Step 1 :Given the sample results are \(\bar{x}_{1}=76.8, s_{1}=11.3, n_{1}=25\) and \(\bar{x}_{2}=62.2, s_{2}=8.0, n_{2}=20\).
Step 2 :The best estimate for the difference in means \(\mu_{1}-\mu_{2}\) is the difference in sample means \(\bar{x}_{1}-\bar{x}_{2}\), which is \(76.8 - 62.2 = 14.6\).
Step 3 :The margin of error can be calculated using the formula for the confidence interval of the difference in means for a t-distribution, which is \(t_{\alpha/2} \cdot \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}\). Here, \(t_{\alpha/2}\) is the t-score for a 90% confidence interval, which we can find using a t-distribution table or a statistical function. In this case, the t-score is approximately 1.729.
Step 4 :Substituting the given values into the formula, we get the margin of error as \(1.729 \cdot \sqrt{\frac{11.3^{2}}{25} + \frac{8.0^{2}}{20}} \approx 4.98\).
Step 5 :The confidence interval is then the best estimate plus or minus the margin of error. So, the 90% confidence interval for \(\mu_{1}-\mu_{2}\) is approximately from \(14.6 - 4.98 = 9.62\) to \(14.6 + 4.98 = 19.58\).
Step 6 :Final Answer: The best estimate for \(\mu_{1}-\mu_{2}\) is \(\boxed{14.60}\). The margin of error is approximately \(\boxed{4.98}\). The 90% confidence interval for \(\mu_{1}-\mu_{2}\) is approximately from \(\boxed{9.62}\) to \(\boxed{19.58}\).