \neq \mu_{2}\)), we need to find the two-tailed P-value. This is twice the probability of observing a t-score as extreme as the observed t-score in the right tail of the distribution.

Step 5 ：Given that \(n = 26\), \(\bar{d} = 7.832\), \(s_{d} = 35.822\), the calculated t-score is approximately 1.1148.

Step 6 ：The calculated P-value is approximately 0.2755.

Step 7 ：Final Answer: The P-Value for the test is \(\boxed{0.2755}\).